Class 11 Physics

Kinetic Theory Of Gases Question and Answers

Question 1. Will the rms speeds of molecules of different gases at the same temperature be the same?
Answer:

If c is the rms speed of 1 mol of a gas, then c = \(\sqrt{\frac{3 R T}{M}} \)

The value of gas constant R is the same for 1 mol of different gases. But different gases have different molecular weights M so at the same absolute temperature T, we have \(c \propto \frac{1}{\sqrt{M}}\)

This means that the rms speed will not be the same for molecules of different gases; gases with higher molecular weight walls have less molecular rms speed.

Question 2. 1 cm³ of hydrogen gas and 1 cm³ of oxygen gas are both at STP. Which one contains more number of molecules?
Answer:

Avogadro’s law states that equal volumes of all gases at the same temperature and pressure contain an equal number of molecules. So, 1cm³ of hydrogen gas and 1cm³ of oxygen gas contain an equal number of molecules at STP.

Question 3. How does the kinetic theory explain the increase of temperature of a gas when heat is supplied from outside?
Answer:

Energy supplied in the form of heat is converted into the kinetic energy of the gas molecules. So the total kinetic energy of the molecules increases. The kinetic theory states that the temperature of a gas is proportional to the kinetic energy of the gas molecules i.e.,  E ∝ T. Thus the tem¬perature of the gas increases.

Question 4. A porous container is filled with a gas mixture. Which gas would leak faster from the container when it is placed in a vacuum?
Answer:

If the temperature is constant, rms speed of gas molecules is inversely proportional to the molecular weight of the gas, i.e., \(c \propto \frac{1}{\sqrt{M}}\)

So, the rms speed of the lighter gas molecules is higher than that of heavier gas molecules. As a result, the lighter gas would leak faster through the pores of the container.

Question 5. An equal number of molecules of an ideal monatomic and an ideal diatomic gas are at the same temperature. Which gas will be more heated if an equal amount of heat is supplied from outside?
Answer:

Let, N = Number of molecules in each of the gases, kept at a temperature T. Degrees of freedom of an ideal monatomic molecule = 3.

So, the average kinetic energy of the monatomic gas, \(E_1=\frac{3}{2} N k T\) then its heat capacity at constant volume, \(C_{v_1}=\frac{3}{2} N k\)

[k = Boltzmann constant]

Similarly, for the diatomic gas, \(C_{v_2}=\frac{5}{2} N k\), as the number of degrees of freedom of a diatomic molecule = 5.

Now, an amount of energy E, in the form of heat, is supplied to each of the gases. If T₁ and T₂ be the increases in tem¬perature, respectively, then

E = \(C_{\nu_1} T_1=C_{\nu_2} T_2\),

or, \(\frac{T_1}{T_2}=\frac{C_{v_2}}{C_{v_1}}=\frac{\frac{5}{2} N k}{\frac{3}{2} N k}=\frac{5}{3}\)

So, T₁ >T₂, Le., this monatomic gas will be more heated.

Question 6. The motion of gas molecules ceases at the temperature of absolute zero. Explain.
Answer:

The average kinetic energy of a gas molecule at an absolute temperature T is 3/2 kT, where k is the Boltzmann constant At T = 0, this kinetic energy becomes zero. So, molecular motion ceases at the temperature of absolute zero.

Question 7. The velocity of a gas molecule is comparable to that of a rifle bullet. Yet a gas molecule spends a much longer time than a bullet does to travel equal distances. Explain.
Answer:

A gas molecule suffers multiple collisions with the other molecules in the gas. As a result, it cannot move straight but travels any finite distance along a random zigzag path. On the other hand, a rifle bullet is much heavier than gas molecules. Collisions with very light air molecules cannot alter the straight path of the bullet. So the bullet travels a finite distance in a much shorter time.

Question 8. How would the rms speed of an ideal gas change if

1. Temperature increases,
2. Density increases at constant pressure,
3. Density increases at constant temperature.

Answer:

Let, the pressure, density, temperature, and molecular weight of an ideal gas are p, ρ, T, and M respectively.

∴ rms speed, c = \(\sqrt{\frac{3 p}{\rho}}=\sqrt{\frac{3 R T}{M}} .\)

1. In the above equation, as c ∝ √T, the rms speed increases with the increase in temperature of the gas.
2. As \(c \propto \frac{1}{\sqrt{\rho}}\) at constant pressure, the rms speed decreases with the increase in density of the gas.
3. At constant temperature, V \(\frac{1}{p}\)

Again, as V \(\frac{1}{p}\)

∴ p ∝ ρ

Hence, \(\frac{P}{\rho}=\mathrm{constant}\)

∴ ρ = \(\sqrt{\frac{3 P}{\rho}}=\mathrm{constant}\)

∴ In this case, the rms speed does not change with an increase in the density of the gas.

Question 9. Why does a real gas obey Boyle’s and Charles’ laws at

1. High temperature and
2. Low pressure?

Answer:

The intermolecular force of attraction is not negligible for real gases. So, a real gas molecule has some potential energy in addition to its kinetic energy.

However, this potential energy can still be neglected in two extreme cases:

1. High temperature: In this case, the molecular kinetic energy is so high that the potential energy is negligibly small.
2. Low pressure: In this case, the distance between molecules is so high that the force of attraction among them is very small so the molecular potential energy may be neglected.

In these two cases, a real gas molecule essentially has a kinetic energy only. So it behaves as an ideal gas and obeys Boyle’s and Charles’ laws.

Question 10. A gas mixture contains 1 mol each of two different gases. Would the average molecular kinetic energy of the two gases be equal? Would the rms speeds be equal?
Answer:

In the mixture, both gases are at the same temperature T. The average molecular kinetic energy is \(\frac{3}{2}\) kT.

So, this is equal for both gases.

But the rms speed is \(\sqrt{\frac{3 A T}{M}}\). As the molecular weight M is different for the two gases, the rms speed of the molecules is different.

Question 11. State the conditions in which a real gas behaves as an ideal gas.
Answer:

Condition 1: Each molecule is effectively a point, i.e., molecular volume is negligible.

Condition 2: Intermolecular attraction is negligible.

Question 12. For a fixed mass of a gas at a constant temperature, the pressure falls when the volume increases, and vice versa. Explain according to the kinetic theory.
Answer:

When volume increases at a constant temperature, the intermolecular distance increases. So, there are less number of molecules in unit volume. As a result, the number of collisions of the molecules per second with the unit area of wall of the container decreases.

Thus, the change of momentum of the molecules i.e., force exerted by the molecules decreases. This is why the pressure of the gas falls. Conversely, the pressure rises due to an increased number of collisions per second when the volume of a gas decreases at a constant temperature.

Question 13. For a fixed mass of a gas at constant volume, pressure rises when temperature increases, and vice versa. Explain the kinetic theory.
Answer:

When the temperature increases at constant volume, the gas molecules move in the container with greater velocities. So the molecules collide with the wall with greater momenta. As a result, they exert greater force on the wall and the pressure of the gas rises.

Conversely, due to the opposite behavior of gas molecules, pressure falls with a decrease in temperature at constant volume.

Question 14. Find out the molecular kinetic energy of 1 mol of an ideal gas. Is it equal for all gases?
Solution:

Let the molecular weight, volume, pressure and temperature of an ideal gas be M, V, p, and T respectively. If the density of the gas is ρ and the rms speed of the molecules is c, then according to the kinetic theory of gases,

p = \(\frac{1}{3} \rho c^2=\frac{1}{3} \frac{M}{V} c^2\)

or, \(\frac{1}{3} M c^2=p V=R T\) (R= universal gas constant)

So, the molecular kinetic energy of 1 mol of the gas

= \(\frac{1}{2} M c^2=\frac{3}{2} \times \frac{1}{3} M c^2=\frac{3}{2} R T\)

In general, real gases do not obey the ideal gas conditions So, the value of molecular kinetic energy differs from the ideal gas value \(\frac{3}{2}\)RT.

Moreover, the value becomes different for different gases, However, when gases obey the Ideal gas conditions, the value of the molecular kinetic energy becomes the same for all gases at the same temperature T.

Question 15. In a closed container, the gas molecules have a highly random motion. Yet, the pressure throughout the container Is uniform at a constant temperature, Explain.
Answer:

The number of gas molecules in a container is extremely large, For example, 1 cm³ of a gas contains nearly 10²³ molecules. So, the individual behavior of the molecules are no longer important rather, the gross statistical behavior dominates, On every unit area anywhere on the wall, the number of collisions of molecules per second, the velocity of impact, etc., are all equal on the average.

The value of the mean velocity does not change if the temperature remains constant, So the pressure remains uniform, as long as the temperature of the gas remains the same.

Question 16. Why does a piece of wood floating on water have no Brownian motion?
Answer:

The piece of wood is very large compared to the dimension of water molecules. At every instant, a very large number of moving water molecules collide with this piece. The force exerted in any direction due to some colliding molecules is canceled by the equal and opposite force due to some other molecules. As a result, the resultant force on the piece of wood becomes zero and it has no Brownian motion.

Question 17. Light gases like hydrogen and arid helium are very rare in the Earth’s atmosphere. Why?
Answer:

The escape velocity from the earth’s surface is 11.2 km · s^-1, approximately. At the upper atmosphere, it is still lower. The atmospheric temperature was very high at the time of the formation of the Earth.

• At that temperature, the rms speed of hydrogen gas molecules was 5 km · s^-1 or higher. As this is an average velocity, a large number of molecules were moving with velocities higher than the escape velocity.
• As a result, those molecules left the earth’s field of gravity forever. This incident occurred over a long period of time. So light gases like hydrogen and helium are rare in the earth’s atmosphere.

Question 18. If n is the number of degrees of freedom of the molecules of an Ideal gas, show that the ratio \(\frac{G_p}{C_p}\) is \(1+\frac{2}{n}\).
Answer:

Average kinetic energy of a gas molecule

= \(n \cdot \frac{1}{2} k T\) (k = Boltzmann constant)

As the molecules have no potential energy, the total Internal energy of 1 mol of a gas Is

E = \(N_0 n \cdot \frac{1}{2} k T=\frac{n}{2} N_0 k T=\frac{n}{2} R T\)

(\(N_0 k=R, \text { where } N_0=\text { Avogadro number }\))

∴ \(C_v=\frac{d E}{d T}=\frac{n}{2} R \text { and } C_p=C_v+R=\left(\frac{n}{2}+1\right) R,\) [for ideal gas]

Then, \(\frac{C_p}{C_p}=\frac{\frac{n}{2}+1}{\frac{n}{2}}=1+\frac{2}{n}\).

Question 19. 1 mol of an ideal monatomic gas \(\left(\gamma=\frac{5}{3}\right)\) is mixed with 1 mol of an ideal diatomic gas \(\left(\gamma=\frac{7}{3}\right)\). Find the value of γ for the mixture.
Answer:

For the monatomic gas, \(C_v=\frac{3}{2} R\) and  \(C_p=\frac{5}{2} R\)

For the diatomic gas, \(C_v=\frac{5}{2} R \text { and } C_p=\frac{7}{2} R \text {. }\).

So, for 2 mol of the mixture,

⇒ \(C_v=1 \times \frac{3}{2} R+1 \times \frac{5}{2} R=4 R ;\)

⇒ \(C_p=1 \times \frac{5}{2} R+1 \times \frac{7}{2} R=6 R\)

or, \(\gamma=\frac{C_p}{C_v}=\frac{6 R}{4 R}=\frac{3}{2}=1.5 .\)

Question 20. The ratio between the specific heats of an ideal gas is γ. Show that the number of degrees of freedom of the gas molecules is n = \(\frac{2}{γ-1}\).
Answer:

⇒ \(C_\nu=\frac{n}{2} R ; C_p=C_\nu+R=\frac{n}{2} R+R=\left(\frac{n}{2}+1\right) R\)

∴ \(\gamma=\frac{C_p}{C_v}=\frac{\frac{n}{2}+1}{\frac{n}{2}}=1+\frac{2}{n} ;\)

or, \(\frac{2}{n}=\gamma-1 \quad$ or, \quad n=\frac{2}{\gamma-1}\)

Question 21 If the absolute temperature of a perfect gas rises to j four times Its initial value, estimate the changes of

1. Molecular rms speed and
2. Total kinetic energy.

Answer:

1. Molecular rms speed, \(c \propto \sqrt{T}\)

∴ \(T_2=4 T_1\)

So, \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2}}\)

or, \(c_2=c_1 \sqrt{\frac{T_2}{T_1}}=c_1 \sqrt{\frac{4 T_1}{T_1}}=2 c_1\)

2. Total kinetic energy, \(E \propto T\).

So, \(\frac{E_1}{E_2}=\frac{T_1}{T_2} \quad or, \quad E_2=E_1 \frac{T_2}{T_1}=4 E_1\)

Question 22. Some gas cylinders are kept on a running vehicle. What will be the change in temperature of the gas molecules inside the cylinders?
Answer:

The motion of the cylinders is an external motion. It does not alter the internal motion of the molecules. So the molecular kinetic energy does not change. As a result, the temperature of the gas remains the same.

Question 23. Find the dimension of the constant a in the van der Waals equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\).
Answer:

The expression \(p+\frac{a}{V^2}\) shows that

⇒ \(\left[\frac{a}{V^2}\right]=[p]\)

So, \(\quad[a]=\left[p V^2\right]=\mathrm{ML}^{-1} \mathrm{~T}^{-2} \cdot\left(\mathrm{L}^3\right)^2=\mathrm{ML}^5 \mathrm{~T}^{-2}\).

Question 24. Find the dimension of the constant b in the van der Waals equation \(\left(p+\frac{a}{V^2}\right)(V-b)=R T\).
Answer:

The expression V- b shows that [b] = [V].

So, [b] = L³.

Question 25. We have a sample of a gas characterized by p, V, T and another sample by 2p, \(\frac{V}{4}\),2T. What is the ratio of the number of molecules in the two samples?
Answer:

For the first sample, pV = n₁RT (n₁ = number of moles)

For the second sample, \(2 p \cdot \frac{V}{4}=n_2 R \cdot 2 T \quad\left(n_2=\text { number of moles }\right)\)

or, \(p V=4 n_2 R T\)

∴ \(n_1 R T=4 n_2 R T \quad \text { or, } \frac{n_1}{n_2}=4 .\)

So, the ratio of the number of molecules in the two samples Is also 4:1.

Question 26. Find out the ratio between the absolute temperatures of two samples of hydrogen and oxygen gases, if their molecular rms speeds are equal.
Answer:

rms speed, c = \(\sqrt{\frac{3 R T}{M}}\)

For hydrogen, molecular weight, M₁ = 2, and for oxygen, M₂ = 32.

∴ \(\frac{c_1}{c_2}=\sqrt{\frac{T_1}{T_2} \cdot \frac{M_2}{M_1}} \text { or, } \frac{T_1}{T_2}=\frac{M_1}{M_2} \cdot\left(\frac{c_1}{c_2}\right)^2=\frac{2}{32} \cdot\left(\frac{1}{1}\right)^2=\frac{1}{16} \text {. }\)

Question 27. At equilibrium, the volume, pressure, and temperature of a gas are V, p, and T, respectively. If the gas is divided into two parts by a partition, what will be the value of these quantities in each part?
Answer:

The rms speed of gas molecules does not depend on the volume of the container. So, speed does not change when the volume of the gas is halved by using a partition.

As a result, the temperature remains the same. As the density of gas does not undergo any change and as p = \(\frac{1}{3} \rho c^2\), pressure will also remain the same. Thus only the volume becomes half, but pressure and temperature remain the same. So, the values of volume, pressure, and temperature will be \([\frac{V}{2}\), p and T, respectively.

Question 28. In a gas-filled container, a molecule of speed 200 m/s collides at an angle of 30° with the horizontal face of this container and rebounds with the same speed. Is the collision elastic or inelastic? In this momentum conserved in this collision?
Answer:

As the molecule rebounds with the same speed, the collision is elastic. The momentum is always conserved in a collision, irrespective of whether the collision is elastic or inelastic.

Question 29. While considering the motion of gas molecules in a container, why do we use rms speed instead of the average speed of molecules?
Answer:

Since a large number of gas molecules is present in a container, therefore for the velocity of any molecule, there exists another molecule with an equal and opposite velocity. As velocity is a vector quantity, the resultant velocity of all the molecules becomes zero. Hence average velocity also vanishes. So we could not derive any conclusion about the velocity of the gas molecules from it.

On the other side, if only the magnitudes of velocities (scalar) are considered to find the average, the average speed does not vanish. But in the kinetic theory of gases, we find that the pressure, temperature, and molecular kinetic energy of a gas are proportional to the rms speed and not with the molecular velocity. Hence, rms speed of a molecule is preferable to the average speed in kinetic theory.

Unit 9 Behavior Of Perfect Gas And Kinetic Theory Chapter 1 Kinetic Theory Of Gases Multiple Choice Questions And Answers

Question 1. If the volume of a body is V₁ and the total volume of the molecules of the body is V₂, then

1. V₁ = V₂
2. V₁ < V₂
3. V₁ >V₂
4. V₁ < V₂ or V₁ > V₂ for different bodies

Answer: 3. V₁ >V₂

Question 2. The molecules of all solids

1. Are relatively closer than those of liquids or gases
2. Are relatively farther than those of liquids or gases
3. Move faster than those of liquids or gases
4. Are stationary as they cannot move inside the solid

Answer: 1. Are relatively closer than those of liquids or gases

Question 3. Which of the following statements is inconsistent with the characteristics of Brownian motion?

1. The velocity of a particle increases as its size decreases
2. The velocity of the particles increases as the temperature increases
3. The velocity of the particles increases as the viscosity of the medium decreases
4. The velocity of the particles increases when the container is shaken

Answer: 4. The velocity of the particles increases when the container is shaken

Question 4. A piece of wood floating on water does not have any Brownian motion, because

1. A part of the wooden piece is above the water
2. The resultant of the applied forces by the water molecules is zero
3. An adhesive force acts between the molecules of wood and water
4. The viscosity of water is comparatively less

Answer: 2. The resultant of the applied forces by the water molecules is zero

Question 5. The velocities of two particles moving towards the east are 4 m · s^-1 and 6 m · s^-1, respectively. The velocities of three other particles moving towards the west are 2 m · s^-1, 3 m · s^-1, and 5 m · s^-1, respectively. The root mean square speed of these 5 particles is

1. 0
2. 4m · s^-1
3. 1.667 m · s^-1
4. 4.242 m · s^-1

Answer: 4. 4.242 m · s^-1

Question 6. The pressure and density of hydrogen gas, kept in a vessel, are 1.013 x 10⁶ dyn · cm^-2 and 0.089 g · L^-1, respectively. The rms speed of the gas molecules will be

1. 18.5m · s^-1
2. 185m · s^-1
3. 1.85 km · s^-1
4. 18.5 km · s^-1

Answer: 3. 1.85 km · s^-1

Question 7. If the mean velocity, rms speed, and maximum probable velocity of gas are c, c, and cm, respectively, then

1. \(c_m<\bar{c}<c\)
2. \(\bar{c}<c<c_m\)
3. \(c_m>\bar{c}>c\)
4. None of these

Answer: 1. \(c_m<\bar{c}<c\)

Question 8. There is a mixture of hydrogen and oxygen gases in a vessel. The root mean square speed of the oxygen molecules is

1. 4 times that of hydrogen molecules
2. 16 times that of hydrogen molecules
3. 1/4 times that of hydrogen molecules
4. 1/16 times that of hydrogen molecules

Answer: 3. 1/4 times that of hydrogen molecules

Question 9. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The ratio of the rms speeds is

1. 0.32
2. 0.45
3. 2.24
4. 3.16

Answer: 4. 3.16

Question 10. At room temperature, the rms speed of the molecules of a certain diatomic gas is found to be 1930 m/s. The gas is

1. H₂
2. F₂
3. O₂
4. Cl₂

Answer: 1. H₂

Question 11. If the volume of a container is V, the pressure on the walls of the container by a gas is p and the internal energy of the gas is U, then

1. U = p V
2. U = \(\frac{1}{3} p V\)
3. U = \(\frac{2}{3} p V\)
4. U = \(\frac{3}{2} p V\)

Answer: 4. U = \(\frac{3}{2} p V\)

Question 12. A certain amount of gas is at 27°C. The rms speed of the gas molecules becomes doubled at

1. 327°C
2. 600°C
3. 927°C
4. 1200°C

Answer: 3. 1200°C

Question 13. If a gas of a particular mass is expanded at a constant temperature, the variable which undergoes a change is

1. Pressure of the gas
2. Internal energy of the gas
3. Rms speed of the gas molecules
4. Kinetic energy of the gas molecules

Answer: 1. Pressure of the gas

Question 14. At equilibrium conditions, the volume, pressure, and temperature of a gas kept in a closed container are V, p, and T, respectively. If the container is divided into two equal parts by a partition, the value of these quantities for each part will be

1. \(\frac{V}{2}, \frac{p}{2}, \frac{T}{2}\)
2. \(\frac{V}{2}, \frac{p}{2}, T\)
3. \(\frac{V}{2}, p, \frac{T}{2}\)
4. \(\frac{V}{2}, p, T\)

Answer: 4. \(\frac{V}{2}, p, T\)

Question 15. According to the kinetic theory of gases, there are no intermolecular attractions, so these molecules do not have

1. Linear momentum
2. Kinetic energy
3. Potential energy
4. Mechanical energy

Answer: 3. Linear momentum

Question 16. If k is Boltzmann constant and T is temperature, the average kinetic energy of each molecule of a gas will be

1. \(\frac{2}{3} k T\)
2. \(\sqrt{\frac{2}{3}} k T\)
3. \(\frac{3}{2} k T\)
4. \(\sqrt{\frac{3}{2}} k T\)

Answer: 3. \(\frac{3}{2} k T\)

Question 17. The rms speed of oxygen molecules at 47°C will be equal to the rms speed of hydrogen molecules at

1. 80K
2. -83K
3. 3K
4. 20K

Answer: 4. 20K

Question 18. The pressure, volume, and temperature in two samples of a gas are p, V, T, and 2p,V/4, 2 T, respectively The ratio of the number of molecules in the two samples is

1. 2:1
2. 4:1
3. 8:1
4. 16:1

Answer: 2. 4:1

Question 19. The rms speed of gas molecules at 0°C will be reduced to half at

1. 0°C
2. -273°C
3. 32°C
4. -204°C

Answer: 4. -204°C

Question 20. A container of 5 L contains 10²⁶ number of molecules of a gas. If the mass and rms speed of each molecule are 2.4 x 10^-25 g and 3.5 x 10⁴ cm · s^-1, respectively the pressure of the gas will approximately be

1. 2 x 10⁶ dyn · cm^-2
2. 10⁶ dyn · cm^-2
3. 3 x 10⁶ dyn · cm^-2
4. 5 x 10⁶ dyn · cm^-2

Answer: 1. 2 x 10⁶ dyn · cm^-2

Question 21. Air is filled in two heat-insulated vessels 1 and 2 having pressure, volume, and temperature p₁, V₁, T₁ and p₂, V₂, T₂ respectively. If the intermediate valve between the two vessels is opened, the temperature of the air at equilibrium will be

1. \(T_1+T_2\)
2. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_2+p_2 V_2 T_1}\)
3. \(\frac{T_1+T_2}{2}\)
4. \(\frac{T_1 T_2\left(p_1 V_1+p_2 V_2\right)}{p_1 V_1 T_1+p_2 V_2 T_2}\)

Answer: 3. \(\frac{T_1+T_2}{2}\)

Question 22. A vessel contains a mixture of one mole of oxygen and two moles of nitrogen at 300 K. The ratio of the average rotational kinetic energy per O₂ molecule to per N₂ molecule is

1. 1:1
2. 1:2
3. 2:1
4. Depends on the moment of inertia of the two molecules

Answer: 1. 1:1

Question 23. 70 cal of heat is required to raise the temperature of 20 moles of an ideal diatomic gas at constant pressure from 30°C. The amount of heat required (in cal) to raise the temperature of the same gas through the same range (30°C to 35°C) at constant volume is

1. 30
2. 50
3. 70
4. 90

Answer: 2. 50

Question 24. Three closed vessels A, B, and C at the same temperature T contain gases that obey the Maxwellian distribution of velocities. Vessel A contains only O₂, B only N₂, and C a mixture of equal quantities of O₂ and N₂. If the average velocity of the O₂ molecule in vessel A is v₂ and that of the N₂ molecule in vessel B is v₂, the average velocity of the O₂ molecule is vessel C is

1. \(\frac{\left(v_1+v_2\right)}{2}\)
2. \(v_1\)
3. \(\left(v_1 \nu_2\right)^{\frac{1}{2}}\)
4. \(\sqrt{\frac{3 k T}{M}}\)

Answer: 2. \(v_1\)

In this type of question, more than one option are correct

Question 25. From the following statements concerning ideal gas at any given temperature T, select the correct one(s).

1. The coefficient of volume expansion at constant pressure is the same for all ideal gases
2. The average translational kinetic energy per molecule of oxygen gas is 3kT, k being the Boltzmann constant
3. The mean free path of molecules increases with a decrease in pressure
4. In a gaseous mixture, the average translational kinetic energy of the molecules of each component

Answer:

1. The coefficient of volume expansion at constant pressure is the same for all ideal gases

3. The mean free path of molecules increases with a decrease in pressure

Question 26. Let \(\bar{v}, v_{\mathrm{rms}} \text { and } v_p\) respectively, denote the mean velocity, root mean square speed, and most probable velocity of the molecules in an feed monatomic gas at absolute temperature T. The mass of a molecule is m. Then

1. No molecule can have a speed greater than \(\sqrt{2} v_{\mathrm{rms}}\)
2. No molecule can have speed less than \(\frac{v_p}{\sqrt{2}}\)
3. \(v_p<\bar{v}<v_{\mathrm{rms}}\)
4. The average kinetic energy of a molecule is \(\frac{3}{4} m v_p^2\)

Answer:

3. \(v_p<\bar{v}<v_{\mathrm{rms}}\)

4. The average kinetic energy of a molecule is \(\frac{3}{4} m v_p^2\)

Question 27. For a jar containing H₂ gases which of the following statements are correct?

1. Both the gas molecules have same average energy
2. Both the gas molecules have same average translational kinetic energy
3. Hydrogen molecules have greater average energy than helium molecules
4. Both the molecules have same average velocity

Answer:

3. Hydrogen molecules have greater average energy than helium molecules

4. Both the molecules have same average velocity

Question 28. The root mean square speed of the perfect gas molecules will be doubled if

1. Pressure is doubled at constant volume
2. Pressure is made 4 times at constant volume
3. Volume is made 4 times at constant pressure
4. Volume is increased by 41.4% at constant pressure

Answer:

2. Pressure is made 4 times at constant volume

3. Volume is made 4 times at constant pressure

Question 29. According to the kinetic theory of gases, which of the following statements are true?

1. Real gas behaves as ideal gas at high temperatures and low pressure
2. The liquid state of an ideal gas is impossible
3. At any temperature and pressure, ideal gas obeys Boyle’s law and Charles’ law
4. The molecules of a real gas do not exert any force on one another

Answer:

1. Real gas behaves as ideal gas at high temperatures and low pressure
2. The liquid state of an ideal gas is impossible
3. At any temperature and pressure, ideal gas obeys Boyle’s law and Charles’ law

Energy

A body can do work by the transfer of some amount of energy.

Energy Definition: The total amount of work that a body can do is the same as its total amount of energy. Hence, energy and work are equivalent physical quantities, having the same units and dimensions. Both are scalar quantities.

Energy is manifested in nature in different forms:

1. Mechanical energy
2. Heat energy
3. Light energy
4. Sound energy
5. Magnetic energy
6. Electrical energy
7. Chemical energy
8. Atomic energy

In all-natural events, energy changes from one form to another. In fact, every natural phenomenon can be interpreted as a sequence of energy transformations.

Energy Examples:

1. In an electric bulb, electrical energy is converted into heat energy at first, and then into light energy.
2. In an electric fan, electrical energy changes into mechanical energy.
3. In a telephone, sound energy is converted into electrical energy (during speaking), and electrical energy into sound energy (during listening).
4. In hydroelectric plants, water stored in a high reservoir flows downwards and its potential energy changes into kinetic energy. This kinetic energy rotates the turbines which generate electrical energy.
5. Water from sea, river and other water bodies evaporate and forms clouds. This water eventually returns to the earth in the form of rain. The potential energy of the clouds converts into the kinetic energy of the raindrops.
6. On burning coal, chemical energy is converted into heat and light energy.

We observe such numerous incidents of energy transformations in daily life. But neither energy is created nor destroyed in such incidents. Energy can only get transferred from one body to another or can change its form. As soon as a body loses energy, another body gains it in equal amounts. Hence the total energy in the universe remains the same. This is the law of conservation of energy.

Law Of Conservation Of Energy: Energy cannot be created or destroyed. It can only be converted from one form to another.

• It means that the total energy in the universe remains constant.
• After the introduction of Einstein’s theory of relativity, the law of conservation of energy was corrected and the law of conservation of mass energy was established.
• This means that mass and energy in the universe are not conserved individually.
• Only mechanical energy is discussed in detail in the following sections.

Work And Energy – Conservation Force

Conservation Force Definition: A system in which the total mechanical energy i.e., the sum of its kinetic and potential energy [remains conserved, is called a conservative system. The forces acting in such a system are called conservative forces.

Conservation Force Example: Gravitational force, restoring force of spring or elastic force, electrostatic force, force between the magnetic poles, etc., are conservative forces.

Work is done to lift a body against gravity and is stored as potential energy. While returning to its initial position, the body uses this stored energy and does work that is exactly equal to the initial work done while lifting. Hence energy is conserved.

Work Done In A Dosed Path Under A Conservative Force: For a conservative system, the work done depends only on the initial and final positions. It does not depend on the path taken in reaching the final position from the initial position.

• This means that the sum of the kinetic and potential energies remains constant throughout.
• It also implies that if a body goes around a complete loop so that its final and initial positions are the same, then the total work done is zero. For any conservative force, the work done is reversible.
• Hence if work done against a force can be restored, the force is called conservative. Also if work done by a force in a closed path equals to zero, the force is called conservative.

Thus in a conservative system, it is possible to restore the initial work done.

Suppose a body gets displaced by dx along the x-axis under the action of a conservative force F (which may vary with position) also acting along the x-axis. Therefore the work done by the conservative force dW = F(x)dx.

Since change in potential energy is negative of work done, then change in potential energy for this displacement dx is given by dU = -F(x)

In a three-dimensional reference frame, this equation is dU = \(-\vec{F} \cdot d \vec{r}\)….(1)

Work And Energy – Conservation Of Mechanical Energy Of A Particle In A Conservative System

Suppose a conservative force \(\vec{F}\), acting on a particle of mass m moves, it from point A to point B.

If the velocity of the particle at any point of its motion in the force field is v, then \(\vec{F} \cdot d \vec{r}=m \frac{d \vec{v}}{d t} \cdot \vec{v} d t=m \vec{v} \cdot d \vec{v}=m d\left(\frac{1}{2} v^2\right)\)

∴ Work done, to take the particle from A to B, by force \(\vec{F}\),

W = \(\int_A^B \vec{F} \cdot d \vec{r}=\left[\frac{1}{2} m v^2\right]_A^B=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2\)…(1)

where v_A and v_B are the velocities of the particle at A and B respectively.

Equation (1), we know \(\vec{F} \cdot d \vec{r}=-d U\)

∴ \(\int_A^B \vec{F} \cdot d \vec{r}=-\int_A^B d U=U_A-U_B\)

= difference in potential energy between the points A and B…(2)

Comparing (1) and (2), \(U_A-U_B=\frac{1}{2} m v_B^2-\frac{1}{2} m v_A^2\)…(2)

Equation (3) establishes that in a conservative field when the kinetic energy of a particle increases, its potential energy decreases.

∴ \(\frac{1}{2} m v_A^2+U_A=\frac{1}{2} m v_B^2+U_B\)…(4)

If K_A and K_B are the kinetic energies of the particle at A and B respectively, then

K_A+U_B = K_B+ U_B or, K + U = constant (E).

Hence in a conservative field i.e., in a conservative system (where no dissipation of energy occurs due to forces like friction, etc.), at each point sum of potential energy and kinetic energy remains constant.

In other words, total mechanical energy (E) of a particle in a conservative system remains constant. This is the statement of the principle of conservation of mechanical energy.

Total mechanical energy In a free fall under gravity remains constant: Let a body of mass m be at rest at point A. DE, the earth’s surface is taken as the reference plane and the height of point A from DE is h. As the body is at rest it has potential energy only and no kinetic energy.

Potential energy at A = mgh

Kinetic energy at A = 0

Total mechanical energy at A = mgh + 0 = mgh

Now the body is released and it starts falling freely. When the body reaches a point B, it acquires a velocity v.

At B, the body has both potential energy (due to its height) and kinetic energy (due to its motion).

Let AB = x

∴ Potential energy at B = mg(h- x)

Kinetic energy at B = \(\frac{1}{2} m v^2=\frac{1}{2} m \cdot 2 g x\)

= mgx (because \(v^2=u^2+2\) as and u=0)

∴ Total mechanical energy at B = mg(h-x) + mgx = mgh

= total energy at A.

Also at C, when the body is about to touch the reference plane, its potential energy becomes zero. If the velocity acquired at that point is V, then kinetic energy at C

= \(\frac{1}{2} m V^2=\frac{1}{2} m \cdot 2 g h\left[because V^2=2 g h\right]\)

= mgh

Total mechanical energy at C = mgh + 0 = mgh

∴ Total energy at A = total energy at B = total energy at C.

Therefore, the total mechanical energy (= potential energy + kinetic energy) of a freely falling body, under the action of i gravity, remains conserved at all positions.

As the body touches the ground, both its potential energy and kinetic energy become zero; the entire mechanical energy transforms into heat, sound, and other forms of energy.

Total Mechanical Energy Of A Body, Falling Under Gravity Along A Frictionless Inclined Surface, Remains Constant: Let a body of mass m be at rest at a point A on a frictionless plane of inclination θ. Earth’s surface CD is taken as the reference plane.

Let the height of point A from the reference plane, GA = h. Being at rest, the body has no kinetic energy at A.

The potential energy at A = mgh Kinetic energy at A = 0

∴ Total mechanical energy at A = mgh + 0 = mgh

On releasing the body, it falls along the incline and reaches a point B such that, AB = x. Let the velocity of the body at B be v.

Acceleration of the body along the incline, a = gsinθ.

∴ \(v^2=0+2 g \sin \theta \cdot x=2 g x \sin \theta \quad \text { [as } v^2=u^2+2 a s \text { ] }\)

∴ Kinetic energy at \(B=\frac{1}{2} m v^2=\frac{1}{2} \cdot m \cdot 2 g x \sin \theta\)

= mgx sinθ

The perpendicular height of B from CD plane = FB = GE = GA-EA = h-x sinθ

(because \(\sin \theta=\frac{E A}{A B} \quad \text { or, }E A=A B \sin \theta=x \sin \theta\))

∴ The potential energy at B = mg(h-x sinθ)

∴ Total mechanical energy at B

= mg(h- xsinθ) + mgx sinθ

= mgh = total energy at A

At point C, i.e., where the body just touches the plane CD, it has no potential energy. If the velocity of the body at that moment is V, then

V² = 0 + 2g sinθ · AC = 2gsinθ · AC

and hence, kinetic energy at C = \(\frac{1}{2}\) m · 2g sinθ · AC = mgh

[sinθ = \(\frac{G A }{A C}\) = \(\frac{h}{A C}\) or, AC sinθ = h]

∴ Total energy at C = 0 + mgh = mgh

Thus, total mechanical energy at A = total mechanical energy at B = total mechanical energy at C.

Hence, the total mechanical energy of a body, moving along a frictionless inclined plane under gravity, is conserved.

Total Mechanical Energy Of A Hydrogen Gas-Filled Balloon Rising Upwards Remains Constant: Let us assume that air and earth form a system. If the earth’s surface is taken as the plane of reference, when the balloon is at rest on the ground, both its potential and kinetic energy = 0, i.e., total mechanical energy = 0.

Now the balloon is released. Upthrust, T due to air on the balloon is more than the weight, mg of the balloon filled with gas and so the balloon rises up. Let the mass of the gas-filled balloon = m and the mean resultant upward force on it = F – T – mg.

∴ Acceleration of the balloon = \(\frac{F}{m}\). If the velocity of the balloon is v at a height h from the earth surface, then

v2² = 2\(\frac{F}{m}\)h [according to the formula v² = u² + 2as]

Hence, kinetic energy at h = \(\frac{1}{2}\)= \(\frac{1}{2}\)m 2 \(\frac{F}{m}\)h = Fh

We know that when a stone falls freely under gravity, the work done is negative. Similarly, when the balloon rises by itself (without the help of any external agent) under the effect of the mean upward resultant force F, the work done is also negative. Here, the upthrust is not applied by an external agent.

So, at height h the change in potential energy = -Fh. As the initial potential energy of the balloon on the ground was zero, the total potential energy at a height h = -Fh.

∴ The total mechanical energy of the balloon at height h = potential energy + kinetic energy = – Fh + Fh = 0 = total energy of the balloon on the ground before its release.

Thus, the total mechanical energy remains conserved for a hydrogen gas-filled balloon when it is rising up. It is to be noted that, with the increase in height h, the kinetic energy increases but the potential energy decreases equally.

Actually, the expression mgh for potential energy should be modified for a balloon as (mg- T)h, where T is the upward thrust exerted on the balloon by the air surrounding it. For a hydrogen-filled balloon, the upward thrust is greater than its weight, i.e., T> mg. So the potential energy (mg-T)h is negative; this negative value goes on increasing with an increase in height h.

Another similar event is observed when a piece of wood is held at the bottom of a bucket full of water. When it is released, it floats up by itself. It can be shown that the total mechanical energy remains conserved. As the piece of wood rises upwards, its kinetic energy gradually increases and the potential energy decreases equally.

Work And Energy – Conservation Of Mechanical Energy Of A Particle In A Conservative System Numerical Examples

Example 1. After falling from a height of 200 m, water flows horizontally with a certain velocity. Ignoring any energy dissipation, find the velocity of flow.
Solution:

Potential energy changes into kinetic energy during the free fall of water.

Let mass of water = m, height = h, final velocity = v

Here, at a height of 200 m

potential energy (P.E.)_i = mgh and kinetic energy (K.E.)_i = 0

when water falls and flows horizontally, potential energy, (P.E.)_f = 0 and kinetic energy (K.E.)_f = \(\frac{1}{2}\)mv²

∴ From the conservation of mechanical energy we,

∴ mgh = \(\frac{1}{2}\)mv

or, \(\nu=\sqrt{2 g h}=\sqrt{2 \times 9.8 \times 200}\)

= \(62.61 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= 62.61 m · s^-1

Example 2. The mass of the bob of a simple pendulum is 10 g and the effective length is 13 cm. The bob is pulled 5 cm away from the vertical and then released. What will be the kinetic energy of the bob when it passes through the lowest point?
Solution:

We observe that, the energy of the bob at point B = potential energy = mgh = 10 x 980 x AC.

Here OA = OB = 13 cm and CB = 5 cm.

∴ OC = \(\sqrt{13^2-5^2}\)

= \(\sqrt{144}=12 \mathrm{~cm}\)

∴ AC = OA-OC = 13 – 12 = 1 cm.

∴ Potential energy of the bob at B = 10 x 980 x 1 = 9800 erg

∴ The kinetic energy of the bob at the lowest point A = potential energy of the bob at B = 9800 erg.

Example 3. After a collision with an ideal spring, a body of mass 8 g, moving with a constant velocity of 10 cm· s^-1 comes to rest Force constant of the spring is 200 dyn · cm^-1. If the total kinetic energy of the body is spent in compressing the spring, find the compression.
Solution:

Here, the kinetic energy of the body transforms into the potential energy of the spring.

∴ \(\frac{1}{2} m v^2=\frac{1}{2} k x^2\), where x = compression of the spring

∴ x = \(\sqrt{\frac{m v^2}{k}}=\sqrt{\frac{8 \times 10^2}{200}}=2 \mathrm{~cm}\) .

Example 4. The effective length of a pendulum is 50 cm and the mass of the bob is 4 g. The bob is drawn to one side until the string is horizontal and is then released. When the string makes an angle 60° with the vertical, what is the velocity and the kinetic energy of the bob?
Solution:

OB is the horizontal position of the string.

At B, the total energy of the bob

= potential energy = mgh = (4 x 980 x 50) erg

At C, the total energy of the bob = kinetic energy + potential energy

= kinetic energy + 4 x 980 x AD

(cos 60° = \(\frac{OD}{OC}\) or, OD = \(\frac{50}{2}\) = 25 cm

∴ AD = OA – OD – 50-25 = 25 cm)

According to the law of conservation of energy, the energy of the bob at C = energy of the bob at B

or, K.E. of the bob at C + P.E. of the bob at C = K.E. of the bob at B + P.E. of the bob at B

or, \(\frac{1}{2} m v^2+4 \times 980 \times 25=0+4 \times 980 \times 50\) (v = velocity of bob at C)

or, \(\frac{1}{2} m v^2=4 \times 980 \times 25=98000\)

∴ \(v^2=\frac{98000 \times 2}{4} \quad \text { or, } v=\sqrt{49000}=222.36 \mathrm{~cm} \cdot \mathrm{s}^{-1}\).

Example 5. The mass of the bob of a pendulum is 100 g and the length of the string is 1 m. The bob is initially held in such a way that the string is horizontal. The bob is then released. Find the kinetic energy of the bob when the string makes an angle of

1. 0° and
2. 30° with the vertical.

Solution:

When the string is horizontal, the height of the bob above its lowest position = 1 m = 100 cm.

The energy of the bob at point P = potential energy of the bob

= mgh = 100 x 980 x 100 = 98 x 10⁵ erg.

1. At 0° angle with the vertical, the string holds the bob at its lowermost position. Hence, energy at B = kinetic energy of the bob = initial potential energy = 98 x 10 erg.

2. When the string makes a 30° angle with the vertical, the height of the bob from its lowermost position

= BD = BA – DA = 100- AC cos30°

= \(100\left(1-\frac{\sqrt{3}}{2}\right)=13.4 \mathrm{~cm}\)

Potential energy of the bob at C = 100 x 980 x 13.4 erg;

kinetic energy at this position = initial potential energy- potential energy at C

= 98 x 10⁵ -(100 x 980 x 13.4) = 8486800 erg.

Example 6. A body of mass 1 kg falls to the ground from the roof of a building 20 m high. Find its

1. Initial potential energy,
2. Velocity when it reaches the ground,
3. Maximum kinetic energy and
4. Kinetic and potential energies at a position 2 m above the earth’s surface.

Solution:

1. Initial potential energy of the body = mgh = 1 x 9.8 x 20 = 196 J.

2. Suppose the body touches the ground with velocity v. Potential energy at roof level = kinetic energy just before touching the ground.

∴ 196 = \(\frac{1}{2}\) x 1 x v²

∴ v² = 392 and v = 19.8 m · s^-1

3. Maximum kinetic energy = initial potential energy = 196 J

4. Potential energy at a height of 2 m = 1×9.8×2 =19.6J

∴  Kinetic energy at that height of 2 m = decrease in initial potential energy = 196 – 19.6 = 176.4 J.

Example 7. A pump lifts 200 L of water per minute through a height of 5 m and ejects it through an orifice 2 cm in diameter. Find the velocity of efflux of water and the power of the pump.
Solution:

Volume of water lifted by the pump in is = \(\frac{200}{60}=\frac{10}{3} \mathrm{~L}=\frac{10^4}{3} \mathrm{~cm}^3 \text {. }\)

Mass of this volume of water = \(\frac{10^4}{3} g=\frac{10}{3} \mathrm{~kg}\)

∴ Increase in potential energy in 1 second

= \(\frac{10}{3}\) x 9.8×5 J · s^-1 = 163.33 J

∴ Power of the pump to raise water up to the height of 5m = 163.33 W.

If the velocity of efflux is v, then \(\pi r^2 \times v=\frac{10^4}{3}\)

or, \(\nu=\frac{10^4 \times 7}{3 \times 22 \times(1)^2}=1061 \mathrm{~cm} \cdot \mathrm{s}^{-1}=10.61 \mathrm{~m} \cdot \mathrm{s}^{-1} \text {. }\)

The kinetic energy of the amount of water thrown out per second

= \(\frac{1}{2}\) x \(\frac{10}{3}\) x(10.61)² J =187.62 J

∴ Power of efflux = 187.62 W

∴ Total power of the pump = 163.33 + 187.62 = 350.95 W.

Example 8. A body of mass 10 kg is raised to a height of 10 m with an upward force of 196 N. Find the work done by the upward force and the work done against gravitation. Show that the total energy in this case is equal to the work done by the upward force. [g = 9.8 m · s^-2]
Solution:

Work done by the upward force, W = force x displacement

= 196 N x 10 m = 1960 N · m = 1960 J.

Upward acceleration of the body in the absence of gravity,

a’ = \(\frac{\text { upward force }}{\text { mass }}=\frac{196}{10}=19.6 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ Effective upward acceleration, a = a’ -g = 19.6-9.8 = 9.8 m · s^-2

If the body starts from rest, and attains a velocity v at the height of 10 m, then from v² = u² + 2 as, v² = 2 x 9.8 x 10 m² · s^-2

∴  Kinetic energy of the body at this height,

K = \(\frac{1}{2}\)mv² = \(\frac{1}{2}\) x 10 x 2 x 9.8 x 10 = 980 J

Work done against gravitational force,

W’ = force due to gravity x displacement = mass x acceleration due to gravity x displacement = 10 x 9.8 x 10 = 980 J

This work done against gravitational pull gets stored as potential energy V of the body. Hence, V = 980 J.

∴ Total mechanical energy of the body at a height of 10 m = V + K= 980 + 980 = 1960 J = W

So, a part of the work done by the upward force changes into kinetic energy of the body, and the other part transforms itself into the stored potential energy. Thus, the total energy and work done by the upward force are equal.

Example 9. A body of mass 10 kg moving with a speed of 2.0 m · s^-1 on a frictionless table strikes a mounted spring and comes to rest. If the force constant of the spring be 4 x 10⁵ N · m^-1, then what will be the compression on the spring?
Solution:

The kinetic energy of the body is E = \(\frac{1}{2}\)mv², where m and v be the mass and the speed of the body respectively.

On striking the spring, the kinetic energy of the spring due to compression is completely converted into the potential energy of the spring.

If the spring is compressed through a distance x then its potential energy is U = \(\frac{1}{2}\)kx²

∴ \(\frac{1}{2}\) mv2 = \(\frac{1}{2}\)kx2

or, x = \(v \sqrt{\frac{m}{k}}=2 \times \sqrt{\frac{10}{4 \times 10^5}}=10^{-2} \mathrm{~m}=1 \mathrm{~cm}\)

Example 10. Shows two blocks of masses m₁ = 3 kg and m₂ = 5 kg, both moving towards the right on a frictionless surface with speeds u₁ = 10 m · s^-1 and u₁ = 4 m · s^-1 respectively. To the back side of m₂ an ideal spring of force constant 1000 N · m^-1 is attached. Calculate the maximum compression of the spring when the blocks collide.

Solution:

m₁ = 3kg, m₂ = 5kg, u₁ = 10 m · s^-1, u₂ = 4 m ·  s^-1

Force constant, k = 1000 N · m^-1

Let v be the speed of the combination.

Using the law of conservation of linear momentum, m₁u₁ + m₂u₂ =(m₁ + m₂)v

∴ v = \(\frac{m_1 u_1+m_2 u_2}{m_1+m_2}=\frac{3 \times 10+5 \times 4}{3+5}=6.25 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Also, let x be the maximum compression of the spring when the blocks collide.

∴ From conservation of mechanical energy, we get \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2}\left(m_1+m_2\right) v^2+\frac{1}{2} k x^2\)

or, \(500 x^2=380-312.5=67.5\)

∴ x = \(\sqrt{\frac{67.5}{500}} \approx 0.367 \mathrm{~m}\)

Work And Energy – Non-Conservative Force

Non-Conservative Force Definition: In the presence of resistive forces in a system, mechanical energy does not remain conserved and gets dissipated. Such a system is called a non-conservative system and the resistive force is called a non-conservative force or dissipative force.

Non-Conservative Force Example: Frictional force is a non-conservative force.

• Frictional force resists the motion of an object. Thus on slid¬ing a body over a rough surface, work done against friction does not get stored in the body as potential energy.
• Frictional force always acts opposite to the direction of motion. To move a body from its initial position to a final position, some amount of work is done to overcome friction.
• If the body is brought back to its initial position along the same path, again some work is done to overcome friction. Thus, each time the body moves, some energy is lost. It can therefore be stated that, it is not possible to restore the initial work done in a non-conservative system.

Work Done In A Closed Path Under A Non-Conservative Force: The total work done to move a body under a non-conservative force along a closed path once completely, is positive or negative but never zero. For example, to slide a body over a rough surface from one point to another, work has to be done against friction.

To return the body to its initial position by sliding it over the same surface following any path, work has to be done against friction again. So, the total amount of work done is not zero in a closed path. Hence, a force is called non-conservative when work done against it cannot be restored. Alternatively, when the work done by a force in a closed path is not zero, the force is called nonconservative.

Work And Energy – Mass Energy Equivalence

The principle of mass-energy equivalence can be obtained from Einstein’s famous theory of relativity.

Mass is a form of energy. Mass can be converted into energy and vice versa. If m amount of mass of a substance is completely converted to energy, then the amount of energy liberated is, E = mc² (c = velocity of light in vacuum = a constant)

According to this equation, the equivalent energy of mass m is E, and the equivalent mass of energy E is m

Mass Energy Equivalence Example:

1. In CGS system c = 3 x 10¹⁰ cm · s^-1

∴ Equivalent energy of 1 g mass

= 1 x (3 x 10¹⁰)²= 9 x 10²⁰ erg = 9 x 10¹³ J

Again in \(\mathrm{SI}\), \(c=3 \times 10^8 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Hence equivalent energy of 1 kg mass = \(1 \times\left(3 \times 10^8\right)^2=9 \times 10^{16} \mathrm{~J} .\)

2. Mass of an electron = \(9.1 \times 10^{-28} \mathrm{~g}\)

Equivalent energy of the mass of 1 electron = \(9.1 \times 10^{-28} \times\left(3 \times 10^{10}\right)^2 \mathrm{erg}\)

= \(\frac{9.1 \times 10^{-28} \times 9 \times 10^{20}}{1.6 \times 10^{-12}} \mathrm{eV}\)

= \(0.511 \times 10^6 \mathrm{eV}=0.511 \mathrm{MeV} .\)

The increase or decrease in energy of an electron while crossing a potential difference of 1V is called 1 electronvolt (eV). 1 eV = 1.6 x 10^-12 erg.

Rest Mass: Einstein’s theory of relativity also informs us that the mass of a substance is not a constant quantity but depends on the velocity of the substance. Especially if the velocity of an object becomes comparable to that of light, then its mass increases significantly.

That is why to measure the true mass of an object as an intrinsic property, the object should be at rest with respect to the observer. The mass of the object thus measured is called its rest mass and the equivalent energy is called rest energy. For example, the rest mass of an electron =9.1 x 10^-28 g, and the rest energy is 0.511 MeV.

Unit Of Mass And Energy: As mass and energy are equivalent to each other, their units are also the same. Sometimes mass is given in units of energy, and energy is given in units of mass. For example, ‘energy of 1g’ denotes 9 x 10²⁰ erg amount of energy; or mass of ‘9 x 10¹⁶ J’ represents a mass of 1 kg. From this equivalence, it can be stated that the rest mass of an electron is 0.511 MeV.

Law Of Conservation Of Mass-Energy: During the mutual transformation of mass and energy, the law of conservation of mass or the law of conservation of energy cannot be applied separately. It becomes the law of conservation of mass energy.

The total amount of mass energy in nature Is constant, it can never be created or destroyed. It can only change from one form to another.

Mass can be transformed into energy only within an atom. The energy thus obtained from mass is the source of atomic energy. When gamma rays with an energy of a few MeV or more enter the electric field of a heavy nucleus, it can be transformed into an electron and a positron (an electron-like particle but of positive charge). This is an example of the transformation of energy to mass.

Two-Dimensional Collisions Numerical Examples

Example 1. Two particles of masses m₁ and m₂, moving with velocities u₁ and u₁, respectively, and making an angle θ between them, collide with each other. After the collision, the 1st particle travels in the initial direction of motion of the 2nd, and vice-versa. Find the velocities of the two particles after collision. Under what condition, would this collision be elastic?
Solution:

Suppose v₁, v₂ are the velocities of the two particles, respectively, after collision. The particles before and after collision move as shown. It also shows the chosen directions of the x and the y-axis.

For momentum conservation along the x-axis, we get, \(m_1 u_1 \cos \frac{\theta}{2}+m_2 u_2 \cos \frac{\theta}{2}=m_1 v_1 \cos \frac{\theta}{2}+m_2 v_2 \cos \frac{\theta}{2},\)

or, \(m_1 u_1+m_2 u_2=m_1 v_1+m_2 v_2\)….(1)

Similarly, along the y-axis, we get, \(-m_1 u_1 \sin \frac{\theta}{2}+m_2 u_2 \sin \frac{\theta}{2}=m_1 v_1 \sin \frac{\theta}{2}-m_2 v_2 \sin \frac{\theta}{2}\)

or, \(-m_1 u_1+m_2 u_2=m_1 v_1-m_2 v_2\)……(2)

Adding equations (1) and (2), \(2 m_2 u_2=2 m_1 v_1\)

or, \(v_1=\frac{m_2}{m_1} u_2\)…(3)

Subtracting equation (2) from (1), \(2 m_1 u_1=2 m_2 v_2\)

or, \(v_2=\frac{m_1}{m_2} u_1\)…..(4)

The kinetic energy before collision is, \(K_1=\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2\)

and that after collision is \(K_2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2=\frac{1}{2} m_1\left(\frac{m_2}{m_1} u_2\right)^2+\frac{1}{2} m_2\left(\frac{m_1}{m_2} u_1\right)^2\)

= \(\frac{1}{2} \frac{m_2}{m_1} m_2 u_2^2+\frac{1}{2} \frac{m_1}{m_2} m_1 u_1^2\)

Here, K₁ ≠K₂; so the collision is inelastic, in general. As a m special case, it would be an elastic collision if K₁ = K₂. It is possible only when m₁ = m₂, i.e., the two particles are of equal masses.

Example 2. A bomb explodes and splits up Into three fragments. Two fragments, each of mass 200 g, move away from each other making an angle of 120°, at a speed of 100 m · s^-1. Find the direction and velocity of the third fragment whose mass is 500 g. Also, find out the energy released in an explosion.
Solution:

The velocity of fragments A and B along OA and OB. A and B have equal mass and speed. From the law of conservation of linear momentum, the third piece must move along OD, in the direction opposite to the resultant of OA and OB.

If the velocity of the third piece is v, then taking the components along the line CD in the CGS system, 500 v = 200 x 10⁴ cos60° + 200 x 10⁴ cos60°

or, v = \(\frac{200 \times 10^4}{500}=4 \times 10^3 \mathrm{~cm} \cdot \mathrm{s}^{-1}=40 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Hence, the velocity of the third fragment is 40 m · s^-1. It moves so as to make an angle 120° with each of OA and OB. The energy released due to the explosion is the kinetic energy of the three fragments.

∴ Energy released = \(\frac{1}{2} \times 200 \times\left(10^4\right)^2\)

+ \(\frac{1}{2} \times 200 \times\left(10^4\right)^2+\frac{1}{2} \times 500 \times\left(4 \times 10^3\right)^2\)

= \(2400 \times 10^7 \mathrm{ergs}=2400 \mathrm{~J} .\)

Example 3. A spaceship while flying in space, splits up into three equal parts, due to an explosion. One fragment keeps moving in the same direction; the other two fly off at 60° to the original direction, on either side. If the energy released due to the explosion is twice the kinetic energy of the spaceship, find the kinetic energy of each of the fragments.
Solution:

Let the mass of the spaceship be 3m and the initial speed be u.

Hence, mass of each fragment = m.

Let their velocities be v₁, v_2, and v₃ after the explosion.

Applying the conservation of momentum law along the direction perpendicular to the original direction of motion,

0 = mv₂ sin60°- mv₃ sin60°

or, mv₂ sin60° – mv₃ sin60°

or, v₂ = v₃ = v (say)

Applying the law of conservation of linear momentum along the original direction of motion,

3 mu = mv₁ + 2mv cos60°

or, 3u = v₁ + v…(1)

The kinetic energy of the spaceship before explosion E= 1/2 x 3mu²

Energy released during the explosion, \(E_r=\frac{1}{2} m\left(v_1^2+2 v^2\right)-\frac{1}{2} \cdot 3 m u^2\)

According to the problem, \(E_r=2 E \quad \text { or, } \frac{1}{2} m\left(v_1^2+2 v^2\right)-\frac{1}{2} \cdot 3 m u^2=2 \times \frac{1}{2} \cdot 3 m u^2\)

or, \(v_1^2+2 v^2-3 u^2=6 u^2\)

or, \(v_1^2+2 v^2=9 u^2\)….(2)

Solving (1) and (2), v₁ =u and v = 2 u

Hence, the kinetic energy of the first fragment after the explosion

= \(\frac{1}{2} m v_1^2=\frac{1}{2} m u^2=\frac{1}{2} \cdot \frac{2}{3} \cdot E=\frac{1}{3} E\)

the kinetic energy of each of the other two fragments

= \(\frac{1}{2} m v^2=\frac{1}{2} m(2 u)^2=2 m u^2=2 \cdot \frac{2}{3} E=\frac{4}{3} E\)

 

Work And Energy Conclusion

If a body undergoes a displacement due to a force acting on it, work is said to be done, and the product of force and displacement is the measure of the work done.

Work Done Against A Force: Consider a particle on which some forces are acting. When an external agent causes a displacement opposite to the direction of the resultant of these forces, work is said to be done against the force.

Work Done By A Force: Consider a particle on which some forces are acting. The resultant of these forces can cause a displacement. This is said to be work done by a force.

• If a force acts at right angles to the direction of displacement of a body, no work is done by the force. This force is called a no-work force.
• The rate of doing work with respect to time is called power. That is work done in unit time is called power.
• Energy is the capacity to do work.
• The ability of a body to do work due to its speed, special position, or special configuration, or all of these, is called its mechanical energy.
• Mechanical energy is of two types— kinetic energy and potential energy.
• The ability of a body to do work due to its speed is called its kinetic energy.
• The ability of a body to do work due to its special position or configuration is called its potential energy.
• The ability of a body to do work, acquired due to its rise against gravity, is called its gravitational potential energy.
• Gravitational potential energy depends on the chosen plane of reference. This energy may also have a negative value.
• The ability of a body to do work, gained due to its special shape, is called elastic potential energy.

Law Of Conservation Of Energy: Energy can neither be created nor destroyed.

Mass-Energy Equivalence: In this universe, the total sum of mass and energy is a constant. This is the law of conservation of mass energy. Mass and energy are equivalent; one can be converted into the other.

• If the total momentum and the total kinetic energy of a system are conserved, the collision is termed as an elastic collision.
• If the total momentum is conserved, but the total kinetic energy is not, it is an inelastic collision.

Coefficient Of Restitution: The coefficient of restitution is defined as the ratio of the velocity with which the two bodies separate after a collision to their velocity of approach before the collision.

1. For elastic collision, e = 1.
2. For perfectly inelastic collision, e = 0.
3. For partially elastic collision, 0 < e < 1.

A system in which the total mechanical energy remains conserved is called a conservative system. Forces acting in such a system are called conservative forces.

In a system where resistive forces are present, the mechanical energy is not conserved. Such systems are called non-conservative systems. Forces of resistance are called dissipative forces.

Work And Energy Useful Relations For Solving Numerical Problems

When a force \(\vec{F}\) acting on a body is associated with a displacement \(\vec{s}\), work done, W = \(\vec{F}\) · \(\vec{s}\) = Fs cosθ, where θ is the angle between \(\vec{F}\) and \(\vec{s}\).

For a variable force \(\vec{F}\), if the displacement of a particle is from A to B, the total work done is

W = \(\int_A^B \vec{F} \cdot d \vec{s}\)

= \(\int_A^B F \cos \theta d s\)

Power (P) = \(\frac{\text { work }(W)}{\text { time }(t)}\) = force (F) x velocity of the body(v)

Kinetic energy = \(\frac{1}{2}\) mv²

For a moving object of mass m, and kinetic energy E, momentum p = √2mE

Gravitational potential energy = mgh

Within the elastic limit, the elastic potential energy of a spring (stretched or compressed by x) = \(\frac{1}{2}\)kx², where k is the spring constant.

Efficiency of a machine

= \(\frac{\text { work output from the machine }}{\text { energy input to the machine }} \times 100 \%\)

In a conservative field, total energy (E) = kinetic energy (K) + potential energy (V) = constant.

The energy equivalent of a mass m is, E = mc², where c is the velocity of light in a vacuum.

Law of conservation of momentum during a linear collision between two bodies: m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

In case of elastic collision

1. \(u_1-u_2=v_2-v_1\) and
2. \(\frac{1}{2} m_1 u_1^2+\frac{1}{2} m_2 u_2^2=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2\)

Coefficient of restitution, e = \(\frac{v_2-v_1}{u_1-u_2}\)

In case of collision of a falling body with a fixed horizontal plane, total distance traveled before coming to rest, d = \(h \frac{1+e^2}{1-e^2}\)

Work And Energy Very Short Answer Type Questions

Question 1. A force is acting on a body in motion, but is not doing any work. Give an example of such a force.
Answer: Centripetal force on a body, in a uniform circular motion

Question 2. Is work a vector or a scalar quantity?
Answer: Scalar quantity

Question 3. What is the amount of work done by a force when the body moves in a circular path?
Answer: Zero

Question 4.In a tug-of-war game, which of the teams does effective work?
Answer: The stronger team

Question 5. A person is carrying a bucket of water and is in a lift moving up with uniform velocity. Is the person doing any work on the bucket of water? Will the energy of the bucket and water remain constant?
Answer: No

Question 6. A motor drives a belt at a constant velocity of v m · s^-1. If m kg of sand falls on the belt per second, what is the rate of work done by the force exerted by the belt on the sand?
Answer: 1/2mv³W

Question 7. A boy tried to lift a bucketful of water but failed. What is the work done by him?
Answer: Zero

Question 8. What is the work done by the tension in the string during the oscillation of a simple pendulum?
Answer: Zero

Question 9. A box was lifted vertically through a height of 6 m in 3 s. If the box had been lifted in a zig-zag way in 5 s, the work done would have been the same. Is the statement true or false?
Answer: True

Question 10. 1 kg · m = ________ J.
Answer: 9.8

Question 11. A force \(\vec{F}=(5 \hat{i}+3 \hat{j}+2 \hat{k}) \mathrm{N}\) acts on a particle, and the particle moves from the origin to a point \(\vec{r}=(2 \hat{i}-\hat{j}) \mathrm{m}\). What will be the work done on the particle?
Answer: 7

Question 12. How many joules are in 1 MeV?
Answer: 1.6 x 10^-13J

Question 13. What is the unit of energy?
Answer: Joule

Question 14. Does the kinetic energy of a ball, thrown inside a moving train, depend on the speed of the train?
Answer: No

Question 15. Which type of energy is lost in doing work against friction?
Answer: Mechanical energy

Question 16. A small car and a lorry are moving with the same kinetic energy. Brakes are applied to produce the same force against the motion. Which one will cover a greater distance before stopping?
Answer: Both cover the same distance

Question 17. When a body falls on the ground from a height, it becomes slightly warm—why?
Answer: Kinetic energy changes to heat energy

Question 18. Is the resistance due to air a conservative force?
Answer: No

Question 19. What happens to internal energy, when the temperature of the body increases?
Answer: Increases

Question 20. What type of energy is stored in the spring of a watch?
Answer: Potential energy

Question 21. The kinetic energies of a heavy and a light object are the same. Momentum of which object will be higher?
Answer: Heavy

Question 22. Momenta of a light and a heavy body are the same. Which body has greater kinetic energy.?
Answer: Lighter

Question 23. If E is the kinetic energy of a body of mass m, what will be its momentum?
Answer: \(\sqrt{2 m E}\)

Question 24. An object breaks up into two masses m₁ and m₂ due to explosion. The two fragments move in opposite directions. What will be the relation between the kinetic energy and the masses?
Answer: Inversely

Question 25. What is the loss of KE of a freely falling body of mass m, during the t th second?
Answer: [1/2 mg²(2t – 1)]

Question 26. The increase in momentum of a body is 100%. What will be the increase in its kinetic energy?
Answer: 300

Question 27. The increase in kinetic energy of a body is 69%. What will be the increase in its momentum?
Answer: 30

Question 28. Which physical quantity in conserved during both the elastic and inelastic collisions?
Answer: Momentum

Question 29. Two objects coalesce after a collision with each other. What is the coefficient of restitution?
Answer: 0

Question 30. The coefficient of restitution between a ball and a horizontal floor is e = 1/2. If the ball falls from a height of 10 m, after its impact with the floor, the ball bounces up to a height of ______.
Answer: 2.5m

Work And Energy Assertion Reason Type Questions And Answers

Direction: These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1.

Statement 1: The absolute PE of a system as measured by two different persons at the same time can be different.

Statement 2: The value of the absolute PE of a system depends upon the reference value chosen.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 2.

Statement 1: Work done by normal contact force can be non-zero.

Statement 2: Normal contact force is always perpendicular to the displacement of the object. (Here displacement is measured with respect to a frame of reference attached to the two surfaces in contact.)

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 3.

Statement 1: In a circular motion, work done by the centripetal force is not always zero.

Statement 2: If the speed of the particle increases or decreases in a circular motion, net force, acting on the particle is not directed toward the center.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 4.

Statement 1: When a body moves uniformly in a circle its momentum goes on changing but its kinetic energy remains constant.

Statement 2: \(\vec{p}=m \vec{v}, \mathrm{KE}=\frac{1}{2} m v^2\). In circular motion \(\vec{v}\) changes, v² does not change.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Question 5.

Statement 1: If a particle of mass m is connected to a light rod and whirled in a vertical circle of radius R, then to complete the circle, the minimum velocity of the particle at the lowest point is \(\sqrt{5 g R}\).

Statement 2: Mechanical energy is conserved and for the minimum velocity at the lowest point, the velocity at the highest point will be zero.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 6.

Statement 1: Work done by constant force is equal to the magnitude of force multiplied by displacement.

Statement 2: Work done is a scalar quantity. It may be positive, negative, or zero.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 7.

Statement 1: If work done by a conservative force is negative then potential energy associated with that force should increase.

Statement 2: This is from the reaction Δu = -W. Here Au is change in potential energy and W is work done by conservative force.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1.

Work And Energy Match Column A With Column B.

Question 1. The displacement time graph of a body is shown.

Answer: 1. A, 2. B, C, D, 3. B, C, D, 4. C

Question 2. A force F = kx (where k is a positive constant) is acting on a particle. In Column A displacements (x) are given and in Column B work done by the force is given.

Answer: 1. B, 2. A, 3. C

Question 3.

Answer: 1. B, 2. D, 3. C

Question 4. The system is released from rest. Friction is absent and string is massless. In time t = 0.3 s (take g = 10 m · s^-2)

Answer: 1. C, 2. A, 3. D, 4. B

Question 5.

Answer: 1. A, B, C, 2. B, C, 3. A, C, 4. A, B, C

Question 6. A particle is suspended from a string of length R. It is given a velocity u = \(3 \sqrt{g R}\) at the bottom.

Answer: 1. C, 2. B, 3. A, 4. E

Work And Energy Question And Answers

Read the following passages carefully and answer the questions at the end of them.

Question 1. A body of mass 2 kg starts from rest and moves with uniform acceleration. It acquires a velocity 20 m · s^-1 in 4 s.

1. Power exerted on the body at 2 s is

1. 50 W
2. 100 W
3. 150 W
4. 200 W

Answer: 2. 100 W

2. Average power transferred to the body in the first 2 s is

1. 50 W
2. 100W
3. 150 W
4. 200 W

Answer: 1. 50 W

Question 2. A ball of mass m is dropped from a height H above a level floor as shown. After striking the ground it bounces back and reaches up to the height h.

1. During the collision, the part of the KE which appears in other forms (other than KE or PE) is (this part of the energy is termed as lost energy as it cannot be utilized properly)

1. mgH
2. mgh
3. mgH – mgh
4. Zero

Answer: 3. mgH – mgh

2. The speed of the ball just after the collision is

1. \(\sqrt{2 g H}\)
2. \(\sqrt{2} g h\)
3. \(\sqrt{2 g(H-h)}\)
4. None of these

Answer: 2. \(\sqrt{2} g h\)

3. If the lost energy in the collision is half of the value computed in Question (1), and H = \(\frac{3 h}{2}\), then the height attained by the ball after the collision is

1. \(\frac{7 h}{4}\)
2. \(\frac{3 h}{4}\)
3. \(\frac{3 h}{2}\)
4. \(\frac{9 h}{5}\)

Answer: 1. \(\frac{7 h}{4}\)

Question 3. A block of 2.5 kg is pulled 2.20 m along a frictionless horizontal table by a constant force of 16 N directed at 45° above the horizontal.

1. Work done by the applied force is

1. 25 J
2. 27 J
3. 24.9 J
4. 22.5 J

Answer: 3. 24.9 J

2. Work done by the normal force exerted by the table is

1. 24.9 J
2. Zero
3. 27 J
4. 27.5 J

Answer: 2. Zero

3. Work done by the force of gravity is

1. 24.9 J
2. 27 J
3. Zero
4. 27.5 J

Answer: 3. Zero

 

Dissipation Of Energy

According to the law of conservation of energy, energy cannot be destroyed. But during the transformation of energy, some energy may change into such a form that has no practical utility and cannot be recovered in any usable form. This is called the dissipation of energy.

Dissipation Of Energy Example: Energy has to be supplied to a machine to make it work. But the work output (i.e., energy) is generally less than the energy supplied. This is because a part of the supplied energy is used to overcome friction and other resistive forces and this part transforms into heat or sound energy that cannot be used for practical purposes and is lost forever.

This is the dissipation of energy. It does not mean the destruction of energy. It simply denotes the transformation of energy into unusable form, also called unavailable energy. Various methods are used to reduce this dissipation of energy. But this could not be minimized to zero yet.

Efficiency Of A Machine: The ratio between the work output of a machine and the energy supplied to it, is called its efficiency. In real life, efficiency is less than 1, and it is often expressed in percentage by multiplying the ratio by 100.

Therefore, the efficiency of a machine = \(\frac{\text { work output of the machine }}{\text { energy supplied (input) }} \times 100 \%\)

For example, 90% efficiency of a machine means that, if 100 units of energy is supplied to the machine, work done by it will be 90 units.

Work Done Against Friction: The total mechanical energy of a body, falling under gravity along a frictionless inclined plane remains conserved. But, a frictionless surface is an ideal one, and cannot be obtained in practice. A frictional force always acts against the motion, and some work has to be done by the body against this force. As a result, some energy is dissipated.

Suppose a body of mass m begins to move from point A under gravity along a rough inclined plane towards C.

The height of point A above the reference plane CD is h. Hence, the potential energy of the body at A = mgh. The body is at rest so its kinetic energy is zero there. Thus, the total mechanical energy of the body at A = mgh + 0 = mgh.

Under the action of the component mg sin# of the weight, the body starts moving down along the incline. Then a frictional force, f = μR = μmg cosθ acts upwards on the body along the inclined plane, where μ = coefficient of friction.

Hence, the resultant downward force along the plane,

F = mg sinθ- μmg cosθ = mg(sinθ – μ cosθ)

the acceleration along the plane, a = \(\frac{F}{m}\) = g(sinθ-μcosθ)

If the velocity of the falling body at B is v, where AB = x, then v² = 2ax = 2xg(sinθ- μcosθ)

Kinetic energy at B = \(\frac{1}{2}\) mv²

= \(\frac{1}{2}\) m · 2xg(sinθ – μcosθ)

= mgx(sinθ-μcosθ)

The potential energy at B = mg · DE = mg(DA-EA) =mg(h- xsinθ)

Hence, total mechanical energy at B

= mg(h- xsinθ) + mgx( sinθ- μcosθ)

= mgh-μmg cosθ · x = mgh – fx…(1)

Equation (1) shows that the mechanical energy at B is less than that at A by fx, which is the work done against the frictional force to cover a distance x along the plane. This amount fx of energy is transformed into unavailable form, in order to overcome the frictional force against motion.

So, the total energy dissipated during the sliding of the body along an inclined plane of length l (=AC) = fl = μmgl cosθ.

The above discussions show that, in the presence of dissipative forces like friction, mechanical energy does not remain conserved for a system. We see that,

Total mechanical energy at A = total mechanical energy at B +fx

The work fx, done against friction, actually transforms into heat energy at the surface of contact of the body with the plane. This heat can never be recovered in any usable form. However, taking this heat into consideration, we see that the total energy is certainly conserved.

Unit 4 Work Energy Power Chapter 1 Work And Energy

Dissipation Of Energy Numerical Examples

Example 1. A block weighing 250 N Is pulled over a horizontal plane at a constant velocity up to a distance of 10 m. The coefficient of kinetic friction Is 0.2 and the force is applied by a string, attached to the block, Inclined at 60° with the vertical. Find the work done against friction.
Solution:

Let the force applied on the block be F

Horizontal component of the applied force along the plane

= \(F \sin 60^{\circ}=\frac{\sqrt{3}}{2} F\) and its vertical component

= \(F \cos 60^{\circ}=\frac{F}{2} .\)

Since there is no vertical acceleration of the block, net force acting vertically is zero.

i.e., R + Fcos60° = W (where R is the normal reaction on the block)

∴ R = W – \(\frac{F}{2}\)

As the body is moving with a uniform velocity, the horizontal component of applied force = frictional force

or, \(\frac{\sqrt{3}}{2} F=\mu R=\mu\left(W-F \cos 60^{\circ}\right)=0.2\left(250-\frac{F}{2}\right)\) = 50-0.1 F

∴ F = \(\frac{50}{0.866+0.1}=\frac{50}{0.966} \mathrm{~N}\)

Hence, work done by the applied force

= \(F \sin 60^{\circ} \times 10=\frac{50}{0.966} \times \frac{\sqrt{3}}{2} \times 10=448.25 \mathrm{~J} .\)

Example 2. A particle Is sliding down along an inclined plane. The frictional force is 0.2 times the normal reaction, and the inclination of the plane is 60J. What Is the acceleration of the particle? If the mass of the particle is 1 g, find the change in the sum of potential and kinetic energies of the particle as it slides down the plane by 1 m.
Solution:

Let the acceleration of the particle along the inclined plane = a and the downward force on the particle along the plane = mg sinθ – f

∴ ma = mgsinθ – μR = mg sinθ – μmg cosθ

or, a = g (sinθ – nμcosθ) =9.8 (sin60° – 0.2cos60°)

= \(9.8\left(\frac{\sqrt{3}}{2}-0.2 \times \frac{1}{2}\right)=7.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Change in mechanical energy

= work done against friction = μmg cosθ · s

= 0.2 x 0.001 x 9.8 x \(\frac{1}{2}\) x 1 = 0.00098 J.

Example 3. A box of mass 12 kg is pushed up by a distance of 10 m on the application of a 100 N force along a plane of inclination 30°. If the coefficient of friction is \(\frac{1}{\sqrt{3}}\), find the work done against friction, [g = 10 m · s^-2]
Solution:

Frictional force, f = μR = μmg cos30°

Hence work done against friction, W = fs

= μmg cos30° · s

= \(\frac{1}{\sqrt{3}} \times 12 \times 10 \times \frac{\sqrt{3}}{2} \times 10=600 \mathrm{~J}\)

Example 4. A car of mass 1000 kg moves up at 40 km · h^-1 along an inclined plane of slope \(\frac{1}{50}\). The coefficient of rolling friction between the road and the wheels of the car is 0.3. Find the power of the car engine.
Solution:

The angle of inclination = θ.

∴ Slope = tanθ = \(\frac{1}{50}\) ≈ sinθ [θ is small]

The resultant downward force on the car along the incline

= mg sinθ + f

= mg sinθ + μR

= mg sinθ + μmg cosθ = mg(sinθ + μcosθ)

= \(1000 \times 9.8\left(\frac{1}{50}+0.3 \times 1\right)\) (as is very small cosθ ≅ 1)

= 9800 x 0.32 = 3136 N

Velocity of the car =40 km · h^-1 = 11.11 m · s^-1

∴ Power of the car =3136 x 11.11 N · m · s^-1 =34840.96 W = 34.84 kW.

Example 5. An engine, working at a constant rate, is pulling a train of mass 500 tonne along a plane of inclination sin^-1\(\frac{1}{100}\).frictional force per metric tonne is 49 N and the train is moving with a velocity of 10 m · s^-1, what is the power of the engine in kilowatt? [1 tonne (metric ton) = 1000 kg]
Solution:

Downward effective force on the train along the inclined plane = mg sinθ + frictional force (f)

= 500 x 1000 x 9.8 x \(\frac{1}{100}\) + 500 x 49 = 7500 x 9.8 N

Velocity of the train = 10 m · s^-1

Since the car is moving upward with a constant velocity, the force applied by the car’s engine, F must exactly balance the net downward force.

∴ F = 3136 N

Hence, power of the engine = 7500 x 9.8 x 10 = 735000 J · s^-1 = 735000 W = 735 kW.

Example 6. A loaded lorry of total mass 5000 kg can come down from the top of a slope (1:40) effortlessly at 18 km · h^-1. What should be the horsepower of its engine so that it can go up with the same speed, from the base to the top? Resistance due to friction may be taken to be the same in both cases.
Solution:

Velocity of the lorry = \(18 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{18 \times 1000}{60 \times 60} \mathrm{~m} \cdot \mathrm{s}^{-1}=5 \mathrm{~m} \cdot \mathrm{s}^{-1}\) \(\sin \theta=\frac{1}{40}\)

As the speed of the lorry remains constant for the downward journey, the friction just balances the downward active force mg sinθ.

∴ Frictional force, f = mg sinθ.

Effective force on the lorry for its upward journey = mg sinθ + f = mg sinθ + mg sinθ = 2mg sinθ

∴ Power of the engine = effective force on the lorry x velocity of the lorry

= 2mgsinθ x 5 = 2 x 5000 x 9.8 x \(\frac{1}{40}\) x 5

= 12250 W = \(\frac{12250}{746}\)hp = 16.42 hp.

Example 7. A car of mass 500 kg is moving up along an inclined surface of slope \(\frac{1}{25}\) at a constant speed of 72 km · h^-1. If the coefficient of friction between the road and the car wheel is 0.1, find the power of the car engine (g = 9.8 m · s^-2).
Solution:

Frictional force, f = μmg cosθ, v = 72 km · h^-1 = 20 m · s^-1, m = 500 kg, g = 9.8m · s^-2, μ = 0.1 and tanθ = \(\frac{1}{25}\)

∴ sinθ  = \(\frac{1}{\sqrt{626}}\) and

cosθ = \(\frac{25}{\sqrt{626}}\)

F = force opposing the motion of the car = f + mg sinθ = mg(μcosθ + sinθ)

∴ Power of the car engine, P = Fv= mg(μcosθ + sinθ)v

= \(500 \times 9.8 \times\left(\frac{0.1 \times 25}{\sqrt{626}}+\frac{1}{\sqrt{626}}\right) \times 20\)

= \(13709 \mathrm{~J} \cdot \mathrm{s}^{-1}=13709 \mathrm{~W} .\)

Example 8. A 1.5 m long chain of mass 0.8 kg is kept on a horizontal table and a part of its length hangs from the edge of the table. When the length of the hanging part is one-third the total length of the chain, it starts sliding off the table. What will be the work done by friction when the whole length of the chain slides off the table?
Solution:

When \(\frac{1}{3}\) of the chain is hanging, it starts sliding.

In this condition, frictional force = weight of the hanging part of the chain

or, μ x \(\frac{2}{3}\)mlg= \(\frac{1}{3}\)mlg [μ = coefficient of friction, m = mass per unit length of the chain, and l = length of the chain]

∴ μ = \(\frac{1}{2}\) = 0.5

When the whole length of the chain slides off the table, the effective frictional force on the chain = 0.

∴ Effective average frictional force on the chain = \(\frac{\frac{2}{3} \mu \mathrm{mlg}+0}{2}=\frac{1}{3} \mu \mathrm{mlg} .\)

The chain moves through a distance of \(\frac{2}{3}\). l against the effective friction. Work done against friction is, therefore,

W = \(\frac{1}{3} \mu m g l \times \frac{2}{3} l=\frac{2}{9} \mu m g l^2\)

= \(\frac{2}{9} \times 0.5 \times \frac{0.8}{1.5} \times 9.8 \times(1.5)^2\)

= \(1.3 \mathrm{~J} .\)

Example 9. A body of mass 10 kg is pushed up 50 cm from the ground, along a plane inclined at 45° to the horizontal. if the coefficient of friction is 0.2, then calculate the work done.
Solution:

Here, h = 50 cm = 0.5 m, m = 10 kg, g = 9.8 m · s^-2, θ = 45°, μ = 0.2

Let the friction acting on the body be f. Then, f = μmg cosθ

The force against which the body is pushed up is F = f + mg sinθ

= mg(μ cosθ + sinθ)

The body is pushed up by a distance \(\frac{h}{\sin \theta}\) along the inclined plane.

Therefore, the work done is, W = \(\frac{F h}{\sin \theta}=\frac{m g h}{\sin \theta}(\mu \cos \theta+\sin \theta)\)

= \(\frac{10 \times 9.8 \times 0.5}{\sin 45^{\circ}} \times\left(0.2 \times \cos 45^{\circ}+\sin 45^{\circ}\right)\)

= \(49 \times \sqrt{2} \times\left(0.2 \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right)=58.8 \mathrm{~J} .\)

Example 10. By application of a force F, a body of mass m Is raised to the top of a hill. F is tangential along the whole path. If the height of the hill Is h, the length of the base of the hill is l, and the coefficient of friction is μ, then find the work done.
Solution:

The total path from the bottom to the top of the hill can be considered as an assembly of a large number of inclined planes of varying angles of inclination. Consider an incline of length Δs.

Its vertical height is Δh and the angle of inclination is α.

Total work done in lifting the body along this inclined surface = work done against gravity + work done against friction.

∴ \(\Delta W=m g \Delta h+\mu m g \cos \alpha \cdot \Delta s\)

= \(m g \Delta h+\mu m g \cos \alpha \frac{\Delta l}{\cos \alpha}=m g(\Delta h+\mu \Delta l)\)

∴ Total work done to lift the body up to the top of the hill,

W = \(\sum \Delta W=m g\left(\sum \Delta h+\mu \sum \Delta l\right)\)

= \(m g(h+\mu l)\)

Work

Work Definition: Work is said to be done when an object is displaced from its initial position under the action of a force. Work is measured as the scalar product of the applied force and the displacement of the object.

Let \(\vec{F}=\overrightarrow{O B}\) = force acting on a particle and \(\vec{s}=\overrightarrow{O A}\) corresponding displacement of the particle.

Then, by definition, the work done is W = \(\vec{F}\) \(\vec{s}\)…..(1)

If θ = ∠BOA = angle between the force and the displacement vectors, then from the property of the scalar product, we have W = Fscosθ……(2)

From the figure we get, \(F \cos \theta=O B \cdot \frac{O C}{O B}=O C\) = component of the force \(\vec{F}\) along the direction of the displacement \(\vec{s}\).

Hence, from equation (2), it may be written that, work done = displacement x component of force along the direction of displacement

If the components of \(\vec{F}\) and \(\vec{s}\) are known, equation (1) can be expressed as,

W = \(\left(F_x \hat{i}+F_y \hat{j}+F_z \hat{k}\right) \cdot\left(s_x \hat{i}+s_y \hat{j}+s_z \hat{k}\right)\)

= \(F_x s_x+F_y s_y+F_z s_z\)

It is very important to note that, force and displacement are both vector quantities, but their product, work, is a scalar quantity.

Work Discussions:

1. Force Without Displacement: If an object is not at all displaced under the action of a non-zero force \(\vec{F}\), then work done is considered to be zero,

i.e., if s = 0 but F≠ 0, then

W = Fscosθ = Fcosθ x 0 = 0

2. Displacement Without Force: If an object is displaced in the absence of any force acting on it, i.e., if s≠0 when F = 0, we find from equation (2), W = 0. For example, an object moving with uniform speed in a straight line maintains its state of motion in the absence of any external force on it (Newton’s 1st law).

But its displacement is not zero, because it is actually moving and as a result changing its position. In this case, there is no work done, only because F = 0.

3. No-Work Force: If the force and the displacement vectors are perpendicular to each other, then θ = 90° or cosθ = 0. So, from the equation (2), we have W = 0. In such cases, the force acting on an object is called a no-work force.

For example, an object in a uniform circular motion is displaced along the tangent of the circle, whereas the active force, called the centripetal force, acts in the radial direction. As the tangential and the radial directions are perpendicular to each other, the centripetal force is a no-work force.

Sign Convention Of Work: Work done (W) is conventionally defined as W = \(\vec{F}\) – \(\vec{s}\) = Fscosθ….(3)

The different values of cosθ give rise to different cases:

1. If θ lies between 0 and 90°, cosθ is positive. So, from equation (3), the work done is positive.
2. If θ lies between 90° and 180°, cosθ is negative. So, from equation (3), the work done is negative.

It should be mentioned that the opposite convention, W = –\(\vec{F}\)– \(\vec{s}\), may also be used. No error will occur if this alternative relation is consistently followed throughout.

In many problems, we need to deal only with the absolute value of work done. Then, the positive or negative sign of W would not be important.

Sign Convention Of Work Discussion: When an object is displaced in a direction opposite to the force acting on it, it is said that the work is done against the force. In this situation, there must exist an external agent that is responsible for the displacement of this object against the force.

Lifting of an object upwards against the downward force of gravity acting on it, the motion of an object on a rough surface against the force of friction, etc., are examples of this type of work. In these cases, usually, some person or some machine acts as the external agent.

Let us take the example of lifting an object against gravity. This can be explained in two ways:

1. The external agent (which lifts the object) does positive work on the object by applying a force \(\vec{F}_2\) or
2. Gravity \(\vec{F}_1\) does negative work on the object.

Similarly, when an object falls under the action of gravity and an external agent opposes the motion, it can also be explained in two ways:

1. The external agent does negative work on the object or
2. Gravity does positive work on the object.

Work Done On A System Of Bodies: The application of a force on a system, comprising a number of bodies, may produce different displacements for different bodies within the system, In such cases, to calculate the work done, the displacement of the point of application of the force is considered.

Hence, in this case, work done = force on the system of bodies x displacement of the point of application

Work Done By A System Of Forces: Since work is a scalar quantity work done due to a number of forces acting on a body is the algebraic sum of the work done due to each of the forces. The resultant of the forces produces the same amount of work.

Let a system comprising three forces act on a body, and the work done due to this system be W. Work done due to the individual forces are, w₁, w_2, and w₃. The resultant of the system of forces is \(\vec{F}\) and the displacement of the body is \(\vec{s}\). Hence W = w₁ + w₂+ w₃ = \(\vec{F}\) · \(\vec{s}\)

Work Done By A Varying Or A Variable Force: So far we have considered the work done by a force that is constant both in magnitude and direction. But often the force doing work is variable. A varying force means, a force whose magnitude or direction or both changes continuously.

• For example, when a rocket is fired upwards from the Earth, the force required to keep the rocket moving away from the Earth is not constant, it continuously decreases.
• Similarly, when a spring, whose end is fixed to a rigid support, is stretched by a force through a small distance x then the restoring force F developed in the spring increases with x.
• Let us consider a variable force F to be acting on a particle. Under the influence of this force, the particle is displaced along the x-axis from the initial position x_i to the final position x_j.
• The x-component of the force is denoted by F_x. In this case, the work done cannot be simply stated by W = F_x(x_f– x_i). This equation is applicable when the applied force is a constant.
• Here the force F varies with x and hence both F and F_x are functions of x. To calculate the work done in this case, the interval x_i to x_f is divided into a large number of small displacements Δx, so that F_x can be assumed to be constant over this displacement Δx.
• Hence for a very small displacement Ax, work done can be considered to be ΔW = F_xΔx.
• This can also be considered to be the area of the shaded region.

Therefore total work done when the particle is displaced from the initial position x_i to the final position x_f is given by the relation

W = \(\sum_{x_i}^{x_f} F_x \Delta x\)…(1)

If \(\Delta x \rightarrow 0\), then \(\lim _{\Delta x \rightarrow 0} \sum_{x_i}^{x_f} F_x \Delta x=\int_{x_i}^{x_f} F_x d x\)….(2)

From (1) and (2) we obtain, \(W=\int_{x_i}^{x_f} F_x d x\)

In this case, the total work done can be considered to be the area enclosed by the curve between x_i and x_f on the x-axis.

Work Done On A Particle Moving Along A Curved Path: Let us consider a particle being moved along the curve AB as shown by a force \(\vec{F}\) which continuously changes in magnitude as well as in direction.

According to the diagram, if \(\vec{ds}\) is a small displacement of the particle under the action of the force, then the total work done in displacing the A particle from A to B is

W = \(\int_A^B \vec{F} \cdot d \vec{s}=\int_A^B F \cos \theta d s\)

It is to be noted that in this integration, neither F nor θ is a constant.

Graphical Representation Of Work: The relationship between forces acting on a body and its displacement can be represented in a force-displacement graph.

For A Constant Force: If the displacement of a body is s₀ under the action of a constant force F₀, we get a straight line graph AB. AB is parallel to the x-axis as F₀ is constant.

Obviously, the area under the graph, ABPO, is equal to F₀S₀ which is the magnitude of the work done.

For A Varying Force: If s₀ is the displacement of a body under a varying force, the force-displacement graph is not a straight line, but a curved line like AC or AD. In these cases also, it can be proved that the area under the curve (AC or AD) gives the work done.

Hence, work done corresponding to graph AC = area ACPO; work done corresponding to graph AD = area ADPO.

Thus, in general, work done = area under the force-displacement graph.

Absolute Units And Dimension Of Work: Work done = force x displacement of the body. Hence, a unit of work depends on units of force and displacement.

A unit work is said to be done when a unit force acting on a body produces a unit displacement along the direction of the force.

Dimension: Dimension of work = dimension of force x dimension of displacement = MLT^-2 x L = M²T^-2

Unit 4 Work Energy Power Chapter 1 Work And Energy

Work Numerical Examples

Example 1. To displace a body by 50 m, 150 J of work is done. What is the force applied in the direction of the displacement?
Solution:

Work done, W = Fs, where W = 150 J, s = 50 m

∴ 150 = 50F or, F = 3 N .

Example 2. A body of mass 10 kg is raised by 5m. What is the work done?
Solution:

Force, F= mg = 10 x 9.8 = 98 N ; displacement, s = 5 m

∴ Work done, W= Fs = 98 x5J = 490 J

Example 3. A cycle with the rider has a total mass of 80 kg; it rolls down 60 m on a plane of inclination 30°. What is the total work done by gravity on the cycle?
Solution:

Force acting vertically downwards, F = 80 kg-wt = 80 x 9.8 N

Displacement, s = 60 m

The angle between force and displacement, θ = 60°

∴ W=Fscosθ

= (80 x 9.8) x 60 x cos60°

= 80 x 9.8 x 60 x 0.5 = 23520 J

Example 4. A man of mass 100 kg climbs up a ladder of length 10 m. The ladder makes an angle 60° with the horizontal. Find the work done by the man against gravity in climbing up the ladder, [g = 9.8 m · s^-2]
Solution:

Work done to climb up the ladder by 10 m is equivalent to the work done to climb up a vertical height h.

Here, \(\sin 60^{\circ}=\frac{h}{10}\)

or, \(h=\frac{10 \sqrt{3}}{2}\)

∴ Work done, \(W=m g \times h\)

= \(100 \times 9.8 \times 5 \sqrt{3}=8487.04 \mathrm{~J}\).

Example 5. A body is constrained to move along the z-axis is subject to a constant force F = \((-\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}\). Calculate the work done by this force in moving the body a distance of 4m along the z-axis.
Solution:

Displacement, \(\vec{s}=4 \hat{k} \mathrm{~m}\)

Force, \(\vec{F}=(-\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}\)

∴ Work done, W = \(\vec{F} \cdot \vec{s}=(-\hat{i}+2 \hat{j}+3 \hat{k}) \cdot 4 \hat{k}\)

= \(-4 \hat{i} \cdot \hat{k}+8 \hat{j} \cdot \hat{k}+12 \hat{k} \cdot \hat{k}\)

= \(0+0+12 \cdot 1=12 \mathrm{~J} .\)

Example 6. A force F-acting on an object varies with distance x as shown here. The 2 force is in N and x in m. What is the amount of work done by the force in moving the object from x = 0 to x = 6m?

Solution:

Work done, W = area of the rectangle OABC + area of the triangle BCD,

= 3 x 3 + 1/2 x 3 x 3 = 13.5J

Example 7. The relationship between the force F and the position x of a body is as shown.  What will be the amount of work done in displacing the body from x = 1 m to x = 5m?

Solution:

Work done,

W = 10 x 1 5 x 1 + (-5 x 1) + 1/2 x 1 x 10

= 10 + 5 – 5 + 5

= 15J

Example 8. A position-dependent force F = (7 – 2x + 3x²) N acts on a body of mass 2 kg and displaces it from x = 0 to x – 5 m. Determine the amount of work done in i joule.
Solution:

Work done,

W = \(\int F d x=\int_0^5\left(7-2 x+3 x^2\right) d x\)

= \(\left[7 x-x^2+x^3\right]_0^5\)

= \(7 \times 5-5^2+5^3=35-25+125=135 \mathrm{~J}\)

Work And Energy

Power

Power Definition: The time rate at which work is done by a force is called power due to that force or simply work done per unit time is called power.

If W is the amount of work done by a force in the time interval t, then power, \(\bar{P}=\frac{W}{t}=\frac{\vec{F} \cdot \vec{s}}{t}=\vec{F} \cdot \vec{v}\)…(1)

In general, work may not be done at a uniform rate throughout the time t. So, equation (1) actually represents the average power.

On the other hand, if work is done at a variable rate i.e., work W( t) done by a force is a function of time (t), then power at any given instant or instantaneous power is more useful. Let dW be the work done in an infinitesimal time interval dt, then the instantaneous power is P = \(\frac{dW}{dt}\)…(2)

If in that interval dt an infinitesimal displacement \(d \vec{s}\) takes place, then dW = \(\vec{F} \cdot d \vec{s}\)

or, P \(=\frac{d W}{d t}=\vec{F} \cdot \frac{d \vec{s}}{d t}=\vec{F} \cdot \vec{v}=F v \cos \theta\); where, \(\vec{v}=\frac{d \vec{s}}{d t}\) = instantaneous velocity and θ = the angle between the direc¬tion of force applied and displacement.

Power Example: Suppose a robot does a work of 1000 erg in 5 s and another robot takes 10 s to do the same work. Hence, the power of the first robot = 1000/5 = 200 erg · s^-1 and that of the second robot = 1000/10 = 100 erg · s^-1.

Though both of them do the same amount of work, the power delivered by the first robot is greater.

Absolute Units And Dimension Of Power: We know, power = \(\frac{work}{time}\). So the unit of power depends on the units of work and time.

The power to do unit amount of work in a unit time is the unit of power.

Abtokito Units:

Abtokito Units Relations:

1W = 1 J · s^-1 = 10⁷ erg ·  s^-1

1 kW = 10³ W, 1 MW = 10⁶ W

The practical unit of power in FPS system is horsepower (hp). When a mass of 550 lb is raised by 1 ft in 1 s against gravity, the power is 1 hp. The power of heat engines or electric motors are usually measured and expressed in horsepower.

So, 1 hp = 550 ft · lb · s^-1

Relation Between Horsepower And Watt: 1 hp =550 ft · lb · s^-1

= (500 x 30.48) cm x (453.6 x 981) dyn · s^-1

= 746 x 107 erg · s^-1 = 746 J · s^-1 = 746 W

∴ 1 kW = \(\frac{1000}{746}\) = 1.34 hp.

Dimension:

Dimension of power = \(\frac{\text { dimension of work }}{\text { dimension of time }}\)

= \(\frac{M L^2 T^{-2}}{T}\)

= \(\mathrm{ML}^2 T^{-3}\)

Concept Of Power: Appliances used in our daily lives like electric bulbs, heaters, and motors are selected on the basis of their respective powers, and not of the total work that such devices can do. The power of a bulb is important as bulbs with higher wattage are more bright. The temperature of water can be raised faster with a high-power heater. The power ratings of appliances play a very important role in our practical lives.

Unit 4 Work Energy Power Chapter 1 Work And Energy

Power Numerical Examples

Example 1. A man of mass 50 kg climbs up 20 steps of a staircase in 5 s. Each step is 30 cm high. Find the power applied by the man.
Solution:

Height of 20 steps =20 x 30 = 600 cm = 6 m

∴ Work done = 50 x 9.8 x 6 = 2940 J [as m = 50 kg, g = 9.8 m · s^-2]

∴ Power = \(\frac{2940}{5}\) J · s^-1 = 588 W

Example 2. Find the power applied by a man of mass 70 kg, carrying a load of 45 kg, moving up at 6.4 km per hour along a plane of inclination \(\frac{1}{10}\).
Solution:

Velocity of the man, \(v=6.4 \mathrm{~km} \cdot \mathrm{h}^{-1}=\frac{6.4 \times 5}{18} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Total weight of the man and the load = (70 + 45)kg = 115 x 9.8 N

While moving up, the man has to work against the force mg sinθ

Hence, the power applied by the man, P = mg sinθ x v

= \(115 \times 9.8 \times \frac{1}{10} \times \frac{6.4 \times 5}{18}\)

[Here, tanθ = 1/10 ≈ sinθ (as θ is small)]

= 200.36 W

Example 3. Water is lifted to a vertical height of 90 m using a 7.46 kW engine. If the efficiency of the engine is 80%, find the amount of water lifted in a minute.
Solution:

Let the mass of water lifted per minute be m.

Effective power of the engine = 7.46 x 10³ x 0.8 W

Now, work done per second = the effective power

∴ \(\frac{F s}{t}=\frac{m g s}{t}=7.46 \times 10^3 \times 0.8\)

or, m = \(7.46 \times 10^3 \times 0.8 \times \frac{t}{g s}\)

= \(\frac{7.46 \times 10^3 \times 0.8 \times 60}{9.8 \times 90}\)

(because \(t=1 \mathrm{~min}=60 \mathrm{~s}, g=9.8 \mathrm{~m} \cdot \mathrm{s}^{-2}, s=90 \mathrm{~m}\)) \(\approx 406 \mathrm{~kg}\)

Work And Energy Long Answer Type Questions

Question 1. A car is moving with a uniform velocity along a horizontal road. Does the engine of the car do any work in this instance?
Answer:

The force of friction acts opposite to the direction of motion of the car. The force applied by the engine of the car against friction maintains its uniform velocity. This force is doing work as displacement occurs against friction. If f = force of friction and v = velocity of the car, work done in 1 s by the engine = force of friction x constant velocity = fv.

Question 2. Earth revolves around the sun (say in a circular path) under the action of the force exerted by the sun on the Earth. Is the sun doing any work? Explain.
Answer:

Centripetal force (which is provided by the gravitational pull of the sun in this case) that acts on a body in a circular motion is perpendicular to the displacement of the body at every point of its motion and hence does not do any work. Component of this force in the direction of the displacement = Fcos90° = 0 (F = applied force). So, the gravitational pull of the sun does not do any work for the revolution of the earth, as the force acts perpendicular to the earth’s direction of displacement.

Question 3. A man is swimming against the current such that his position with respect to the river bank remains unaltered. Is the man doing any work?
Answer:

The relative displacement of the man with respect to the river bank is zero. So, no work is done by the man. The resistive force of river water just balances the force applied by the man. So this case is similar to a force applied against static friction.

Question 4. When a weightlifter over his head, remains to how much stationary work does he do?
Answer:

To lift a weight mg above his head through a height h, work mgh has to be done against gravity. But when the weight is held stationary above his head, no displacement occurs and no work is done by the lifter at that stage.

Question 5. A bullet is fired from a rifle. The rifle is free to recoil. Compare the kinetic energy of the rifle with that of the bullet.
Answer:

K.E., \(K=\frac{1}{2} m v^2=\frac{1}{2} \frac{(m v)^2}{m}=\frac{p^2}{2 m}\) [p = momentum]

Both the rifle and the bullet were at rest before firing. After firing, the momenta of the rifle and the bullet must be equal in magnitude and opposite in direction in order to conserve linear momentum.

As p is the same for both, K ∝ \(\frac{1}{m}\). As the rifle is heavier, its K.E. is less than that of the bullet.

Question 6. If a car is driven along a straight path by an engine of constant power, find the displacement-time relation.
Answer:

Let the power of the engine be P (constant).

But P = velocity x force = \(\nu m a=v m \frac{d v}{d t}\)

or, \(v d v=\frac{P}{m} d t\).

Integrating, \(\frac{v^2}{2}=\frac{P}{m} t+C\), where C= integration constant.

or, \(v=\sqrt{2} \sqrt{\frac{P}{m} t+C}\)

or, \(d s=\sqrt{2} \sqrt{\frac{P}{m} t+C} d t\left[because v=\frac{d s}{d t}\right]\)

Let \(\frac{P}{m} t+C=z\)

∴ \(\frac{P}{m} d t=d z \quad \text { or, } d t=\frac{m}{P} d z\)

∴ ds = \(\sqrt{2} \frac{m}{P} z^{\frac{1}{2}} d z\)

By further integration we get,

s = \(\frac{\sqrt{2} m \frac{2}{P}}{3} z^{3 / 2}+C^{\prime}\)

= \(\frac{\sqrt{2} m}{P} \cdot \frac{2}{3}\left(\frac{P}{m} t+C\right)^{3 / 2}+C^{\prime}\)

[C’ = integration constant]

= \(\frac{2 \sqrt{2}}{3} \sqrt{\frac{P}{m}}\left(t+\frac{m C}{P}\right)^{3 / 2}+C^{\prime}\)

This is the required relationship between time and displacement.

Question 7. A car and a lorry are moving with the same kinetic energy. Both are brought to rest with the same opposing force applied through brakes. Which one will stop within a short distance?
Solution:

As soon as the lorry and the car come to rest, their kinetic energies become zero. Hence, the change in kinetic energy (= work done by the brakes) will be the same for both vehicles. Now, work done = force x displacement. As the force applied is the same, both the car and the lorry will cover the same distance before coming to rest.

Question 8. The magenta of a lighter and a heavier mass are equal. Which one of them has a greater kinetic energy?
Answer:

Let the mass of the lighter body = m and its velocity = v, the mass of the heavier body = M and its velocity = V

Given, mv = MV.

Now, the kinetic energy of the lighter body, \(K_l=\frac{1}{2} m v^2\) and the kinetic energy of the heavier body, \(K_h=\frac{1}{2} M V^2\)

∴ \(\frac{K_l}{K_h}=\frac{\frac{1}{2} m v^2}{\frac{1}{2} M V^2}=\frac{m^2 v^2}{m} \times \frac{M}{M^2 V^2}=\frac{M}{m}>1\) [because m v=M V]

So, the kinetic energy of the lighter body is greater than that of the heavier body.

Question 9. The kinetic energy of a lighter body is equal to that of a heavier body. Which one of them has greater momentum?
Solution:

Let the mass and velocity of the lighter body be m and v, and those of the heavier body be M and V, respectively.

Given, \(\frac{1}{2} M V^2=\frac{1}{2} m v^2\) or, \(\frac{M^2 V^2}{M}=\frac{m^2 v^2}{m}\)

or, \(\frac{m^2 v^2}{M^2 V^2}=\frac{m}{M}\)

or, \(\frac{m v}{M V}=\sqrt{\frac{m}{M}}\) or, \(\frac{\text { momentum of the lighter body }}{\text { momentum of the heavier body }}\)

= \(\sqrt{\frac{m}{M}}<1\)

Hence, the momentum of the heavier body is greater.

Question 10. A body has mechanical energy but no momentum, or has momentum but no mechanical energy—is it possible?
Answer:

A body at rest at a height from the ground has potential energy, a form of mechanical energy, but no momentum.

A body with momentum must have some velocity and thus some kinetic energy associated with it. Since kinetic energy is a form of mechanical energy, a body with momentum must have some mechanical energy.

Question 11. A man with a bucket of water in his hand is standing inside a lift moving upwards. Write

1. Whether the man does any work on the bucket of water
2. Whether the energy of the bucket remains unchanged.

Answer:

1. As there is no displacement of the bucket with respect to the man, no work is done by the man on it.
2. The height of the bucket of water, with respect to the ground, increases as the lift moves up. So, the potential energy of the bucket increases. Hence, the total energy of the bucket increases.

Question 12. One gets less hurt when one jumps from a height on sand than on hard floor. Why?
Answer:

When a man jumps on any surface, the surface in contact with the man gets a little depressed. This depression is more in the case of a sandy surface than of a hard floor. The initial potential energy of the man is the same in both cases. So the kinetic energy just before hitting the ground is also the same.

Suppose, the mass of the man = m and he jumps from a height h. The average reaction force of the ground on him is R. He stops after depressing the ground by x. So his kinetic energy is used up in work done against the net reaction force R – mg.

As kinetic energy just before hitting the ground = potential energy at height h = mgh

(R- mg)x = mgh or, R = (1 + \(\frac{h}{x}\) mg)

As x is more for a sandy surface, R is less and hence, the man is less hurt.

Question 13. Does work done against gravity depend on the speed of lifting a body? Explain.
Answer:

Magnitude of work done to raise a body through a height h,

W = weight of the body x height

Thus work done does not depend on the time of rise, i.e., on the speed of doing the work.

Question 14. Inside a compartment of a train running with a uniform velocity, a boy throws a ball. Does the kinetic energy of the ball depend on the velocity of the train?
Answer:

Here, the train runs with a uniform velocity. So, if we consider the train as the frame of reference the velocity and the kinetic energy of the ball becomes zero and does not depend on the velocity of the train. When the kinetic energy is calculated, taking the earth’s surface as the inertial frame of reference, it will depend on the velocity of the train.

Question 15. To reach the same height, why is it easier to follow a path of lower inclination?
Answer:

To reach the same height, the work done is equal for all paths since the potential energy attained is the same. Now, work done = force x displacement. For a path of lower inclination, this displacement is higher; so the force required is less. Thus it is easier to follow such a path.

Question 16. Show that for a simple pendulum, the work done by the tension of the string becomes zero during its oscillation.
Answer:

Work done by a force is the product of the force with the component of the displacement along the force. In the case of an oscillating simple pendulum, the tension in the string always acts at right angles to the displacement of the bob. Hence, the component of displacement of the bob is zero in the direction of tension; therefore, the tension does not do any work.

Question 17. Prove that, a freely falling body of mass m loses its potential energy by \(\frac{1}{2}\)mg(2t-l) during its fall in the tth second.
Answer:

Suppose the body starts from rest. The displacement in t seconds,

X = \(\frac{1}{2}\)gt

The displacement in (t – 1) seconds, y = \(\frac{1}{2}\)g(t-1)²

∴ Displacement in the t th second,

h = \(x-y=\frac{1}{2} g t^2-\frac{1}{2} g(t-1)^2=\frac{1}{2} g(2 t-1)\)

∴ Decrease in potential energy in the t th second = \(m g h=m g \cdot \frac{1}{2} g(2 t-1)=\frac{1}{2} m g^2(2 t-1)\)

Question 18. Gravitational force is a conservative force, but fric¬tional force is non-conservative —why?
Answer:

A conservative force is one for which the work done can be restored. To lift a bodywork has to be done against gravity. This work gets stored in the body as its potential energy. The body, while returning to its initial state, does the same amount of work using that stored potential energy. Hence, gravitational force is a conservative force.

If the work done against a force cannot be restored, the force is called non-conservative. To pull a body over a rough surface, work has to be done against friction. This work can never be recovered. To return the body to its initial state, again work has to be done against friction. So friction is a non-conservative force.

Question 19. Can the kinetic energy of a body be negative?
Answer:

A body of mass m moving with a velocity v has kinetic energy \(\frac{1}{2}\)mv². m cannot be negative, and v² being the square of a real quantity, cannot also be negative. Hence, the kinetic energy of a body cannot be negative.

Question 20. The momentum of a body is increased by 100%. What is the percentage increase in its kinetic energy?
Answer:

Let the mass of the body be m.

Initial momentum = p₁; hence, final momentum after a 100% increase, p₂ = 2p₁.

If K₁ and K₂ are initial and final kinetic energies, \(K_1=\frac{p_1^2}{2 m} \text { and } K_2=\frac{p_2^2}{2 m}\)

∴ \(\frac{K_2}{K_1}=\frac{p_2^2}{p_1^2}=\frac{\left(2 p_1\right)^2}{p_1^2}=4=\frac{4}{1}\)

or, \(\frac{K_2-K_1}{K_1} \times 100=\frac{4-1}{1} \times 100=300\)

So, the kinetic energy increases by 300 %.

 

 Work And Energy Multiple Choice Questions And Answers

Question 1. A constant force \(\vec{F}=-\hat{i}+2 \hat{j}+3 \hat{k} \mathrm{~N}\) acts on a body, and shifts it 4 m along the z-axis and then 3 m along the y-axis. Work done by \(\vec{F}\) will be

1. 6J
2. 12 J
3. 18 J
4. 24 J

Answer: 3. 18 J

Question 2. A force acts on a particle of mass 3 kg, such that the position of the particle changes with time as per the equation x = 3t – 4t² + t³ if we express x in m and t in s, work done in 4 s will be

1. 570 mJ
2. 450 mJ
3. 490 mJ
4. 576 mJ

Answer: 4. 576 mJ

Question 3. A chain is on a smooth horizontal table with 1/3 of its length hanging off the edge. If the mass and length of the chain are M and l respectively, work done to pull up the hanging part of the chain will be [g = acceleration due to gravity]

1. Mgl
2. \(\frac{M g l}{3}\)
3. \(\frac{M g l}{9}\)
4. \(\frac{M g l}{18}\)

Answer: 4. \(\frac{M g l}{18}\)

Question 4. As an object revolves in a circular path of radius r, a force F is acting on it such that its direction is perpendicular to that of the instantaneous velocity v of the object. Work done by the force in one complete revolution is

1. F · v
2. F · r
3. F · 2πr
4. 0

Answer: 4. 0

Question 5. A particle moving on xy -plane is acted upon by a force \(\vec{F}=-K(y \hat{i}+x \hat{j})\), where AT is a constant. Starting from the origin, the particle is brought to the point (a, 0) along the positive x-axis and then to the point (a, a) parallel to the y-axis. Work done by the force on the particle will be

1. -2Ka²
2. 2 Ka²
3. -Ka²
4. Ka²

Answer: 3. -Ka²

Question 6. A force is acting on a mass of 6 kg. Displacement x of the mass is related to time t as x = \(\frac{t^2}{4}\) m. Work done by the force in 2 s is

1. 12 J
2. 9J
3. 6 J
4. 3 J

Answer: 4. 3 J

Question 7. Work done by a force \(\vec{F}=(\hat{i}+2 \hat{j}+3 \hat{k}) \mathrm{N}\) acting on a particle in displacing it from the point \(\overrightarrow{r_2}=\hat{i}-\hat{j}+2 \hat{k}\) to the point \(\vec{r}_1=\hat{i}+\hat{j}+\hat{k}\) is

1. -3 J
2. -1J
3. Zero
4. 2J

Answer: 2. -1J

Question 8. A particle could be taken from point A to point B following three paths, 1,2, and 3, as shown Work done in these three cases are W₁, W₂, and W₃ respectively. If these works are done in the gravitational field of a point mass m, W₁, W₂, W₃ are related as

1. W₁ > W₃> W₂
2. W₁ = W₂ = W₃
3. W₁ < W₃ <W₂
4. W₁ < W₂ < W₃

Answer: 2. W₁ = W₂ = W₃

Question 9. A mass M is lowered with the help of a string by a distance x at a constant acceleration The magnitude of work done by the string will be

1. Mgx
2. 1/2 Mgx²
3. 1/2 Mgx
4. Mgx²

Answer: 3. 1/2 Mgx

Question 10. Force acting on a particle is (\(2\hat{i}+3 \hat{j}\))N. Work done by this force is zero when a particle is moved on the line 3y+ kx = 5. Here value it is

1. 2
2. 4
3. 6
4. 8

Answer: 1. 2

Question 11. A particle of mass m accelerating uniformly has velocity v at time t₁. What is work done in time t?

1. \(\frac{1}{2} \frac{m v^2}{t_1^2} t^2\)
2. \(\frac{m v^2}{t_1^2} t^2\)
3. \(\frac{1}{2}\left(\frac{m}{t_1}\right)^2 t^2\)
4. \(\frac{2 m v^2}{t_1^2} t^2\)

Answer: 1. \(\frac{1}{2} \frac{m v^2}{t_1^2} t^2\)

Question 12. A mass of 2 kg falls from a height of 40 cm on a spring with a force constant of 1960 N/m. The spring is compressed by (take g = 9.8 m/s²)

1. 10 cm
2. 1.0 cm
3. 20 cm
4. 5 cm

Answer: 1. 10 cm

Question 13. A gardener pushes a lawn roller through a distance 20 m. If he applies a force of 20 kg-wt in a direction inclined at 60° to the ground, the work done by him is

1. 1960 J
2. 196 J
3. 1.96 J
4. 196 KJ

Answer: 1. 1960 J

Question 14. The power required to raise a mass of 120 kg vertically upwards at a velocity of 4.5 m · s^-1 is

1. 5kW
2. 5.3 kW
3. 8 kW
4. 11.2 kW

Answer: 2. 5.3 kW

Question 15. A machine, applying a constant power, is driving an object along a straight line. Displacement of the object in time t is

1. Directly proportional to √t
2. Directly proportional to \(\sqrt[4]{t^3}\)
3. Directly proportional to \(\sqrt{t^3}\)
4. Directly proportional to t²

Answer: 3. Directly proportional to \(\sqrt{t^3}\)

Question 16. A windmill generates electrical energy. Suppose, the windmill converts a fixed part of the wind energy linked with the blades, to electrical energy. If the wind velocity is v, the output electric power is directly proportional to

1. v
2. v²
3. v³
4. v⁴

Answer: 3. v³

Question 17. A particle moves with a velocity \(5 \hat{i}-3 \hat{j}+6 \hat{k}\) m/s under the influence of a constant force, \(\vec{F}=10 \hat{i}+20 \hat{k} \mathrm{~N}\). The instantaneous power applied to the particle is

1. 200 J/s
2. 40J/S
3. 140 J/s
4. 170 J/s

Answer: 4. 170 J/s

Question 18. A quarter-horsepower motor runs at a speed of 600 rpm. Assuming 40% efficiency, the work done by the motor in one rotation will be

1. 7.46 J
2. 7400 J
3. 7.46 erg
4. 74.6 J

Answer: 1. 7.46 J

Question 19. The energy of a body thrown upward is

1. Maximum at the highest point
2. Minimum at the highest point
3. Same at all points
4. Can be maximum at any point

Answer: 3. Same at all points

Question 20. A particle is moving in a straight line such that its retardation is directly proportional to its displacement. A decrease in the kinetic energy of the body is directly proportional to

1. x²
2. e^x
3. x
4. log_ex

Answer: 1. x²

Question 21. Starting from rest a car moves on a plane. The coefficient of friction (μ) between the tyres and the plane increases linearly with the distance (x). The kinetic energy (K) of the car depends on x as

1. \(K \propto \frac{1}{x^2}\)
2. \(K \propto \frac{1}{x}\)
3. \(K \propto x\)
4. \(K \propto x^2\)

Answer: 4. \(K \propto x^2\)

Question 22. A particle falls from rest under the action of gravity. Variation of kinetic energy (KE) and potential energy (PE) of the particle, with time t is represented by the graph

Answer: 2

Question 23. A long spring is stretched by 2 cm. Its potential energy is U. If the spring is stretched by 10 cm, its potential energy would be

1. \(\frac{U}{25}\)
2. \(\frac{U}{25}\)
3. 5 U
4. 25 U

Answer: 4. 25 U

Question 24. In which of the following cases the potential energy is defined

1. Both conservative and non-conservative forces
2. Conservative force only
3. Non-conservative force only
4. Neither conservative nor non-conservative forces

Answer:

Question 25. A body of mass m thrown vertically upwards attains a maximum height h. At what height will its kinetic energy be 75% of its initial value?

1. \(\frac{h}{6}\)
2. \(\frac{h}{5}\)
3. \(\frac{h}{4}\)
4. \(\frac{h}{3}\)

Answer: 3. \(\frac{h}{4}\)

Question 26. For a moving particle (mass m, velocity v) having a momentum p, which one of the followings correctly describes the kinetic energy of the particle?

1. \(\frac{p^2}{2 m}\)
2. \(\frac{p}{2 m}\)
3. \(\frac{v^2}{2 m}\)
4. \(\frac{v}{2 m}\)

Answer: 1. \(\frac{p^2}{2 m}\)

Question 27. The potential energy of a system increase if work is done

1. By the system against a conservative force
2. By the system against a non-conservative force
3. Upon the system by a conservative force
4. Upon the system by a non-conservative force

Answer: 1. By the system against a conservative force

Question 28. Two bodies of masses 4 kg and 5 kg are moving with equal momentum. Then, the ratio of their respective kinetic energies is

1. 4:5
2. 2:1
3. 1:3
4. 5:4

Answer: 4. 5:4

Question 29. A particle of mass m at rest is acted upon by a force P for a time t. Its kinetic energy after an interval t is

1. \(\frac{p^2 t^2}{m}\)
2. \(\frac{p^2 t^2}{2 m}\)
3. \(\frac{p^2 t^2}{3 m}\)
4. \(\frac{P t}{2 m}\)

Answer: 2. \(\frac{p^2 t^2}{2 m}\)

Question 30. If the linear momentum of a body is increased by 50%, then the kinetic energy of that body increases by

1. 100%
2. 125%
3. 225%
4. 25%

Answer: 2. 125%

Question 31. One end of a thread of length h has a stone tied to it. Taking the other end as the center, it revolves in a vertical plane. When the stone reaches the lowest point of its path, it attains the speed u. When the thread is horizontal, the speed of the stone is

1. \(\sqrt{u^2-2 g h}\)
2. \(\sqrt{u^2-g h}\)
3. \(\sqrt{2 g h}\)
4. \(2 \sqrt{u^2-g h}\)

Answer: 1. \(\sqrt{u^2-2 g h}\)

Question 32. Along the surface of a hemispherical container, a small ball is pushed down from a height of h, such that the ball rises up to the opposite edge. If the height of the container is R, the ball must be pushed with a velocity

1. √2gh
2. √2gR
3. √2g(R+h)
4. √2g(R-h)

Answer: 4. √2g(R-h)

Question 33. A bucket full of water is rotated in a vertical circle of radius r. If the water does not split out, the minimum speed of the bucket at top most point will be

1. √rg
2. √5rg
3. √2rg
4. √r/g

Answer: 1. √rg

Question 34. A ball with a momentum p falls on a floor vertically and bounces a number of times. If the coefficient of restitution is e, momentum transferred to the floor is

1. p(1 + e)
2. \(\frac{p}{1-e}\)
3. \(p\left(1+\frac{1}{e}\right)\)
4. \(\left(p\left(\frac{1+e}{1-e}\right)\right.\)

Answer: 4. \(\left(p\left(\frac{1+e}{1-e}\right)\right.\)

In this type of question, more than one option are correct.

Question 35. The potential energy of a particle is given by the formula U = 100 – 5x+ 100x², U and x are in SI units. If the mass of the particle is 0.1 kg then the magnitude of its acceleration

1. At 0.05 m from the origin is 50 m · s^-2
2. At 0.5 m from the mean position is 100 m· s^-2
3. At 0.05 m from the origin is 150 m · s^-2
4. At 0.05 m from the mean position is 200 m · s^-2

Answer:

Question 36. With reference to the graphs, which of the following are correct?

1. The particle has stable equilibrium at points 3 and b
2. The article is in neutral equilibrium at points b and 2
3. No power is delivered by the force on the particle at points 1, 3, and b
4. The particle has the least kinetic energy at position 1

Answer:

1. The particle has stable equilibrium at points 3 and b

3. No power is delivered by the force on the particle at points 1, 3, and b

4. The particle has the least kinetic energy at position 1

Question 37. When a bullet is fired from a gun

1. The kinetic energy of a bullet is more than that of a gun
2. The acceleration of a bullet is more than that of a gun
3. The momentum of the bullet is more than that of the gun
4. The velocity of a bullet is more than that of gun

Answer:

1. The kinetic energy of a bullet is more than that of gun

2. The acceleration of the bullet is more than that of gun

4. The velocity of the bullet is more than that of the gun

Question 38. Suppose two particles 1 and 2 are projected in the vertical plane simultaneously. Their angles of projection are 30° and θ, respectively with the horizontal. Suppose they collide after a time t in the air. Then

1. \(\theta=\sin ^{-1}\left(\frac{4}{5}\right)\) and they will have same speed just before the collision
2. \(\theta=\sin ^{-1}\left(\frac{4}{5}\right)\) and they will have different speeds just before the collision
3. x < (1280√3 – 960)m
4. It is possible that the particles collide when both of them are at their highest points

Answer:

2. \(\theta=\sin ^{-1}\left(\frac{4}{5}\right)\) and they will have different speeds just before the collision

3. \(\theta=\sin ^{-1}\left(\frac{4}{5}\right)\) and they will have different speeds just before the collision

4. It is possible that the particles collide when both of them are at their highest points

Rotation Of Rigid Bodies Angular Momentum

The rotational analogues of the mass (m) of a body and its linear velocity (v) are moment of inertia (I) and angular velocity (ω), respectively. Hence, the rotational analogue of the linear momentum (mv) of the body is Iω. This physical quantity is called the angular momentum (L) of the body.

Angular Momentum Definition: The dynamical property generated in a body under rotational motion, due to the moment of inertia about an axis and angular velocity, is called the angular momentum of the body about that axis.

Angular momentum is measured by the product of moment of inertia and angular velocity, i.e., L = Iω.

Since I is a scalar and ω is an axial vector, angular momentum L is also an axial vector whose direction is along the axis of rotation, and in the direction of ω.

Unit And Dimension Of Angular Momentum:

CGS System: g · cm² · s^-1

SI: kg · m² · s^-1

Dimension of L = dimension of I x dimension of ω = ML² x T^-1 = ML²T^-1

Relation Between Linear Momentum And Angular Momentum: Suppose a body is revolving with an angular velocity ω about an axis. If m₁, m₂, m₃,…. are the constituent particles of that body and they are at distances r₁, r₂, r₃,…. respectively from the axis of rotation, then the moment of inertia of the body,

I = \(m_1 r_1^2+m_2 r_2^2+m_3 r_3^2+\cdots=\sum_i m_i r_i^2\)

In the case of pure rotation, the angular velocity of each particle becomes equal to the angular velocity of the body.

So, the angular momentum of the body,

L = \(I \omega=\sum_i m_i r_i^2 \cdot \omega=\sum_i m_i r_i v_i\) (because \(v_i=\omega r\))

= \(\sum_i r_i \times m_i v_i=\sum_i r_i \times p_i\)

[p_i = m_iv_i = linear momentum of i-th particle]

For the particles, the quantities r₁ x m₁v₁, r₂ x m₂v₂,…… etc., can be called the moments of linear momentum, or in brief, moments of momentum (in analogy with the moment of force).

So, the angular momentum of a body about an axis is the algebraic sum of the moments of linear momentum about the same axis, of all particles constituting the body.

Thus, for a particle rotating about a circle of radius r and having a linear momentum p, the angular momentum will be L = rp.

Vector Representation: The vector representation for the relation between linear and angular momentum is \(\vec{L} = \vec{r} \times \vec{p}\). This is often referred to as the defining equation of \(\vec{L}\).

We know the vector representation for the relation between linear velocity and angular velocity is \(\vec{v}=\vec{\omega} \times \vec{r}\).

If \(\vec{v}\) and \(\vec{\omega}\) are replaced by \(\vec{p}\) and \(\vec{L}\), respectively, the geometric form for the relation of \(\vec{L}\), \(\vec{p}\) and \(\vec{r}\) is obtained.

Relation Between Angular Momentum And Torque: In case of rotational motion, when a torque is applied to a body, an angular acceleration is produced in it. If the initial angular velocity of the body is ω₁ and its angular velocity after time t is ω₂, then the angular acceleration of the body,

α = \(\frac{\omega_2-\omega_1}{t}\)

Again, torque = moment of inertia x angular acceleration

or, \(\tau=I \alpha=I \times \frac{\omega_2-\omega_1}{t}=\frac{I \omega_2-I \omega_1}{t}\)

or, \(\tau t=I \omega_2-I \omega_1\)

Hence, torque x time = change in angular momentum of the body during that interval

This is the relation between torque and angular acceleration. From this relation, it is evident that a change in angular momentum takes place about the axis along which the torque acts on the body.

We know that in the case of translational motion, Ft = mv – mu and the rotational analogue of this equation is τt = Iω₂  – Iω₁. The quantity Ft is known as the impulse of force. Similarly, the quantity τt is known as the angular impulse or the impulse of torque.

Law Of Conservation Of Angular Momentum: Suppose the moment of inertia of a body changes from I₁ to I₂ in time t. In this case, the equation τt = Iω₂  – Iω₁ changes to τt = I₂ω₂  – I₁ω₁ Now, if no external torque acts on the body, i.e., if τ = 0, then from the equation, τt = I₂ω₂  – I₁ω₁ we get, I₂ω₂  – I₁ω₁ = 0, or, τt = I₂ω₂  = I₁ω₁

It means that the final angular momentum of the body is equal to its initial angular momentum, i.e., the angular momentum is conserved.

Law: if the net external torque on a body is zero, the angular momentum of the body rotating about an axis always remains conserved.

So, this law is nothing but the rotational analogue of the law of conservation of linear momentum.

Again we know, \(\frac{dL}{dT}\) = τ_ext

From this, it is clear that, if total external torque acts on a body is zero; its angular velocity decreases with the increase of its moment of inertia and vice versa i.e., angular momentum remains constant.

Related Experiments And Practical Examples:

1. A man is sitting on a turntable holding a pair of dumbbells of equal mass, one in each hand with his arms out-stretched while the turntable rotates with a definite angular velocity, If the man suddenly draws the dumbbells towards his chest, the speed of rotation of the turntable is found to increase.

• This is due to the fact that when the man draws the dumbbells towards his chest, the moment of inertia of the man about the axis of rotation decreases and his angular velocity increases due to conservation of angular momentum.
• If the man again stretches his arms, his angular velocity decreases due to an increase in moment of inertia, and the turntable consequently rotates slowly.

2. In a diving event, when a competitor dives from a high platform or springboard into water, he keeps his legs and arms outstretched and starts descending with less angular velocity, After that he curls his body by rolling the legs and arms inwards, his moment of inertia decreases.

• As angular momentum is conserved, his angular velocity goes on increasing rapidly. As a result, his body begins to spin rapidly and before reaching the surface of the water, he can perform a good number of somersaults.
• In the case of skating on the surface of ice or during the performance of acrobatics, the principle of conservation of angular momentum can be applied in a similar way.

Rotation Of Rigid Bodies Angular Momentum Numerical Examples

Example 1. If the radius of the earth decreases by \(\frac{1}{2}\)%, then what will be the change in the length of a day? Assume that the earth is a uniform sphere and its moment of inertia, I = \(\frac{2}{5}\)MR², where M and R are the mass and the radius of the earth.
Solution:

If the mass of a solid sphere remains unaltered, then its moment of inertia ∝ (radius)².

Here, the changed radius \(=\frac{100-\frac{1}{2}}{100} R=\frac{199}{200} R\).

So, if the moment of inertia of the earth for its present radius R is I and the moment of inertia for its changed radius is I’, then

⇒ \(\frac{I}{I^{\prime}}=\frac{R^2}{\left(\frac{199 R}{200}\right)^2}=\left(\frac{200}{199}\right)^2\)

If the present angular velocity of the earth is ω and its changed angular velocity is ω’, then according to the principle of conservation of angular momentum,

⇒ \(I \omega=I^{\prime} \omega^{\prime}\)

or, \(\omega^{\prime}=\frac{I \omega}{I^{\prime}}\)

or, \(\frac{2 \pi}{T^{\prime}}=\frac{I}{I^{\prime}} \times \frac{2 \pi}{T}\)

or, \(T^{\prime}=\frac{I^{\prime}}{I} \cdot T=\left(\frac{199}{200}\right)^2 \times 24=23.76 \mathrm{~h}\)

∴ The length of the day will decrease by (24-23.76) = 0.24 h = 14 min 24 s

Example 2. A solid sphere of mass 1 kg and of radius 10 cm is rotating about one of its diameters with an angular; velocity of π rad · s^-1. Calculate the kinetic energy of the sphere by using the relevant formula.
Solution:

Let the moment of inertia of the sphere about its diameter I = \(\frac{2}{5}\)MR², M = mass of the sphere and R = radius of the sphere.

The kinetic energy of the body = rotational kinetic energy of the body

= \(\frac{1}{2} I \omega^2=\frac{1}{2} \times \frac{2}{5} M R^2 \cdot \omega^2\)

= \(\frac{1}{5} \times 1000 \times(10)^2 \times \pi^2\)

= \(197392.09 \mathrm{erg} .\)

Example 3. A thin rod of length l and mass m per unit length is rotating about an axis passing through the midpoint of its length and perpendicular to it. Prove that its kinetic energy \(\frac{1}{24}\) mω²l³ = ω = angular velocity of the rod.
Solution:

Kinetic energy of the rod = \(\frac{1}{2}\) mω²

According to the problem,

I = \(\frac{1}{12}\)Ml² [M = mass of the rod = ml]

= \(\frac{1}{12}\) x ml x l² = \(\frac{m l^3}{12}\)

∴ Kinetic energy of the rod = \(\frac{1}{2} \times \frac{m l^3}{12} \times \omega^2=\frac{1}{24} m \omega^2 l^3 .\)

Example 4. Calculate the moment of inertia of a solid cylinder of I length 10 cm and of radius 20 cm about its own axis. The density of the material of the cylinder = 9 g · cm^-3.
Solution:

L= length of the cylinder, R = radius of the cylinder and M = mass of the cylinder

= volume of the cylinder x density

=  πR²L X density

= π x (20)² x 10 x 9 g

Moment of inertia of a solid cylinder about its own axis,

I = \(\frac{1}{2} M R^2\)

I = \(\frac{1}{2} \times \pi \times(20)^2 \times 10 \times 9 \times(20)^2\)

= \(22.6 \times 10^6 \mathrm{~g} \cdot \mathrm{cm}^2\)

Example 5. A solid sphere of diameter 2 cm and of mass 20 g is rolling with a velocity of 3 cm · s^-1. What is the total kinetic energy of the sphere?
Solution:

Let M = mass of the sphere, R = radius of the sphere, V = linear velocity of the sphere, I = \(\frac{2}{5}\)MR² (moment of inertia of the sphere about its diameter), ω = \(\frac{V}{R}\)

Total kinetic energy of the sphere = translational kinetic energy + rotational kinetic

= \(\frac{1}{2} M V^2+\frac{1}{2} I \omega^2=\frac{1}{2} M V^2+\frac{1}{2} \times \frac{2}{5} M R^2\left(\frac{V}{R}\right)^2\)

= \(\frac{1}{2} M V^2+\frac{1}{5} M V^2=\frac{7}{10} M V^2=\frac{7}{10} \times 20 \times(3)^2\)

= \(126 \mathrm{erg}\)

Example 6. A stone of mass m tied with a thread Is rotating along a horizontal circular path (force of gravity is neglected). The length of the thread decreases gradually in such a manner that the angular momentum of the stone remains constant with respect to the centre of the circle. If the tension in the thread Is T = Arⁿ, where A = constant, r = instantaneous radius of the circle, then find the value of n.
Solution:

If the instantaneous angular velocity of the stone is w, then angular momentum,

L = Iω = mr²ω = constant (according to the problem)

or, ω = \(\frac{L}{m r^2}\)

Here the tension in the thread provides the necessary centripetal force for rotation.

So, T = \(A r^n=m \omega^2 r=m \cdot \frac{L^2}{m^2 r^4} r=\frac{L^2}{m} r^{-3}\)

= \(A r^{-3} \quad\left(A=\frac{L^2}{m}=\text { constant }\right)\)

∴ n=-3 .

Example 7. Two ends of a uniform rod weighing W, are placed on supports so that the rod remains horizontal. If a support at one end is suddenly removed, what will be the force exerted on the horizontal rod by the support at the other end?
Solution:

Let the length of the rod = l cm, its weight = W = Mg, where M is the mass of the rod. When the support at one end is removed suddenly, the centre of gravity of the rod falls downwards with an acceleration a. Let R = reaction force at the end with the support. Hence, if the C.G. now falls with an acceleration a, the rod will turn about the point P.

The torque on the rod = Mg · \(\frac{l}{2}\)

Also, Mg – R = Ma or, \(a=\frac{M g-R}{M}\)

Here moment of inertia, I =  \(\frac{1}{3}\)Ml² = moment of inertia of the rod about the perpendicular axis passing through the end of the rod and the angular acceleration, α = \(\frac{a}{V / 2}=\frac{2 a}{l}\)

∴ \(\frac{1}{3} M l^2 \alpha=M g \frac{l}{2}\) (because \(\tau=I \alpha\))

or, \(\frac{1}{3} M R^2 \cdot \frac{2 a}{l}=M g \frac{l}{2} \text { or, } \frac{2}{3} a=\frac{g}{2} \text { or, } \frac{2}{3}\left(\frac{M g-R}{M}\right)=\frac{g}{2}\)

R = \(\frac{M g}{4}=\frac{W}{4}\)

Therefore, when one support is removed, the support at the other end will exert a reaction force of \(\frac{W}{4}\)

Example 8. A rod of length L and M is attached with a hinge on a wall at point O. After releasing the rod from its vertical position OA, when it comes to position OA’, what is the reaction on point O of the rod by the hinge?

Solution:

Let, the angular velocity of the rod at the horizontal position OA’ is ω.

∴ At that instant its kinetic energy = \(\frac{1}{2} I \omega^2=\frac{1}{2} \cdot \frac{M L^2}{3} \cdot \omega^2=\frac{M L^2 \omega^2}{6}\)

The centre of mass of the rod shifts down by \(\frac{L}{2}\) from OA to OA’.

So, decrease in potential energy of the rod = Mg\(\frac{L}{2}\)

According to the kinetic energy conservation law, \(M g \frac{L}{2}=\frac{M L^2 \omega^2}{6} \quad \text { or, } \omega=\sqrt{\frac{3 g}{L}}\)…(1)

Two forces act on the rod at position OA’

1. Gravitational force (Mg) vertically downward direction and
2. Reaction force (n) of the hinge

Let, the horizontal and the vertical n components of n are n_x and n_y respectively; the horizontal and the vertical components of the acceleration of the centre of mass of the rod area a_x and a_y respectively.

∴ According to \(M g-n_y=M a_y\)….(2)

and \(n_x=M a_x=M \omega^2 \cdot \frac{L}{2}\)

(because \(a_x=\) centripetal acceleration)

= \(M \cdot \frac{3 g}{L} \cdot \frac{L}{2}=\frac{3}{2} M g\)

[putting the value of ω from equation (1)]

The rod starts to rotate due to the action of torque created by n_y and Mg.

If the angular acceleration of the rod is α, \(M g \cdot \frac{L}{2}=I \alpha=\frac{M L^2}{3} \alpha\)

∴ \(\alpha=\frac{3 g}{2 L}\)

The acceleration along the vertical direction, \(a_y=\frac{L}{2} \alpha=\frac{3 g}{4}\)

Putting the value of ay in equation (2) we get, \(M g-n_y=\frac{3 M g}{4} \text { or, } n_y=\frac{M g}{4}\)

∴ n = \(\sqrt{n_x^2+n_y^2}=\sqrt{\left(\frac{3}{2} M g\right)^2+\left(\frac{M g}{4}\right)^2}=\frac{\sqrt{37}}{4} M g\)

 Rotation Of Rigid Bodies Long Answer Type Questions

Question 1. Is there any change in the angular velocity of the earth when a body strikes the earth’s surface from outside?
Answer:

We know that, the angular momentum of a body = moment of inertia of that body x angular velocity of the body.

• The velocity of anybody coming from outside and striking the surface of the earth is usually directed towards the centre of the earth. For this reason, the body possesses no angular momentum with respect to the axis of rotation of the earth.
• As a result, the angular momentum of the earth remains conserved. However due to the inclusion of the body, the mass of the earth increases, and consequently, the moment of inertia of the earth also increases.
• For this reason, the angular velocity of the earth decreases slightly. But actually, the mass of such a foreign body is so small that there is no appreciable change in either the moment of inertia or the angular velocity of the earth.

Question 2. Why is It easier to rotate an object tied to the end of a short string than that of a long string?
Answer:

We know that the moment of inertia of an object about the axis of rotation is, I = mr², where m is the mass of the object and r is the perpendicular distance of the object from the axis of rotation.

So, with the increase in the length of the string, the moment of inertia of the object also increases. The object is rotated along a circular path by applying a torque against the resistive force of the air.

Now, torque = moment of inertia x angular acceleration.

Hence, with the increase in moment of inertia, the magnitude of required torque also increases.

Moreover, the centrifugal reaction generated at the centre of rotation is mω²r, where ω is the angular velocity of the stone. So, keeping the value of ω constant, if the value of r is increased, centrifugal reaction also increases.

Hence, if the string is short, then less torque is required to rotate the stone tied to the string. Also, it has to withstand a smaller centrifugal reaction and as a result, it is easier to rotate the object.

Question 3. A solid and a hollow sphere of the same mass have the same outer radius. Which one has a larger radius of gyration?
Answer:

The mass of the solid sphere is distributed uniformly from its centre. On the other hand, the mass of the hollow sphere is distributed over a comparatively further distance.

The more the distance of the mass distribution from the axis of rotation of two bodies of identical mass having the same outer radius, the more the radius of gyration.

In this sense, the mass of the hollow sphere is distributed over a comparatively larger distance from the axis of rotation than that of the solid sphere, and hence the radius of gyration of the hollow sphere will be greater.

Question 4. If the ice at the poles of the earth melts, how would this affect the length of the day?
Answer:

If the polar ice melts, a part of the water thus produced will shift from the poles towards the equatorial region, and hence, this water will shift away from the axis of rotation of the earth. Consequently, the moment of inertia of the earth will increase.

Now, according to the principle of conservation of angular momentum, with the increase in moment of inertia of the earth, its angular velocity will decrease, and hence, the length of the day will increase.

Question 5. A cricket ball sometimes rebounds from the cricket pitch with a velocity greater than which it was bowled with by a bowler. How can it be possible?
Answer:

If the cricket ball spins just before it hits the ground, then this spinning kinetic energy is added to the kinetic energy of the translation of the ball. As a result, the ball rebounds from the cricket pitch with a greater velocity by virtue of this spuming or rotational kinetic energy.

Question 6. Can the moment of Inertia of a body be different about different axes?
Answer:

The moment of inertia of a body depends on the mass of the body, the position of the axis of its rotation and the distribution of mass of the body about its axis of rotation. So, the moment of inertia of a particular body may be different in different axes.

Question 7. A man is standing on a rotating table and he drops a heavy mass from his hand outside the table. How will the angular speed of the table change?
Answer:

When the mem drops the heavy mass from his hand outside the table, the moment of inertia of the system about the axis of rotation decreases. We know that angular momentum = moment of inertia x angular velocity. Since no torque is applied from outside, according to the principle of conservation of angular momentum, the angular velocity of the system will increase due to a decrease of its moment of inertia.

Question 8. When a body of mass m slides down from the top of an inclined plane and reaches the bottom, its velocity becomes v. When a circular disc of the same mass is rolled down the inclined plane, it acquires a velocity v₁. Show that, \(v_1=\sqrt{\frac{2}{3}} v\).
Answer:

Let the vertical height of the inclined plane be h. In case of the first body, \(\frac{1}{2} m v^2=m g h \quad \text { or, } \quad v=\sqrt{2 g h}\)…(1)

If the body is a circular disc, then it possesses both translational kinetic energy and rotational kinetic energy while rolling down the inclined plane.

If I am a moment of inertia of the circular disc and ω be angular velocity of the disc at the bottom of the inclined plane,

⇒ \(\frac{1}{2} m v_1^2+\frac{1}{2} I \omega^2=m g h\)

or, \(\frac{1}{2} m v_1^2+\frac{1}{2} \cdot \frac{1}{2} m r^2 \cdot \frac{v_1^2}{r^2}=m g h \)

r = (radius of the disc, \(I=\frac{1}{2} m r^2, \omega=\frac{v_1}{r}\))

or, \(\left(\frac{1}{2}+\frac{1}{4}\right) v_1^2=g h \text { or, } \frac{3}{4} v_1^2=g h \text { or, } v_1=\sqrt{\frac{4}{3} g h}\)

∴ \(\frac{v_1}{v}=\sqrt{\frac{4}{2} g h}=\sqrt{\frac{2}{3}} \text { or, } v_1=\sqrt{\frac{2}{3}} v .\)

Question 9. Prove that the length of a day becomes T’ = 6h instead of T = 24 h if the earth suddenly contracts to half its present radius (consider the earth as a spherical body), without having any change in its mass.
Answer:

Since the earth is a solid sphere, its moment of inertia, I ∝ R²(R = radius of the earth)

(moment of inertia of solid sphere = \(\frac{2}{5}\)MR²)

So, if the present radius is R and the changed radius is \(\frac{R}{2}\) then,

⇒ \(\frac{I}{I^{\prime}}=\frac{R^2}{\left(\frac{R}{2}\right)^2}=4\)

Again, if the present angular velocity is ω and the changed angular velocity is ω’, then according to the principle of conservation of angular momentum,

⇒ \(I \omega=I^{\prime} \omega^{\prime} \text { or, } \omega^{\prime}=\frac{I}{I^{\prime}} \omega=4 \omega\)

∴ \( \frac{2 \pi}{T^{\prime}}=4 \cdot \frac{2 \pi}{T} \quad \text { or, } T^{\prime}=\frac{T}{4}=\frac{24}{4}=6 \mathrm{~h} \text {. } \)

Question 10. Show that the torque acting on a body is equal to the rate of change of angular momentum of the body.
Answer:

We know that angular momentum, \(\vec{L}=\vec{r} \times \vec{p} \quad \text { or, } \vec{L}=\vec{r} \times m \vec{v}\)

⇒\(\frac{d \vec{L}}{d t}=\frac{d \vec{r}}{d t} \times m \vec{v}+\vec{r} \times \frac{d}{d t}(m \vec{v})\)

= \(\vec{v} \times m \vec{v}+\vec{r} \times \vec{F}=\vec{r} \times \vec{F}\) (because \(\vec{v} \times \vec{v}=0\))

= \(\vec{r}\)

So, the rate of change of angular momentum of a body is equal to the torque acting on the body

Question 11. What is the relation between torque and angular acceleration?
Answer:

Torque, \(\vec{\tau}= \frac{d \vec{L}}{d t}=\frac{d}{d t}(I \vec{\omega})=I \frac{d \vec{\omega}}{d t}\)

(\(\vec{L}\)= angular momentum = \(I \vec{\omega}\))

= \(I \vec{\alpha}[\vec{\alpha}\) = angular acceleration

Question 12. Keeping the radius of the earth unchanged, If the mass of the earth is doubled, then what will be the length of a day?
Answer:

If the present angular velocity is ω and the changed angular velocity is ω’, then according to the principle of conservation of angular momentum, Iω = I’ω’

[here, I and I’ are the present and the changed moments of inertia of the earth respectively]

∴ \(\frac{2}{5} M R^2 \times \frac{2 \pi}{24}=\frac{2}{5} \times 2 M \times R^2 \times \frac{2 \pi}{T^{\prime}}\)

[here, the present mass of the earth is M, its radius is R and the changed length of a day is T’]

or, \(\frac{1}{24}=\frac{2}{T^{\prime}} \quad \text { or, } \quad T^{\prime}=48 \text { hours }\)

So, the length of a day will be 48 h.

Question 13. A circular disc of mass m and radius r is rolling over a horizontal table top with angular velocity ω. Prove that the total energy of the disc, K = \(\frac{3}{4}\)mω²r²
Answer:

The total kinetic energy of the disc,

K = translational kinetic energy+rotational kinetic energy

= \(\frac{1}{2} m v^2+\frac{1}{2} I \omega^2\)

Here, v = linear velocity of the disc = ωr

I = moment of inertia of the disc about the perpendicular axis passing through its centre

= \(\frac{1}{2} m r^2\)

K = \(\frac{1}{2} m(\omega r)^2+\frac{1}{2} \cdot \frac{1}{2} m r^2 \cdot \omega^2\)

= \(\frac{1}{2} m \omega^2 r^2+\frac{1}{4} m \omega^2 r^2=\frac{3}{4} m \omega^2 r^2\)

Question 14. Between two circular discs of equal mass and equal thickness but of different densities, which one would have a greater moment of Inertia about its central perpendicular axis?
Answer:

Suppose the mass of each disc is M, thickness d, densities of their materials ρ₁ and ρ₂ (ρ₁ > ρ₂), radii r₁ and r₂ respectively.

∴ \(M=\pi r_1^2 d \rho_1=\pi r_2^2 d \rho_2 \quad \text { or, } r_1^2 \rho_1=r_2^2 \rho_2\)

or, \(\frac{r_1^2}{r_2^2}=\frac{\rho_2}{\rho_1}\)

∴ \(\rho_1>\rho_2, \quad \frac{\rho_2}{\rho_1}<1\)

∴ \(\frac{r_1^2}{r_2^2}<1\)

The moment of inertia of the two discs about their central perpendicular axes are \(I_1=\frac{1}{2} M r_1^2 \text { and } I_2=\frac{1}{2} M r_2^2\)

∴ \(\frac{I_1}{I_2}=\frac{r_1^2}{r_2^2}<1\)

∴ \(I_1<I_2\)

So, the disc having a lower density will have a greater moment of inertia about its centred perpendicular axis.

Question 15. Given the moment of inertia of a disc of mass M and radius R about any of its diameters to be \(\frac{M R^2}{4}\). Find its moment of inertia about an axis normal to the disc and passing through a point on its edge.
Answer:

Given,

According to the parallel-axes theorem, the moment of inertia of the disc about the axis AB, normal to the disc and passing through a point on its edge,

Question 16. ‘Moment of inertia plays the same role in rotational motion as mass plays in translational motion’explain the statement.
Answer:

When a force acts on a body, some linear acceleration is produced in that body. Similarly, angular acceleration is generated in a body due to the application of a torque on it. So the rotational analogues of force and linear acceleration are torque and angular acceleration, respectively.

Again, in the case of linear motion, force = mass x acceleration; and in the case of circular motion, torque = moment of inertia x angular acceleration. Comparing these two relations it can be inferred that the rotational analogue of mass is the moment of inertia.

So, moment of inertia in rotational motion plays the same role as mass plays in linear motion.

Question 17. Find the moment of inertia of a sphere about a tangent to the sphere. Given the moment of inertia of the sphere about any of its diameters to be \(\frac{2 M R^2}{5}\), where M is the mass of the sphere and R is the radius of the sphere.
Answer:

The centre of mass (cm) of C the sphere is on its diameter AB.

So, I_cm = \(\frac{2}{5}\)MR²

According to the parallel-axes theorem, the moment of inertia of the sphere about the tangent CD.

I = \(I_{\mathrm{cm}}+M R^2=\frac{2}{5} M R^2+M R^2=\frac{7}{5} M R^2\)

Question 18. A solid sphere of mass m and radius R rolls down from the top of a table. With how much angular speed will it touch the ground?
Answer:

At position 2, the resultant of mgcosθ and n supplies the necessary centripetal force to roll on, i.e., mω²R = mgcosθ – n

[ω = angular speed of the spehere]

When n = 0, the sphere will not be in contact more with the table.

Then, \(m \omega^2 R=m g \cos \theta \quad \text { or, } \cos \theta=\frac{\omega^2 R}{g}\)

The centre of mass of the sphere is lowered down by R(1-cosθ).

Let I be the moment of inertia of the sphere about the tangent at the contact point between the table and the sphere. From the conservation law of mechanical energy,

⇒ \(m g \cdot R(1-\cos \theta)=\frac{1}{2} \times I \omega^2=\frac{7}{10} m R^2 \omega^2\) (because \(I=\frac{7}{5} m R^2\))

∴ \(m g R\left(1-\frac{\omega^2 R}{g}\right)=\frac{7}{10} m R^2 \omega^2\)

or, \(m g R-m \omega^2 R^2=\frac{7}{10} m R^2 \omega^2\)

∴ \( \omega=\sqrt{\frac{10 g}{17 R}}\)

Considering a free fall of the sphere, as no torque acts on it, this angular speed remains unaltered. It means that the sphere touches the ground with an angular speed \(\sqrt{\frac{10 g}{17 R}}\)

Question 19. A uniform rod AB of mass M and length L is hung from a ceiling in such a way that the rod can rotate freely in the vertical plane around point A. An object of mass m coming horizontally with velocity v₀ hits the rod at point B and sticks to the rod. What will be the maximum angle with the vertical that the rod makes in this type of collision? (Here, M = 6m)

Answer:

In the case of the composite system of the rod and the object of mass m, applying the law of conservation of angular momentum about A,

⇒ \(m v_0 L=I \omega=\left[\frac{M L^2}{3}+m L^2\right] \omega=\left(\frac{M}{3}+m\right) L^2 \omega\)

∴ \(\omega=\frac{m v_0}{\left(\frac{M}{3}+m\right) L}=\frac{m v_0}{(2 m+m) L}=\frac{v_0}{3 L}\)

The distance of the centre of mass of the composite system from point A when the object sticks to the rod

= \(\frac{6 m \times \frac{1}{2}+m \times L}{6 m+m}=\frac{4}{7} L\)

The rod makes the maximum angle θ with the vertical, and at that position, the centre of mass of the composite system raises at a height h from its initial position.

h = \(\frac{4}{7} L(1-\cos \theta)\)

According to the law of conservation of mechanical energy, \(\frac{1}{2} I \omega^2=(m+6 m) g h\)

or, \(\frac{1}{2}\left(\frac{1}{3} M L^2+m L^2\right) \times \frac{\nu_0^2}{9 L^2}=4 m g L(1-\cos \theta)\)

or, \(1-\cos \theta=\frac{v_0^2}{24 g L} therefore \theta=\cos ^{-1}\left[1-\frac{\nu_0^2}{24 g L}\right]\)

Question 20. A spherical object of mass m is released on a smooth inclined plane which is inclined at an angle θ with the horizontal. State whether it will roll or slip. Give reasons in support of your answer.
Answer:

No frictional force acts on a smooth plane. The only downward force acting on the centre of mass of the object along the inclined plane is mgsinθ. But there is no torque about the centre of mass due to the absence of the frictional force. So, the object will slip down without rolling with acceleration gsinθ.

Unit 5 Motion Of System Of Particles And Rigid Body Chapter 2 Rotation Of Rigid Bodies Synopsis

Two equal, parallel but oppositely directed forces acting simultaneously along two different lines of action constitute a couple.

• The vector quantity formed by the combination of the couple applied on a body and the arm of the couple, which produces a rotational tendency in that body is called the moment of the couple or torque.
• The moment of the couple or torque is expressed by the product of the magnitude of any of the forces of the couple and the arm of the couple.
• The torque about a point = the algebraic sum of the moments of the two forces of the couple with respect to that point. The torque acting on a body with respect to any axis of rotation divided by the corresponding angular acceleration generated is called the moment of inertia of that body with respect to that axis of rotation.

If the whole mass of a body is assumed to be concentrated at a point such that the moment of inertia of the whole body equals the moment of inertia of that point, then the radial distance of the point from the axis of rotation is called the radius of gyration.

Parallel-axes Theorem: The moment of inertia (I) of a rigid body about any axis is equal to the sum of its moment of inertia about a parallel axis through its centre of mass (I_cm) and the product of the mass of the body (M) with the square of the perpendicular distance between the two axes (r²).

Mathematical expression: I = I_cm + Mr²

Perpendicular-axes Theorem: The moment of inertia of a plane lamina about an axis perpendicular to its plane (I_z) is equal to the sum of the moments of inertia of the lamina about two mutually perpendicular axes (I_x +I_y) lying on the plane of the lamina and intersecting each other at the point through where the perpendicular axis passes. Mathematical expression: I_x+I_y= I_z

The dynamical property generated in a rotating body by the moment of inertia of the body about an axis and its angular velocity together is called the angular momentum of the body about that axis.

Principle Of Conservation Of Angular Momentum: If the net external torque on a body is zero, the angular momentum of the body rotating about an axis always remains conserved.