WBCHSE Class 11 Physics Newton Law Of Motion Short Answer Questions

Unit 3 Laws Of Motion Chapter 1 Newton Law Of Motion Short Answer Type Questions

WBBSE Class 11 Short Answer Questions on Newton’s Laws

Question 1. A block of mass M is pulled along a frictionless horizontal surface by a rope of mass m by applying a horizontal force F at the other end of the rope. The force exerted by the rope on the block is

1. \(\frac{m F}{m+M}\)
2. \(\frac{m F}{M-m}\)
3. \(\frac{M F}{M-m}\)
4. \(\frac{M F}{m+M}\)

Answer:

Given

A block of mass M is pulled along a frictionless horizontal surface by a rope of mass m by applying a horizontal force F at the other end of the rope. The force exerted by the rope on the block is

Acceleration of the system, a \(=\frac{\text { total force }}{\text { total mass }}=\frac{F}{m+M}\)

The force exerted by the rope on the block, \(M a=\frac{M F}{m+M}\)

The option 4 is correct.

Question 2. Two blocks A and B are standing side-by-side touching each other, on a smooth horizontal table. The masses of the blocks are 3 kg and 2 kg respectively. Block A is pushed towards block B with 10 N horizontal force.

1. How much force does block A apply on block B?
2. If the 10 N horizontal force were applied on block B towards block A, how much force would block B have applied on block A

Answer:

Given

Two blocks A and B are standing side-by-side touching each other, on a smooth horizontal table. The masses of the blocks are 3 kg and 2 kg respectively.

1. Let force applied on B by A is F_AN.

Here, acceleration of the system due to force 10 N = acceleration of B due to force F_A

So, \(\frac{10}{3+2}=\frac{F_A}{2} \quad therefore F_A=4 \mathrm{~N}\)

2. Force applied on A by B, \(F_B=\frac{10 \times 3}{3+2}=6 \mathrm{~N}\)

Question 3. A gun is mounted on a platform fitted with frictionless wheels. The mass of the platform with the gun, shells, and the operator is M. The gun fires shells one after another with a velocity of v in the horizontal direction. If another with a velocity v in the horizontal direction. If the mass of each shell is m, show that the velocity of recoil of the platform after N shells are fired is \(V_N=\frac{N m v}{(M-m N)}\)
Answer:

Given

A gun is mounted on a platform fitted with frictionless wheels. The mass of the platform with the gun, shells, and the operator is M. The gun fires shells one after another with a velocity of v in the horizontal direction. If another with a velocity v in the horizontal direction. If the mass of each shell is m

Let the velocity of the recoil of the platform after 1st shell is fired = v₁.

Therefore, from the law of conservation of momentum, \((M-m) v_1=m v\) or, \(v_1=\frac{m v}{M-m}\)

Again from the law of conservation of momentum, after 2nd shell is fired, we get mv-(M-2m)v₂ = -(M- m)v₁

[v₂ = The velocity of recoil of the platform after 2nd shell is fired]

or, \(v_2=\frac{2m v}{M-2m}\)

Similarly the velocity of recoil of the platform after 3rd shell is fired, is \(v_3=\frac{3m v}{M-3m}\)

Therefore, the velocity of recoil of the platform after N shells are fired is \(v_N=\frac{N m v}{M-Nm}\)

Key Concepts in Newton’s Laws: Short Answers

Question 4. Which is easier to lift in the air, 1 kg of steel or 1 kg of wool?
Answer:

The volume of 1 kg of wool is much greater than that of 1 kg of steel. Hence, wool faces more resistance while moving through air. Therefore, it is easier to lift 1 kg of steel.

Question 5. A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 36 km/h. What is the impulse imparted to the ball? Given, the mass of the ball is 0.157 kg.
Answer:

Given

A batsman deflects a ball by an angle of 45° without changing its initial speed which is equal to 36 km/h.

Let initial velocity of the ball is \(\vec{v}_1\) and final velocity is \(\vec{v}_2\).

Impulse, \(\vec{I}\)= m(\(\vec{v}_2\) – \(\vec{v}_1\)) [m = mass of the ball]

The magnitude of initial and final velocities are the same,

i.e., \(\left|\vec{v}_1\right|=\left|\vec{v}_2\right|=36 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

= \(\frac{36 \times 10^3}{3600} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

= \(10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

From figure, \(\left|\vec{v}_2-\vec{v}_1\right|=\sqrt{v_1^2+v_2^2+2 v_1 v_2 \cos 45^{\circ}}\)

= \(\sqrt{10^2+10^2+2 \times 10 \times 10 \times \frac{1}{\sqrt{2}}}\)

∴ \(18.5 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ \(|\vec{I}|=m\left|\vec{v}_2-\vec{v}_1\right|=0.157 \times 18.5 \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}=2.9 \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}\)

The direction of imparted impulse is along the bisector of the angle between \(\vec{v}_2\) and \(\vec{-v}_1\)

Applications of Newton’s First Law: Short Answer Questions

Question 6. In which frame of reference Newton’s first law of motion is applicable?
Answer:

Newton’s first law of motion is applicable in an inertial frame of reference.

Question 7. A mass of 1 kg is suspended by a thread. It is

1. Lifted up with an acceleration of 4.9m · s^-2,
2. Lowered with an acceleration of 4.9m · s^-2. The ratio of the tensions or the thread is

1. 3:1
2. 1:2
3. 1:3
4. 2:1

Answer:

1. Weight of the body = mg (downwards)

Tension on the thread = T (upwards)

∴ Resultant force acting on the body = T- mg (upwards)

If a is the upward acceleration then from F = ma we get, T- mg = ma or, T = m(g+ a)

2. In case of downward acceleration a, resultant force acting on the body = mg- T

∴ mg- T = ma or, T = m(g- a)

∴ The ratio of tensions on the thread in the two cases

= \(\frac{m(g+a)}{m(g-a)}=\frac{g+a}{g-a}=\frac{9.8+4.9}{9.8-4.9}=\frac{14.7}{4.9}=\frac{3}{1}\)

The option 1 is correct.

Question 8. A bullet is fired from a gun. Which one will possess greater momentum—gun or bullet?
Answer:

A bullet is fired from a gun

Before firing the bullet, both the gun and the bullet were at rest. Hence, initially, the total linear momentum was zero. After firing the bullet, the gun will have an equal but opposite momentum to that of the bullet so as to conserve the total linear momentum of the system.

Understanding Newton’s Second Law: Short Answers

Question 9. A smooth massless string passes over a smooth fixed pulley. Two masses m₁ and m₂ (m₁> m₂) are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest. The total external force acting on the two masses is

1. \(\left(m_1+m_2\right) g\)
2. \(\frac{\left(m_1-m_2\right)^2}{m_1+m_2} g\)
3. \(\left(m_1-m_2\right) g\)
4. \(\frac{\left(m_1+m_2\right)^2}{m_1-m_2} g\)

Answer:

Given

A smooth massless string passes over a smooth fixed pulley. Two masses m₁ and m₂ (m₁> m₂) are tied at the two ends of the string. The masses are allowed to move under gravity starting from rest.

If a is the acceleration of the two masses and T is the tension in the two sides of the strings, then m₁g-T=m₁a….(1)

and T- m₂g = m₂a…(2)

Now from equations (1) and (2), we get, a = \(\frac{\left(m_1-m_2\right) g}{m_1+m_2}\)

Therefore, total external force = \(\left(m_1+m_2\right) a=\left(m_1-m_2\right) g\)

The option 3 is correct.

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Question 10. A mass of 1 kg is suspended by means of a thread. The system is

1. Lifted up with an acceleration of 4.9 m/s²
2. Lowered with an acceleration of 4.9 m/s². The ratio of tension in the first and second case is

1. 3:1
2. 1:2
3. 1:3
4. 2:1

Answer:

1. Tension while lifting the mass, \(T_1=g+a=g+\frac{g}{2}=\frac{3 g}{2}\left[because a=\frac{g}{2}\right]\)
2. Tension while lowering the mass, \(T_2=g-a=g-\frac{g}{2}=\frac{g}{2}\)

∴ \(T_1: T_2=\frac{3 g}{2}: \frac{g}{2}=3: 1\)

The option 1 is correct.

Question 11. A block of mass 1 kg starts from rest at x = 0 and moves along the x-axis under the action of a force F = kt, where t is time and k = 1 N/s. The distance the block will travel in 6 seconds is

1. 36m
2. 72m
3. 108m
4. 18m

Answer:

Given

A block of mass 1 kg starts from rest at x = 0 and moves along the x-axis under the action of a force F = kt, where t is time and k = 1 N/s.

F = kt or, \(m \frac{d v}{d t}=k t \text { or, } \frac{d v}{d t}=\frac{k}{m} t\)

or, \(\int_0^v d v=\frac{k}{m} \int_0^t t d t\)

or, \(v=\frac{1}{2} \frac{k}{m} \cdot t^2 or, \frac{d x}{d t}=\frac{1}{2} \frac{k}{m} t^2\)

or, \(\int_0^x d x=\frac{1}{2} \frac{k}{m} \int_0^6 t^2 d t\)

or, x = \(\frac{1}{2} \frac{k}{m} \frac{6^3}{3}=\frac{1}{2} \times \frac{1}{1} \times \frac{216}{3}[because k=1 \mathrm{~N} / \mathrm{s}, m=1 \mathrm{~kg} \mid\)

= 36 m

The option 1 is correct.

Question 12. The velocity v of a particle (under a force F) depends on its distance (x) from the origin (with x > 0) \(v \propto \frac{1}{\sqrt{x}}\). Find how the magnitude of the force (F) on the particle depends on x.

1. \(F \propto \frac{1}{x^{3 / 2}}\)
2. \(F \propto \frac{1}{x}\)
3. \(F \propto \frac{1}{x^2}\)
4. \(F \propto x \mid 1]\)

Answer:

Given

The velocity v of a particle (under a force F) depends on its distance (x) from the origin (with x > 0) \(v \propto \frac{1}{\sqrt{x}}\).

⇒ \(\quad v \propto \frac{1}{\sqrt{x}} \text { or, } v=\frac{k}{\sqrt{x}}\)

∴ \(\frac{d \nu}{d t}=\frac{d}{d x}\left(k x^{-\frac{1}{2}}\right) \frac{d x}{d t}=\nu \frac{d}{d x}\left(k x^{-\frac{1}{2}}\right)\)

= \(k^{\prime} \times \frac{1}{x^2}\left[\text { where } k^{\prime}=-\frac{k^2}{2}\right]\)

∴ \(\propto \frac{1}{x^2}\)

So, \(F \propto \frac{1}{x^2}\)

The option 3 is correct.

Short Answer Questions on Action and Reaction Forces

Question 13. The force F acting on a particle of mass ni is indicated by the force-time graph shown below. The change in momentum of the particle over the time interval from zero to 8 s is

1. 24 N • s
2. 20 N • s
3. 12 N • s
4. 6 N •s

Answer:

Given

The force F acting on a particle of mass ni is indicated by the force-time graph shown above.

F = ma = \(m \frac{d v}{d t}\)

∴ P = \(\int m d v=\int F d t\)

So, momentum is the total area of the forcetime graph.

Therefore, change in momentum,

ΔP =(1/2 x 6 x 2)-(3 x 2) + (3 x 4)

= (6-6+12) = 12N · s

The option 3 is correct.

Question 14. A balloon with mass m is descending down with an acceleration a (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration a?

1. \(\frac{2 m a}{g+a}\)
2. \(\frac{2 m a}{g-a}\)
3. \(\frac{m a}{g+a}\)
4. \(\frac{m a}{g-a}\)

Answer:

Given

A balloon with mass m is descending down with an acceleration a (where a < g).

Let upthrust of air be \(F_a\), then for downward motion \(m g-F_a=m a \text { or, } F_a=m(g-a)\)

For upward motion, \(F_a-(m-\Delta m) g=(m-\Delta m) a\) or, \(\Delta m(g+a)=m a+m g-F_a=2 m a\)

∴ \(\Delta m=\frac{2 m a}{g+a}\)

The option 1 is correct.

Question 15. Three blocks A, B, and C, of masses 4 kg, 2 kg, and 1 kg respectively, are in contact on a frictionless surface, as shown. If a force of 14 N is applied on the 4 kg block, then the contact force between A and B is

1. 2N
2. 6N
3. 8N
4. 18N

Answer:

Given

Three blocks A, B, and C, of masses 4 kg, 2 kg, and 1 kg respectively, are in contact on a frictionless surface, as shown.

Acceleration of the whole system, a = \(\frac{14}{4+2+1}=2 \mathrm{~m} / \mathrm{s}^2\)

Hence, the required force for blocks B and C to have this acceleration =(2 + l)x2 = 6N.

So, the A block will apply 6 N force in the plane of contact of A and B.

Therefore, the contact force between A and B = 6 N.

The option 2 is correct.

Question 16. A girl jumps down from a moving bus, along the direction of motion of the bus, tilting slightly forward. She falls on

1. A sheet of ice
2. A patch of glue.

1. In case (1) she falls backward and in case (2) she falls forward
2. In both cases (1) and (2) she falls forward
3. In both cases (1) and (2) she falls backward
4. In case (1) she falls forward and in case (2) she falls backward

Answer:

1. Due to super smoothness of the sheet of ice, her leg cannot move forward while her head moves forward due to inertia of motion. Hence, she falls forward.
2. Her leg becomes still at the patch of glue but her leg cannot move forward while her head moves forward due to inertia of motion. Hence, she falls forward.

The option 2 is correct

Short Answer Questions on Force and Mass Relationship

Question 17. A block of mass m is placed on a smooth inclined wedge ABC of inclination 9 as shown in the figure. The wedge is given an acceleration towards the right. The relation between a and 9 for the block to remain stationary on the wedge is

1. a = \(g \cos \theta\)
2. \(a=\frac{g}{\sin \theta}\)
3. a = \(\frac{g}{{cosec} \theta}\)
4. a = \(g \tan \theta\)

Answer:

For the mass m to remain still in the non-inertial reference frame,

Rsinθ = ma…(1)

Rcosθ = mg…(2)

(1) + (2) \(\tan \theta=\frac{a}{g} \quad \text { or, } a=g \tan \theta\)

The option 4 is correct

Question 18. Two masses 8 kg and 12 kg are connected to the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses and the tension in the string when the masses are released.
Solution:

Given

Two masses 8 kg and 12 kg are connected to the two ends of a light inextensible string that goes over a frictionless pulley.

Let m₁ = 12 kg and m₂ = 8 kg

Then, a = downward acceleration of m₁ = upward acceleration of m₂

So, the force equations are, for mass m₁: m₁g-T= m₁a….(1)

and for mass m₂: T- m₂g = m₂a…(2)

Now adding equations (1) and (2), we get, \(g\left(m_1-m_2\right)=a\left(m_1+m_2\right)\)

or, \(a =\frac{m_1-m_2}{m_1+m_2} g=\frac{12-8}{12+8} \times 9.8\)

= \(1.96 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Then from (2), \(T=m_2 g+m_2 \frac{m_1-m_2}{m_1+m_2} g\)

= \(m_2 g \cdot \frac{2 m_1}{m_1+m_2}=\frac{2 m_1 m_2}{m_1+m_2} g\)

= \(\frac{2 \times 12 \times 8}{12+8} \times 9.8=94.08 \mathrm{~N}\)

Question 19. Define acceleration. When does a body possess uniform acceleration?
Answer:

The acceleration of a body is its rate of change of velocity with time.

From Newton’s 2nd law, the acceleration \(\vec{a}\) of a body of mass m is related to an applied force \(\vec{F}\) as, \(\vec{F}\) = m\(\vec{a}\). So, when the body is acted upon by a constant force \(\vec{F}\), we get \(\vec{a}\) = constant. Thus, the body moves with a uniform acceleration.

Real-Life Examples of Newton’s Laws: Short Answer Questions

Question 20. A monkey of mass 40 kg climbs on a rope that can stand a maximum of 600 N. In which case the rope will break: the monkey

1. Climbs up with an acceleration of 6 m · s^-2
2. Climbs down with an acceleration of 4 m · s^-2
3. Climbs up with a uniform speed of 5 m · s^-1
4. Falls down the rope nearly under gravity

[Ignore the mass of the rope. Take g = 10 m · s^-2]

Answer:

Real weight of the monkey = 40 kg · wt = 40 x 10 N = 400 N; So the rope can withstand his real weight.

Here g = 10 m · s^-2 = actual acceleration due to gravity, acting downwards.

1. Acceleration, a = -6 m s^-2; the negative sign comes as the acceleration is upwards.

Thus, relative downward acceleration, g’ = g- a = 10- (-6) = 16 m · s^-2

Hence, apparent weight = mg’ = 40 x 16 = 640 N

As it is greater than 600 N, the rope will break.

2. Here, acceleration a = + 4 m · s^-2

Thus, relative downward acceleration, g’ = g- a = 10- 4 = 6 m · s^-2

Hence, apparent weight = mg’ = 40 x 6 = 240 N

This force does not break the rope.

3. The speed is uniform, i.e., there is no acceleration.

Hence, Apparent weight = real weight = 40 kg · wt = 400 N

Clearly, the rope does not break due to this force.

4. When the monkey falls freely under gravity, then a = g

Thus relative downward acceleration, g’ = g – g= 0

Hence, apparent weight = mg’ = 0 In this case, the rope does not break.

Question 21. A cricket ball of mass 150 g moving with a speed 12 m · s^-1 is hit by a bat so that the ball is turned back with a velocity of 20 m · s^-1. Calculate the impulse received by the ball.
Answer:

Given

A cricket ball of mass 150 g moving with a speed 12 m · s^-1 is hit by a bat so that the ball is turned back with a velocity of 20 m · s^-1.

The force on the ball acts in the backward direction; this direction is taken to be positive.

So, initial velocity, v₁ = -12 m • s^-1 and final velocity, v₂ = + 20 m • s^-1

From Newton’s second law of motion, with proper choice of the unit of force,

force = \(\frac{\text { change of momentum }}{\text { time interval }}\)

So, impulse  = force x time interval

= change of momentum

= \(m v_2-m v_1=m\left(\nu_2-v_1\right)\)

= \(\frac{150}{1000} \mathrm{~kg} \times\{20-(-12)\} \mathrm{m} \cdot \mathrm{s}^{-1}\)

= \(4.8 \mathrm{~kg} \cdot \mathrm{m} \cdot \mathrm{s}^{-1}\)

Question 22. Show that Newton’s second law of motion is the real law of motion.
Answer:

From Newton’s 2nd law, the force \(\vec{F}=\frac{d}{d t}(m \vec{v})\) with proper choice of the unit of force.

1. If the external force \(\vec{F}\)=0; we have, \(\frac{d}{d t}(m \vec{v})\)= 0

∴ m\(\vec{v}\) = constant, so, \(\vec{v}\) = constant This means that a body at rest would remain at rest and a body in motion would continue to maintain its uniform velocity. This is Newton’s 1st law of motion.

2. We consider a system of two bodies, 1 and 2.

∴ \(\vec{F}_21\) = force on body 1, exerted by body 2;

and \(\vec{F}_12\) = force on body 2, exerted by body 1.

In the absence of any other external forces, Newton’s 2nd law gives,

For body 1: \(\vec{F}_{21}=\frac{d}{d t}\left(m_1 \vec{v}_1\right)\)

For body 2: \(\vec{F}_{12}=\frac{d}{d t}\left(m_2 \vec{v}_2\right)\)

Now, if the combination of the two bodies is considered, \(\vec{F}_21\) and \(\vec{F}_12\) are internal forces only and do not contribute to any external force. If the external force is zero, we have from the 2nd law,

0 = \(\frac{d}{d t}\left(m_1 \vec{v}_1+m_2 \vec{v}_2\right)=\frac{d}{d t}\left(m_1 \vec{v}_1\right)+\frac{d}{d t}\left(m_2 \vec{v}_2\right)\)

0 = \(\vec{F}_{21}+\vec{F}_{12}\)

∴ \(\vec{F}_{21}=-\vec{F}_{12}\) This is Newton’s 3rd law.

Thus, Newton’s 2nd law of motion is sometimes called the “real” law of motion—it encompasses both the 1st and the 3rd laws.

However, Newton’s wisdom dictated him to propose the 1st and the 3rd laws separately in order to introduce, respectively,

1. The Notion Of The Inertial Frames Of Reference, And
2. The Concept Of Action-Reaction Pair Of Forces.

Question 23. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m · s^-1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
Answer:

Given

A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m · s^-1. If the mass of the ball is 0.15 kg

The impulse is imparted towards the bowler from the position of the batsman. If that direction is taken to be positive, then

The initial velocity of the ball = -12 m · s^-1

The final velocity of the ball = +12 m · s^-1

So, the impulse = change in momentum

= mass x change in velocity

= 0.15 x{12-(-12)} = 0.15×24

= 3.6 kg m · s^-1

Question 24. A bullet of mass 0.04 kg moving with a speed of 90 m · s^-1 enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
Answer:

Given

A bullet of mass 0.04 kg moving with a speed of 90 m · s^-1 enters a heavy wooden block and is stopped after a distance of 60 cm.

Let mass of the bullet, m = 0.04 kg

The initial velocity of the bullet, u = 90 m · s^-1

The final velocity of the bullet, v = 0

Distance traversed by the bullet, s = 60 cm = 0.6 m

If a be the retardation of the bullet, v² = u²-2as or, 0 = (90)² – 2a x 0.6

∴ a = \(\frac{(90)^2}{2 \times 0.6}=6750 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

So the resistive force, F = ma = 0.04 x 6750 = 270 N

The actual resistive force, and therefore, the retardation of the bullet may not be uniform. The answer, therefore, only indicates the average resistive force.

Question 25. A car and a truck are moving on a level road so that their linear momenta are equal. Which one is moving faster?
Answer:

A car and a truck are moving on a level road so that their linear momenta are equal.

The car is moving faster as its mass is less than that of the truck.

Question 26. Why a cricket player lowers his hands while catching a cricket ball? Explain
Answer:

We know, impulse = force x time

= change in linear momentum.

Force not only depends on the change in momentum but also on how fast the change is brought about. The same change in momentum brought about in a shorter time needs greater force.

By lowering his hands while catching the cricket ball, the player allows a longer time for the momentum to change, thereby preventing his hands from getting injured.

Question 27. Calculate the net force acting on a body of mass 10 kg moving with a uniform velocity of 2 m/s.
Answer:

Since, velocity is uniform, so, acceleration, a = 0.

Hence, net force, F = ma = 0

Question 28. A lift of mass 400 kg is hung by a wire. Calculate the tension in the wire when the lift is

1. At rest,
2. Moving upward with a constant velocity of 2.0 m/s,
3. Moving upward with an acceleration of 2.0 m/s² and
4. Moving downward with an acceleration of 2.0 m/s².

Answer:

1. Lift at rest T – W = mg = 4000 N
2. Moving up with 2m/s, a = 0; T = W = mg = 4000 N
3. Moving up with a – 2m/s²; T = mg+ ma – 4000 + 400 x 2 = 4800 N
4. Moving down with a = -2m/s²; T = mg- ma = 4000 – 400 x 2 = 3200N

Question 29. Write its two applications. The linear momentum of a body can change in the direction of applied force. Comment.
Answer:

• While firing a gun, the gun must be held tightly to the shoulder.
• When a man jumps out of a boat, the boat is pushed away, which pushes the man forward.
• The statement is correct. It is in accordance with Newton’s second law of motion.