WBBSE Solutions For Class 9 Physical Science Chapter 2 Force And Motion Topic B Newtons First And Second Laws Of Motion

Chapter 2 Topic B Newton’s First And Second Laws Of Motion Synopsis

Newton’s First Law Of Motion:

Everybody continues to be in its state of rest or of uniform motion in a straight line unless an external force is applied to it.

Inertia is the property of a body existing in a state of rest or of uniform motion due to which it continues to remain in that state and opposes any effort or attempt to change that state.

The tendency of a moving body to continue to exist in its state of motion with uniform velocity in a straight line is called inertia of motion.

The tendency of a stationary body to continue its state of rest is called inertia of rest.

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A single force, which represent the result of the summation of a number of forces both in magnitude and in direction, is called the resultant of those forces.

The Law Of Parallelogram Of Forces:

If a particle is acted on by two forces represented in magnitude and direction by the two sides of a parallelogram drawn from a point simultaneously then they are equivalent to a force represented in magnitude and direction by the diagonal of the parallelogram passing through the point.

Explanation: If P and Q be two forces acting simultaneously on point A are represented in magnitude and direction by two adjacent sides AB and AD of a parallelogram ABCD respectively, then their resultant force R is given in magnitude and in direction by the diagonal of the parallelogram.

Resolution of a force refers to the division of a force in two given directions in such a way that the resultant of the two components is equal to the given force.

Resolution Of A Force Into Components At Right Angles To Each Other:

Resolution of a force into components at right angles to each other means that the force is resolved into two components which are at right angles to each other, so that the resultant of the two components is equal to the given force.

Explanation: A force R is resolve into two components P and Q perpendicular to each other.

Here, \(R^2=p^2+Q^2 \text { or, } R=\sqrt{p^2+Q^2}\)

Where 9 is the angle between R and P.

When more than one force acts on a body such that the resultant of the forces is zero, then the forces are called balanced forces. Balanced forces acting on a body do not produce acceleration of the body but the body becomes strained.

When one or more than one force acts on a body, the resultant of the acting forces is non-zero, then these forces are called effective forces or unbalanced forces. Acceleration of a body arises mainly due to the influence of effective force.

During application of force on a body, if the applicator is a part of the whole system, then the applied force cannot change the velocity of the particle but only attempts to change it. This force is called internal force.

During application of force on a body, if the applicator of the force and the body are different systems, then the applied force changes the velocity of the body or attempts to change the velocity. This force is called external force.

The ratio of the applied force on a body and the acceleration of the body is called, its inertial mass.

Linear Momentum is defined as the property of motion of a body that is produced due to the combination of velocity and mass of the moving body. Linear momentum is a vector quantity. Unit of linear momentum in SI is kg • m • s^-1 and its dimensional formula is MLT^-1.

Newton’s Second Law Statement:

The time rate of change of momentum of a body is proportional to the applied force and takes place in the direction in which the force acts.

Mathematical Expression: Suppose an external constant force F is applied on a body of constant mass m.

If a be the acceleration of the body, then according to Newton’s second law of motion

F = ma, i.e., Force = mass x acceleration

The force applied on a body of unit mass and produces an acceleration of one unit is called unit force.

Newton’s Second Law Of Motion For A Constant Mass:

If the mass unchanged, then the effective force applied on the body is equal to the product of its mass and acceleration. Acceleration takes place in the direction of the applied force.

Variable Mass System is a system whose mass increases or decreases during its state of motion but its mass is never created nor destroyed.

Newton’s Second Law Of Motion For A Variable Mass:

The rate of change of momentum of a body is equal to the external working force applied on the body. Change of momentum takes place in the direction in which the force is applied.

Units Of Force in CGS system and SI are dyne and newton respectively.

The amount of force that acts on a mass of 1 g and produces an acceleration of 1 cm/s² is called 1 dyne.

The amount of force that acts on mass of 1 kilogram and produces an acceleration of 1 m/s² is called 1 newton.

Relation Between Newton And Dyne:

I N = 1 kg x 1 m • s^-2

= 1000 g x 100 cm • s^-2  = 10⁵ g • cm • s^-2 = 10⁵ dyn

Dimensional Formula Of Force:

Dimensional formula of force = Dimensional formula of mass x Dimensional formula of acceleration

= M x LT^-2 = MLT^-2

Chapter 2 Topic B Newtons First And Second Laws Of Motion Short And Long Answer Type Questions

Question 1. What is resultant force?

Answer:

A resultant force is a single force obtained by combining several active forces. The important describing feature of a resultant force is that it has the same effect on the body as the original system of forces.

Question 2. What are the effects of an applied force on a body.

Answer:

By applying a force,

1. A static body may be set in motion,
2. Magnitude or direction or both of a moving body may be changed,
3. A moving body may be brought to a stationary state or
4. Form or shape of a body may be changed.

Question 3. What is balanced force?

Answer:

When more than one force acts on a body and the resultant of the forces is zero, then the forces are called balanced forces.

Question 4. State Newton first law of motion.

Answer:

Every body continues to be in its state of rest or of uniform motion in a straight line unless the body is compelled to change its state by a net external force i.e. there will be no acceleration of the object.

Question 5. Give an example of balanced force.

Answer:

In a tug-of-war game, four boys on the left side and four boys on the right side pull the rope in such a way that the rope does not move in either direction. Here, a total of eight forces are working on the rope.

But as the forces applied by the boys on the left side are equal to the forces applied by the boys on the right side and are in the opposite directions, resultant force on the rope is zero.

Hence, the rope does not move in any direction. These eight forces working on the rope are called balanced forces.

Question 6. What do you mean by effective force? Which property of the body changes due to its influence?

Answer:

When the resultant of the forces due to the application of one or more than one force on a body is not zero, then those forces are called effective forces. The property of motion (rest or motion) changes due to the influence of effective forces.

Question 7. Two friends are pulling a table without being able to move it. How is it possible?

Answer:

If two friends are Ruling a table from two sides by applying two opposite and equal forces, then it will not move. This type of application of force means that balanced force is being applied on the table.

Due to this reason, the value of the resultant force on the table is zero.

Question 8. Two friends are moving a table by pulling it with equal forces. Application of forte in which direction does make this possible?

Answer:

If the table is to be moved by the two friend by pulling it with equal force, the forces can never be applied opposite to each other.

Applying the forces in any other direction budges the table.

Question 9. What is the influence of effective force on a body?

Answer:

The action of effective force on a body generates acceleration, i.e., change of velocity.

Change of velocity may take place in different ways:

1. A static body may start moving,
2. There may be a change in the velocity of a moving body,
3. There may be a change of direction of the moving body or
4. Both the magnitude and the direction of the moving body may change.

Question 10. The action of balanced force on e body creates strain—explain with example.

Answer: A spring shrinks when it is compressed with equal force by two hands. In this case, equal and opposite forces are applied on the spring by the hands.

This is a balanced force. No acceleration is produced in the spring due to this. But the spring gets compressed. This compression or shrinkage of the spring is known as strain.

Question 11. What is internal force and external force?

Answer:

Internal Force:

During application of a force on a body, if the applicator is a part of the system applying force, then the applied force cannot change the velocity of the particle but only attempts to change it. This force is called internal force.

External Force:

During application of a force on a body, if the applicator of the force and the body are different systems, then applied force changes the velocity of the body or attempts to change the velocity. This force is called external force.

Question 12. Which force is responsible for change of the state of a body? Answer with the help of Newton’s first law of motion.

Answer: It is obvious from Newton’s first law of motion that it is possible to change the state of rest or state of motion of a body with the help of only external effective force. Internal force or external balanced force cannot do it.

Question 13. Give an idea of force from Newton’s first law of motion.

Answer:

We get a qualitative definition of force from Newton’s first law of motion. The velocity of a body does not change if any external effective force is not applied on the body.

This means that there is no change in velocity of the body if it is zero (body at rest) or of any definite magnitude (with uniform velocity).

Question 14. What are the things known from Newton’s first law of motion?

Answer:

The following things are known from Newton’s first law of motion:

1. Property of inertia of matter.
2. Qualitative definition of force.
3. Change of velocity of a body is possible only by external unbalanced force and not by any internal force.
4. A clear statement that there is some sort of similarity or resemblance between the state of rest and state of motion with uniform velocity.

Question 15. What is inertia? How many types of inertia are there and what are those?

Answer:

The property of a body due to which it exists in a state of rest or of uniform motion and continues to remain in that state and opposes any attempt to change that state, is called inertia.

It is of two types, inertia of rest and inertia of motion.

Question 16. What Is inertia of rest and inertia of motion?

Answer:

Inertia Of Rest:

The tendency of a stationary body to always continue in its state of rest is called inertia of rest.

Inertia Of Motion:

The tendency of a moving body to always continue in its state of motion with uniform velocity in a straight line is called inertia of motion.

Question 17. A stationary bus starts moving suddenly. Why do the passengers standing inside lean in the backward direction?

Answer:

When the bus is at rest, the bodies of the passengers in contact with it also remain at rest.

When the bus starts moving suddenly, the lower parts of the bodies of the passengers in contact with the floor of the bus move forward while the upper parts tend to remain at rest due to inertia and the passengers lean in the backward direction.

Question 18. What happens if a bucket full of water is suddenly pushed?

Answer:

When a force is applied suddenly to a bucket full of water in the forward direction, the bucket moves forward. But water inside the bucket continues to remain at rest due to inertia of rest. So, water splashes in the backward direction.

Question 19. If a running bus stops suddenly, passengers standing inside lean forward. Why? Or, It is not prudent to get down from a running bus, because if one does it carelessly he may fall in the forward direction. Explain.

Answer:

If a running bus with passengers stops suddenly, passengers within the bus stumble or lean in the forward direction. The reason is that when the bus was in motion, the passengers were also moving with the same velocity.

When the bus stops by the application of brakes, lower parts of the bodies of passengers remaining in contact with the floor stop immediately.

But the upper parts move forward due to inertia of motion and as a result, passengers lean forward. This is the reason why one should not get down from a running bus.

Question 20. Why cannot the participant of a sprinting competition stop after reaching the end point?

Answer:

In a sprinting competition, the goal of the competitor is to reach the end point before all.

With this goal, he does not reduce his speed even at the moment before reaching the end. Now, if he stops suddenly after reaching the end point, the upper part of his body continues to move forward due to inertia of motion.

As a result, there are chances of his stumbling in the forward direction. In order to save himself from injury due to the fall, he does not stop suddenly. Rather, he gradually reduces his speed and runs for some distance before stopping.

Question 21. When a blanket is to be dusted, why is rope and then beaten with a stick?

Answer:

To dust a blanket, it is hung on a support (say, a rope) and then beaten vigorously with a stick. The reason is that when beaten vigorously, the blanket moves in the direction of the applied force.

But the dust particles attached lightly to the blanket prefer to continue in stationary state due to inertia of rest. As a result, dust particles get separated from the blanket, thereby cleaning it.

Question 22. Is it possible for a passenger sitting inside a stationary car to move it by pushing from the inside?

Answer:

No, it is not possible for a passenger sitting inside a stationary car to move it by pushing from the inside. Because a passenger sitting inside may be considered to be a part of th system.

Without application of force from outside, it is not possible to move it by overcoming the frictional force between the car and the road.

Question 23. A passenger sitting inside a bus moving on a horizontal plane with uniform velocity throws a ball upward. After some time, how does the ball come back in the hands of the passenger?

Answer:

The velocities of the passenger and the ball inside the bus moving with uniform velocity are always equal to the velocity of the bus. When the ball is thrown up, an upward motion is also created along with horizontal motion.

Horizontal velocity and perpendicular velocity are in perpendicular direction to each other. So, the velocities do not affect each other. This is the reason why the ball tries to maintain its inertia of motion in the horizontal direction even when it is in air.

As a result, the distance traversed by the passenger in the forward direction is equal to the distance traversed by the ball in the forward direction during the same period.

On the other hand, motion of the ball in the perpendicular direction is accompanied by acceleration due to gravity. So, its velocity gradually decreases to zero and then it falls downwards to reach the hands of the passenger.

Question 24. Write down Newton’s second law of motion without introducing the concept of momentum. Or, Write down Newton’s second law of motion in case of unchanged mass.

Answer:

If the mass of a body remains unchanged, then the magnitude of external applied effective force on the body is the product of its mass and acceleration. Acceleration takes place in the direction of the applied force.

This is Newton’s second law of motion without the concept of momentum or in case of unchanged mass.

Question 25. Explain the equation F = m x o of Newton’s second law of motion with examples.

Answer:

A qualitative definition of force is obtained from Newton’s first law of motion. But a quantitative definition of force is obtained from Newton’s second law of motion. A driver starts his car.

As the engine of the car applies force on the car, it starts moving. The driver increases the velocity with the help of the accelerator to raise it to a certain value. Accelerator is responsible for the acceleration of the car.

Force applied by the engine is responsible for this acceleration. Now, if the applied force by the engine increases, acceleration also increases. So, there is a direct relationship between the applied force and the acceleration produced.

According to the second law of motion, force applied on a body of definite mass and its acceleration are directly proportional and the direction of the acceleration generated is in the direction of applied force, i.e., applied force = mass x acceleration

If applied force = F, mass of the body = m, and acceleration = a, then F = ma.

Question 26. How do you infer that mass is an inherent property of matter? Or, A hanging sand bag and an oscillating pendulum are displaced by the same distance. Which one requires more force to displace and why?

Answer:

A sand bag requires more force for its displacement by the same distance compared to that for an oscillating pendulum. Both of them were initially at rest.

Now, while producing the same amount of acceleration in them by application of force, both the bodies oppose the forces due to their properties of inertia.

This property of inertia of a body is measured by its mass. Since the mass of the sand bag is many times greater than the mass of the oscillating pendulum, hence more force is required in this case.

Question 27. From Newton’s second law of motion, explain mass as an inherent property of a body.

Answer:

According to Newton’s second law of motion, force (F) applied on a body = mass of the body (m) x acceleration (a) due to the application of force.

According to this, we can infer that applying same amount of force on different stationary bodies do not produce the same magnitude of acceleration in them.

The reason due to which this difference arises is the mass of the body. So, mass is an inherent property of a body. It is not possible to change the mass of a body unless quantity of matter in the body is increased or decreased.

Question 28. According to Newton’s second law of motion, if force is taken as the cause then acceleration may be treated as the effect.Explain the above statement with reasons.

Answer:

From our ordinary experience, we observe that more the amount of force applied on a body, more is the amount of change of its state of motion, that is, the acceleration or deceleration of the body will be more.

For example, a rope is tied to a solid brick kept on the ground and one starts walking by holding the rope in his hand. The brick starts moving with a definite velocity from its state of rest.

This means that the brick has got acceleration. Instead of walking, if one starts running, the brick also starts moving with the same velocity. Acceleration is greater in the second case.

Hence, we may conclude that increase of force on the body increases its acceleration. So, if force is taken as the cause, acceleration may be regarded as the effect.

Question 29. Mass of a body is the measure of its inertia—explain.

Answer:

Suppose, forces F₁ and F₂ are applied on two bodies of masses m₁ and m₂ respectively to produce same acceleration on them.

According to Newton’s second law of motion,

\(F_1=m_1 a \text { and } F_2=m_2 a\)

∴ \(\frac{F_1}{F_2}=\frac{m_1}{m_2}\)

Now as m₁ > m₂, F₁ is greater than F₂. Therefore, to produce same amount of acceleration (that is, to change state of motion by same amount at the same time) in a heavier body, a comparatively greater amount of force has to be applied.

Resistance against the force, i.e., property of inertia will be greater for a heavier body.

Question 30. How is inertial mass measured?

Answer:

Let the acceleration of a body be a when a force of F is applied on it. From Newton’s second law of motion, we may say that mass of the body,

\(m=\frac{F}{a}\)

So, the ratio of the force applied on a body and the acceleration of the body is the inertial mass (m,).

∴ inertial mass, \(m_i=\frac{F}{a}\)

Question 31. Draw applied force-acceleration graph for a body of constant mass.

Answer:

If a body of constant mass m is moving under the action of an external applied force F. Then according to Newton’s second law of motion,

F = ma, where a is the acceleration of the body.

If a is considered along x-axis (horizontal axis) and F is considered along y-axis (vertical axis) then F-a graph is a straight line that passes through the origin and the gradient of the straight line is the mass (m) of the body.

[gradient (m) = tanθ = \(\frac{F}{a}\) = m]

Question 32. Define unit force from Newton’s second law of motion.

Answer:

If a force F is applied on a mass of m to produce acceleration a, then according to Newton’s second law of motion, F = ma

Now, if m = 1 and a = 1, then F = 1.1 = 1.

So, the amount of force applied on a body of unit mass to produce unit acceleration is called a unit force.

Question 33. Write down the definition of units of force in CGS system and SI. Establish a relationship between them.

Answer:

Units of force in CGS system and SI are dyne (dyn) and newton (N), respectively.

The amount of force that is applied on a body of mass of 1 g to produce an acceleration of 1 cm/s² is called 1 dyn.

∴ 1 dyn = 1 g • cm • s^-2

Again, the amount of force that is applied on a body of mass 1 kg to produce an acceleration of 1 m/s^-2 is called 1 N.

∴ 1 N = 1 kg • m • s^-2

Relation between Newton (N) and dyn:

∴ 1 N = 1 kg. m.s^-2

= 1000 g x 100 cm/s²= 10⁵ dyn

Question 34. With your heavy school bag full of books kept in the carrier of the bicycle, you are riding the bicycle to go to the school with uniform velocity. The bag suddenly falls from the carrier.

1. What is the change of motion of the cycle at that moment due to this fall?
2. What is the change of this system-cycle with the bag?
3. Give an idea of momentum through this incident.

Answer:

1. The velocity of the cycle increases at the moment when the bag falls.
2. When the bag falls, the mass of the system decreases and as a result, its velocity increases.
3. It is clear from this incident that the velocity of the body may change simply by a change of mass of the moving body without applying any force from outside. Therefore, a new property of the body emerges by combination of the velocity of the moving body with its mass. This property is called the momentum of the moving body.

Question 35. A cycle and a rickshaw moving with the same uniform velocity are to be stopped within the same interval of time. Greater force has to be applied on which vehicle to achieve the desired result?

Answer: One has to apply a greater force to stop the rickshaw compared to the moving cycle within the same interval of time. This is because the mass of the rickshaw is more than that of the cycle.

Hence, the momentum of the rickshaw is greater even though both of them have the same velocity. To stop within the same interval of time, rate of change of momentum of the rickshaw or force applied on the rickshaw is greater.

Question 36. Define linear momentum. How do you measure the momentum of a body? What are the units of linear momentum in CGS system and SI? Write down the dimensional formula of linear momentum.

Answer:

Linear momentum is defined as the property of motion of a body that is manifested due to the combination of velocity and mass of the moving body.

Momentum of a body is measured by the product of its mass and velocity. If v is the velocity of a body of mass m, then its momentum, p = mv.

Units of linear momentum in CGS system and SI are g • cm • s^-1 and kg • m • s^-1, respectively.

Dimensional formula of linear momentum is MLT^-1.

Question 37. What do you mean by variable mass system? Give examples.

Answer:

Variable mass system is such a system whose mass keeps on increasing or decreasing during its state of motion, but its mass is neither created nor destroyed.

Examples:

1. A rain drop falls towards the earth’s surface when it is attracted by gravity. When it passes through moist air, its mass keeps on increasing due to the blending of water particles with this drop.
2. When a rocket is launched, combustion of rocket’s fuel takes place and the gas produced is emitted with very high velocity. Due to this burning of fuel, the mass of the rocket decreases gradually.

Question 38. Using the concept of linear momentum, write down Newton’s second law of motion.

Answer:

The rate of change of linear momentum of a body is equal to the force applied on it. The direction of change of momentum of a body takes place in the direction of the applied force.

Question 39. Several phenomena are described here. In which, case, the formula \(F=\frac{m_1-m_2}{t} \times u\) applied, and in which case, the formula x if is applied?

1. A body is falling freely from a height.
2. A piece of iron kept in front of a magnet ; is attracted to it.
3. Brakes are applied to maintain the uniform velocity of a cycle when a heavy bag falls from the carrier of a running cycle.
4. A moving wagon is filled up with coal and its uniform velocity maintained with the help of an engine.
5. A bucket full of water, with a hole in its bottom, is tied to a rope and is towered from the second floor with uniform velocity.

Answer:

1. There is no change in the mass of a freely falling body. But due to gravitational attraction, its velocity gradually increases. In this case, the formula F = ma is applicable. Here, F = gravity and a = acceleration due to gravity.

2. A piece of iron moves towards the magnet with uniform acceleration. The formula F = ma is applicable as there is no change of mass.

3. The mass of a cycle is reduced when a heavy bag falls off from the carrier of a moving cycle. Brakes have to be applied to maintain its velocity.

In this case, the formula \(F=\frac{m_1-m_2}{t} \times u\) is applicable.

Here, m₁ = mass of the cycle with the bag,

m₂ = mass of the cycle without the bag,

u = uniform velocity of the cycle,

F = force applied on the wheel of the cycle due to application of brakes,

t = duration of application of brakes at the moment the bag falls off.

4. The mass of a moving wagon gradually increases when it is filled up with coal. As a result, its velocity tends to decrease. Its uniform velocity is maintained by application of force by the engine.

Here, the formula \(F=\frac{m_2-m_1}{t} \times u\) is applicable.

Here, m₂ = mass of the wagon with additional coal, and m₁ = mass of the wagon before coal was dumped in it.

5. Water leaks continuously through the hole at the bottom of the bucket. As a result, mass of the water filled bucket is reduced. To maintain a uniform velocity, magnitude of tension force on the rope in the upward direction is to be increased.

Here, the formula \(F=\frac{m_1-m_2}{t} \times u\) is applicable.

Question 40. Establish the equation F = ma from the concept of linear momentum. Here, m = mass of the body, F = applied force and a = acceleration of the body.

Answer:

Suppose, a body of mass m is moving with a velocity u. Due to the application of a constant force F in the direction of motion of the body for a time t, its velocity becomes v.

Initial linear momentum of the body = mu and after time f, its linear momentum = mv.

∴ Change of linear momentum during time t = mv – mu = m(v – u)

∴  Rate of change of linear momentum

= \(\frac{m(v-u)}{t}=m a\)

Where, a = \(\frac{m(v-u)}{t}\) = acceleration of the body.

According to Newton’s second law of motion, F ∝ ma

or, F = Kma ……(1)

(where K is a constant)

If we assume that unit force produces unit acceleration on a body of unit mass, then m = 1, F= 1, results in F= 1.

From Equation (1), we get

1 = K • 1 • 1 or, K = 1

∴ F = ma

This is Newton’s second law of motion in case of a stationary mass.

Question 41. Explain Newton’s first law of with the help of second law of motion.

Answer:

If an acceleration a is produced by the application of a force F on a body of a mass m, then according to Newton’s second law of motion, F = ma. If no external force is applied on the body, then F = 0 and thus, o = 0 (because m cannot be zero).

This means the body does not have any acceleration. A body is without acceleration only if it is stationary or is moving with uniform velocity.

Hence, if no external force is applied on a body, then a stationary body always remains stationary and a moving body always keeps moving with uniform velocity. This is Newton’s first law of motion.

Question 42. Write down the law of parallelogram of forces and explain it.

Answer:

A particle is acted on by two forces represented in magnitude and direction by the two sides of a parallelogram drawn from a point. The resultant is equivalent to a force represented in magnitude and direction by the diagonal of the parallelogram passing through that point. This is the law of. parallelogram of forces.

Suppose, two forces P and Q are working simultaneously from point A. Forces P and Q have been represented in magnitude and direction by the two adjacent sides AB and AD of the parallelogram ABCD, respectively.

Then, diagonal of the parallelogram through the point A or the line AC will represent the resultant of the two forces.

Question 43. A car and a lorry are moving with the same momentum. Equal and retarding forces are applied to bring them to rest. Determine which one will take less time to stop?

Answer:

Mass of lorry is greater than that of car.

Suppose, mass of car = m₁,

Mass of lorry = m₂,

Initial velocity of car = v₁

And initial velocity of lorry = v₂.

As per the question, both had the same initial momentum,

∴ m₁v₁ = m₂v₂

Suppose, same opposing force F is applied in both the cases to stop them. This means final velocity is zero for both of them.

∴ \(F=m_1 a_1=m_1 \frac{v_1}{t_1}\)

[a₁ = deceleration of car and t₁ = duration of application of force on the car]

and \(F=m_2 a_2=m_2 \frac{v_2}{t_2}\)

[a₂ = deceleration of lorry and t₂ = duration of application of force on the lorry]

∴ \(m_1 \frac{v_1}{t_1}=m_2 \frac{v_2}{t_2}\)

or, \(m_1 v_1 t_2=m_2 v_2 t_1\)

or, \(\frac{t_2}{t_1}=\frac{m_2 v_2}{m_1 v_1}=1\)

or, \(t_2=t_1\)

∴ equal time is required to stop both of them.

Question 44. How does a bird fly in the sky? Can a bird fly In an air-free atmosphere?

Answer:

When a bird intends to fly from ground to sky, it exerts force on air through its wings. As a reaction, air also exerts equal and opposite force on the bird. It flies along the resultant of these two reaction forces.

In an air-free atmosphere, this type of reaction force cannot be produced. So, a bird cannot fly in air-free atmosphere.

Question 45. Explain resolution of forces with an example.

Answer:

As two forces may be added, similarly a single force may be resolved into two forces. But the resolution should be such that these resolved forces when added, would result in the given force.

Suppose, a force of 5 N is to be divided along two angles 30° and 45° on two opposite sides. By taking length 1 cm as 1 N, a straight line OA of length 5 cm is drawn. Lines OD and Of are drawn in two opposite sides of OA with angles 30° and 45°, respectively.

From point A, straight lines parallel to Of and OD are drawn which intersect OD and Of at points B and C, respectively.

As per the diagram, OBAC is a parallelogram. Lengths of OB and OC are measured by a scale.

It is seen that OB = 3.7 cm (approx.) and OC = 2.6 cm (approx).

This means that magnitude of resolved part of force along OD is 3.7 N and magnitude of resolved part of force along Of is 2.6 N.

Question 46. Resolve a force R along two opposite angles a and fi.

Answer:

Force R is represented by OC. R is to be resolved along two angles α and β on the opposite sides. Two lines OA and OB are drawn in two opposite sides along angles α and β.

Two straight lines are drawn from point C parallel to OA and OB which intersect OB and OA at points E and D, respectively. ODCE is a parallelogram. If OD = P and OE=Q, then P and Q are two components of R along two opposite angles α and β.

Question 47. What is resolution of a force into components at right angles to each other?

Answer:

The resolution of a force into components at right angles to each other means that the force is resolved into two components in such a way that the resultant of the two components is equal to the given force.

Question 48. Resolve a force R into components Which are at right angles to each other along an angle 0 and its perpendicular.

Answer:

The force R is represented by OC. R is to be divided along an angle θ or along OX and its perpendicular, that is along / into components at right angles to each other.

From point C, two perpendiculars CD and CE are drawn on OX and OY, respectively. So, ODCE is a parallelogram. Hence, OD = P and OE = Q are the two rightangled components of force R.

Question 49. Explain orthogonal resolution Of a force with examples.

Answer:

Suppose a force 4 N has to be resolved into components at an angle 30° and its perpendicular. By taking a length of 1 cm as 1 N, a straight line OA of length 4 cm is drawn. An angle of 30° is drawn at one side of OA along straight line OX and also along OY, its perpendicular direction is drawn.

Two perpendiculars AB and AC are drawn from point A on OX and OY, respectively. OBAC is a rectangle. OB and OC are two right angled components of a force 4 N along OX and OY. Measurement with a scale gives OB = 3.4 cm (approx) and OC = 2 cm

Hence, the orthogonal components of 4 N force along OX and OY are 3.4N (approx.) and 2N, respectively.

Question 50. Resolution of a force Is the opposite process of addition of forces is this statement wholly true?

Answer:

A definite force is obtained by the addition of two forces. Again, resolution of a force means its division in two different directions such that the resultant of these two resolved forces is equal to the given force. So we may get a lot of forces by resolution of a force. Hence, the given statement is not wholly true.

Question 51. A brick attached to a rope is dragged through the ground when one starts walking by holding the other end of the rope. Explain this phenomenon by orthogonal components of a force.

Answer:

The weight of a brick acts downwards in a perpendicular way. If tension force is applied to the rope in a slightly oblique way to the surface of the earth, the vertical component of the force acts in an upward direction.

This vertical component reduces the natural weight of the weight of the brick body to some extent. In this case, the brick is dragged on the ground due to the horizontal component of the tension force.

Question 52. Why is it easier to pull a roller than to push it?

Answer:

Suppose, a roller on the ground is pulled along the line OA with a force F. The force F is resolved into its parallel and perpendicular components. Parallel component H acting along the line OB helps the roller to move forward.

The perpendicular component V along OC that works against the weight (W) of the roller acts in an opposite direction to reduce its weight.

Again, when a roller is pushed with a force F along OA, the force F is resolved into parallel and perpendicular directions at right angle.

The parallel component H working along OB helps the roller to move forward and the perpendicular component V working along OC towards the direction of the weight of the roller increases its effective weight.

For this reason, it is easier to pull a roller than to push it.

Chapter 2 Topic B Newtons First And Second Laws Of Motion Very Short Answer Type Questions Choose The Correct Answer

Question 1. Which law of motion of Newton defines force qualitatively?

1. First law of motion
2. Second law of motion
3. Third law of motion
4. Law of gravitation

Answer: 1. First law of motion

Question 2. Dimensional formula of momentum is

1. MLT^-1
2. MLT^-2
3. ML^-1T^-2
4. ML^-1T²

Answer: 1. MLT^-1

Question 3. Amount of force acting on a body of mass 1 kg moving with an acceleration of lm • s~2 is

1. 1 dyn
2. 1 N
3. 5 dyn
4. 10 N

Answer: 2. 1 N

Question 4. If a force of 50 dyn acts on a body of mass 10 g, its acceleration is

1. 3 cm • s^-2
2. 7cm • s^-2
3. 8 cm • s^-2
4. 5cm • s^-2

Answer: 4. 5cm • s^-2

Question 5. Compared to a light body, inertia of a heavy body is

1. More
2. Less
3. Same
4. Not possible to determine

Answer: 1. More

Question 6. The momentum of a body of mass 100 kg is 8000 kg • m • s^-1. Its velocity is

1. 80 cm • s^-1
2. 8 x 10⁵m • s^-1
3. 8 x 10⁵cm • s^-1
4. 80 cm • s^-1

Answer: 4. 80 cm • s^-1

Question 7. An example of inertia of rest is

1. Dusting of a blanket by a stick
2. Fixing of a nail by a hammer
3. Recoiling of a gun after firing of a bullet
4. Circling of a fan after switching off

Answer: 1. Dusting of a blanket by a stick

Question 8. When a force of 75 N acts on a body to produce an acceleration of 3m• s^-2, mass of the body is

1. 20 kg
2. 30 kg
3. 35 kg
4. 25 kg

Answer: 4. 25 kg

Question 9. An iron balL rolls and strikes another ball. This is an example of a force called

1. Frictional force
2. Colliding force
3. Normal force
4. Tension force

Answer: 2. Colliding force

Question 10. A book which is pushed on the table, goes some distance and stops. The book stops due to a force called

1. Frictional force
2. Colliding force
3. Normal force
4. Tension force

Answer: 1. Frictional force

Question 11. Which of the following is an example of normal force?

• A stone is attached to one end of a rope and is hung from hand
• A man is standing on the floor
• An iron nail is moving towards a magnet
• A man stumbles forward while getting down from a car

Answer: 2. A man is standing on the floor

Question 12. A body is moving with constant speed. Force is not required in which of the following cases?

1. To reduce speed of the body
2. To increase speed of the body
3. To change the direction of velocity of the body
4. To keep the direction of velocity of the body same

Answer: 4. To keep the direction of velocity of the body same

Question 13. A motor-driven belt is in motion with a uniform velocity of 8 m/s. If sand is poured on the belt at the rate of 2 kg/s, then how much force does the motor apply to keep the same motion?

1. 8N
2. 12N
3. 32N
4. 16N

Answer: 4. 16N

Question 14. The value of a force which has two rectangular components of 15N and 8N is

1. 17N
2. 23N
3. 20N
4. 46N

Answer: 1. 17N

Question 15. A force x has two rectangular components, one of them being y. The other one is

1. x – y
2. x² – y²
3. \(\sqrt{x^2-y^2}\)
4. \(\frac{x}{2}-y\)

Answer: 3. \(\sqrt{x^2-y^2}\)

Question 16. A particle moving with uniform acceleration has an initial velocity of 10 m/s and a final velocity of 20 m/s. Velocity of the particle when it finishes half the total journey time is

1. 15 m/s
2. 18 m/s
3. 12 m/s
4. 16 m/s

Answer: 1. 15 m/s

Question 17. A particle moving with uniform acceleration has an initial velocity of 30 cm/s and a final velocity of 40 cm/s. Velocity of the particle when it crosses half the total distance is

1. 35 m/s
2. 25√2 cm/s
3. 32 cm/s.
4. 28√2cm/s

Answer: 2. 25√2 cm/s

Question 18. Starting from rest, a particle traversed a certain distance in 100 s with uniform acceleration. How much time does it take to traverse half the distance?

1. 50s
2. 70s
3. 70.7 s
4. 71s

Answer: 3. 70.7 s

Question 19. A bullet loses half of its velocity after entering 3 cm in a wooden block. The distance that the bullet goes further before coming to rest is

1. 3 cm
2. 2 cm
3. 1.5 cm
4. 1cm

Answer: 2. 2 cm

Question 20. A bullet moving with a velocity of 100 m/s can penetrate a 1 cm thick wooden plank. The velocity required by the bullet to penetrate a plank having a thickness of 16 cm is

1. 200 m/s
2. 400 m/s
3. 800 m/s
4. 150 m/s

Answer: 2. 400 m/s

Question 21. Dimensional formula of force is

1. LT^-2
2. MLT^-2
3. MLT^-1
4. ML

Answer: 2. MLT^-2

Question 22. In inertial frame of reference

1. Only Newton’s first law of motion is valid
2. Only Newton’s second law is valid
3. Only Newton’s.third law is valid
4. All the laws of motion are valid

Answer: 4. All the laws of motion are valid

Question 23. SI unit of force is

1. Newton
2. Fermi
3. Pascal
4. Dyne

Answer: 1. Newton

Question 24. Measure of inertia is

1. Mass
2. Velocity
3. Acceleration
4. Displacement

Answer: Mass

Question 25. Momentum of a body of mass 2 kg and velocity 20 cm/s is

1. 0.04 kg • m • s^-1
2. 0.4 kg • m • s^-1
3. 4 kg • m • s^-1
4. 40 kg • m • s^-1

Answer: 2. 0.4 kg • m • s^-1

Question 26. Unit of linear momentum is

1. kg • m • s^-2
2. N/s
3. N • s
4. kg2 • m • s^-1

Answer: N • s

Question 27. From Newton’s first law of motion we get

1. Qualitative definition of force
2. Definition of inertia
3. Measurement of weight
4. Idea about kinetic energy

Answer: 2. Definition of inertia

Question 28. By applying a force 1 N, one can hold a body of mass (g = 10 m • s^-2)

1. 1 kg
2. 0.1 kg
3. 10 kg
4. 0.01 kg

Answer: 2. 0.1 kg

Question 29. Which of the following has the largest inertia?

1. A pencil
2. A needle
3. A brick
4. Your body

Answer: 4. Your body

Question 30. The momentum of a body of given mass is proportional to its

1. Density
2. Velocity
3. Volume
4. Displacement

Answer: 2. Velocity

Question 31. A body of mass 1 kg moves along a straight line with a velocity 1 m • s^-1. Magnitude of the external applied force on the body is

1. 1 N
2. kgf
3. 0
4. 1 dyn

Answer: 3. 0

Question 32. Which of the following relation is not correct?

1. 1 N = 10⁵ dyn
2. 1 N = 1 kg • m/s²
3. 1 N = 981 dyn
4. 1 N = 10⁵ g. cm/s²

Answer: 3. 1 N = 981 dyn

Question 33. Mathematical representation of Newton’s second law is

1. \(F=\frac{m}{a}\)
2. F = ma
3. \(\frac{F}{a}=m^2\)
4. a = Fm

Answer: 2. F = ma

Question 34. Which of the following relation is correct?

1. \(\text { acceleration }=\frac{\text { displacement }}{\text { time }}\)
2. \(\frac{\text { mass } \times \text { displacement }}{\text { time }}=\text { momentum }\)
3. \(\text { velocity }=\frac{\text { distance }}{\text { time }}\)
4. \(\frac{\text { mass } \times \text { displacement }}{\text { time }}=\text { force }\)

Answer: 2. \(\frac{\text { mass } \times \text { displacement }}{\text { time }}=\text { momentum }\)

Question 35. When a bus suddenly starts moving from its rest position, the passengers standing on it lean backwards in the bus. This is because

1. Inertia of rest
2. Inertia of motion
3. Angular momentum
4. Conservation of mass

Answer: 1. Inertia of rest

Question 36. A car is moving with uniform velocity. Resultant force acts on the car F is such that

1. F > 0
2. F ≥ 0
3. F = 0
4. F ≤ 0

Answer: 3. F = 0

Question 37. Velocity of a body of mass 1 kg changes from 20 m/s to 30 m/s in 2 s. Amount of force applied on the body is

1. 25N
2. 4N
3. 2N
4. 5N

Answer: 3. 2N

Question 38. N • kg^-1 is unit of the physical quantity

1. Acceleration
2. Velocity
3. Time rate of change of velocity
4. Speed

Answer: 3. Time rate of change of velocity

Question 39. Resultant force applied on a body under the action of balanced forces is

1. Zero
2. Unit force
3. Undefined
4. None of these

Answer: 1. Zero

Question 40. Under the action of a constant force a body moves with

1. Constant velocity
2. Constant momentum
3. Constant acceleration
4. Constant kinetic energy

Answer: 3. Constant acceleration

Chapter 2 Topic B Newtons First And Second Laws Of Motion Answer In Brief

Question 1. What is the shape of the path of a particle when it is in uniform motion?

Answer: The shape of the path of a particle when it is in uniform motion is straight line.

Question 2. What is the change in total linear momentum of two bodies due to their collision?

Answer: In this case, if there is no applied force from the outside, the total linear momentum of the two bodies does not change.

Question 3. Two friends are displacing a table by application of unequal forces. Application of force in which direction does make this possible?

Answer: When two friends displace the table by application of unequal forces, then application of force in any direction makes this possible.

Question 4. Under the influence of which force on a mass at rest, no acceleration is generated?

Answer: No acceleration is generated on a mass at rest under the influence of balanced force.

Question 5. Under the influence of which force on a mass at rest, acceleration is generated?

Answer: Acceleration is generated on a mass at rest under the influence of effective force.

Question 6. Strain of a body is produced under the influence of which force?

Answer: Strain of a body is produced under the influence of balanced force.

Question 7. Which force can change the state of rest or state of motion of a body?

Answer: External effective force can change the state of rest or the state of motion of a body.

Question 8. Write down Newton’s first law of motion.

Answer: Everybody continues to be in its state of rest or uniform motion in a straight line unless an external force is applied to it.

Question 9. What are the forces active on a body?

Answer: Two types of forces can be active on a body.

These are:

1. Internal forces and
2. External forces.

Question 10. Which law of motion of Newton gives a qualitative definition of force?

Answer: Newton’s first law of motion gives a qualitative definition of force.

Question 11. A bucket full of water is suddenly pushed. State whether the water splashes forward or backward.

Answer: If a bucket full of water is suddenly pushed, water splashes backward.

Question 12. Why do the blades of a rotating electric fan continue to rotate for some time even after switching it off?

Answer: Due to inertia of motion, the blades of a rotating electric fan rotates for a couple of times before it comes to rest.

Question 13. Which law of motion of Newton gives a quantitative definition of force?

Answer: Newton’s second law of motion gives a quantitative definition of force.

Question 14. What is the relationship between Newton and Dyne?

Answer: 1 N = 10⁵ dyn

Question 15. What is the dimensional formula of linear momentum?

Answer: Dimensional formula of linear momentum is MLT^-1.

Question 16. Which one is more convenient—pushing a roller or pulling a roller?

Answer: Pulling a roller is more convenient than pushing it.

Question 17. What is attraction?

Answer: If two bodies come nearer to each other due to action-reaction, then that is called attraction.

Question 18. What do you understand by resolution of a force into two components?

Answer: A force is said to be resolved into two components in two given directions such that the resultant of these two components is equal to the applied force.

Question 19. What is the magnitude of a force if its two rectangular components are 3 N and 4 N?

Answer: Magnitude of the force = \(\sqrt{3^2+4^2}\) = 5 N

Question 20. A force of 13 N has two rectangular components 5 N and x. What is the magnitude of x?

Answer: The component,

\(x =\sqrt{13^2-5^2}\)

= \(\sqrt{169-25}=\sqrt{144}\)

= \(12 \mathrm{~N}\)

Question 21. Write the relationship between force applied on a body and change of momentum of the body due to this force.

Answer: The relationship is force = rate of change of momentum i.e. change of momentum

\(\begin{gathered}
\text { Force }=\frac{\text { change of momentum }}{\text { duration of time of application }} \\
\text { of the force }
\end{gathered}\)

Question 22. In which case Newton’s laws of motion are not applicable?

Answer: In non-inertial frame of reference Newton’s law of motion is not applicable.

Question 23. In which type of reference frame Newton’s laws of motion are applicable?

Answer: Newton’s laws of motions are applicable in inertial frame of reference.

Question 24. Find the mass of the body on which force of attraction of the earth is 1 N. [g = 10 m.s-2]

Answer:

Mass of the body = \(\frac{\text { force }}{\text { acceleration }}\)

= \(\frac{1 \mathrm{~N}}{10 \mathrm{~m} \cdot \mathrm{s}^{-2}}=\frac{1}{10} \mathrm{~kg}\)

= \(\frac{1000}{10} \mathrm{~g}=100 \mathrm{~g}\)

Question 25. Is motion possible without force?

Answer: Yes, uniform motion is possible without external force.

Question 26. What will be the direction of motion of a body if more than one forces are applied on it?

Answer: The body will move in the direction of the resultant of those applied forces.

Question 27. Force = mass x y. Write unit of x in CGS system.

Answer: Here y is acceleration. Therefore CGS unit of y is cm/s-2.

Question 28. lgf = ______ N.

Answer: 1 gf = 980 dyn = 980 x 10-5 N = 0.0098 N

Question 29. A particle is moving with uniform velocity and another particle is moving with uniform acceleration. On which particle force is acting?

Answer: Force is acting on that particle which is moving with uniform acceleration.

Question 30. Under the influence of balanced force can a body have acceleration?

Asnwer: No. A body under the influence of balanced force have no acceleration.

Question 31. Velocity of a body of variable mass is fixed. Is there any force acting on the body?

Answer: Here velocity is fixed. But mass of the body is variable. Therefore momentum of the body changes with time. Thus force is acting on the body.

Question 32. Two bodies X and Y have mass 20 kg and 90 kg respectively. Which have greater inertia?

Asnwer: Mass is the measurement of inertia. Here Y is heavier than X and hence it have greater inertia.

Chapter 2 Topic B Newtons First And Second Laws Of Motion Fill In The Blanks

Question 1. In spite of application of a ______ force on a moving body of fixed mass, it may move with uniform velocity.

Answer: Balanced

Question 2. In case of addition of ______ quantities, law of parallelogram is applied.

Answer: Vector

Question 3. The two resolved parts of any force are the two _____ of that force.

Answer: Components

Question 4. Any body at _____ cannot move on its own and any _____ body can not stop on its own.

Answer: Rest, moving

Question 5. Mass is the ________ property of a body.

Answer: Inherent

Question 6. Mass is the measure of __________ of a body.

Answer: Inertia

Question 7. 1 N = ________ dyn.

Answer: 10⁵

Question 8. The total linear momentum of a system of bodies is conserved if the total external force is ________

Answer: Zero

Question 9. Dimensional formula of force is ______

Answer: MLT^-2

Question 10. An electric fan rotates due to _______, even after the current is switched off.

Answer: Inertia of motion

Question 11. More the mass of a body is increased, more its ________ increases.

Answer: Property of inertia

Question 12. Force = mass x ______

Answer: Acceleration

Question 13. We know about the property of inertia from Newton’s ________ law of motion.

Answer: First

Question 14. If the amount of _______ applied to a body increases, its acceleration also increases.

Answer: Force

Question 15. Absolute unit of force in CGS system is _______

Answer: dyn

Question 16. Unit of momentum in SI is ________

Answer: kg.m.s^-1

Question 17. When a bus at rest suddenly starts moving, its passengers fall backward due to ________

Answer: Inertia of rest

Question 18. A body moves with uniform velocity when no ________ act on it.

Answer: Force

Question 19. Direction of momentum is directed along the ________ of the body.

Answer: Velocity

Question 20. Momentum of two bodies of masses m and 4m are equal ratio of their velocities is ________

Answer: 4: 1

Question 21. To accelerate a body an _________ must act on it.

Answer: Unbalanced force

Question 22. Acceleration is not produced under the action of ________ force.

Answer: Balanced

Chapter 2 Topic B Newtons First And Second Laws Of Motion State Whether True Or False

Question 1. If more than one force acts on a body such that the resultant of the forces is non-zero, then the forces are called balanced forces.

Answer: False

Question 2. Variable mass system is a system whose mass remains constant during its state of motion.

Answer: False

Question 3. The ratio of the force applied on a body and the acceleration of the body is called its inertial mass.

Answer: True

Question 4. Dimensional formula of linear momentum is MLT^-2.

Answer: False

Question 5. Newton’s second law of motion states that to every action, there is an equal and opposite reaction.

Answer: False

Question 6. The tendency of a stationary body to continue to exist in its state of rest is called inertia of rest.

Answer: True

Question 7. Amount of force can be calculated from Newton’s second law of motion.

Answer: True

Question 8. Inertia of a body increases with the increase of acceleration of the body.

Answer: False

Question 9. Amount of force required to stop a body depends on the momentum of the body.

Answer: True

Question 10. Force is necessary to maintain the uniform velocity of a moving object having variable mass.

Answer: True

Question 11. CGS unit of linear momentum is g • cm/s.

Answer: True

Question 12. Qualitative definition of force is known from Newtons first law of motion.

Answer: True

Question 13. 1 N = 10⁶ dyn .

Answer: False

Question 14. Two rectangular components of a force are 5 N and 12 N. The magnitude of the force is 13 N.

Answer: True

Question 15. According to Newton’s second law of motion the direction of change of linear momentum is along the direction of applied force.

Answer: True

Question 16. Definition of unit force is known from Newton’s first law.

Answer: False

Chapter 2 Topic B Newtons First And Second Laws Of Motion Numerical Examples

Linear momentum of a body of mass m and velocity v is p = mv.

If initial and final velocity of a body of mass m are u and v respectively and time interval be t, then

Initial momentum = mu, final momentum = mv. Change of linear momentum = mv-mu.

Rate of change of linear momentum \(p=\frac{m(v-u)}{t}=m \cdot a\)

If a be the acceleration of a body of mass m under the influence of force F then,

\(F=m a \quad \text { or, } a=\frac{F}{m}\)

If two rectangular components of a force F be F₁ and F₂ then,

∴ \(F^2=F_1^2+F_2^2 therefore F=\sqrt{F_1^2+F_2^2}\)

Question 1. A motor-driven belt is moving with a uniform velocity of 10 m/s. If sand from above falls on the belt at the rate of 2 kg/s, then what amount of force has to be applied by this motor to maintain the same motion?

Answer:

Here, velocity of the belt remains unchanged but the mass of the belt increases by 2 kg per second due to the sand falling at the rate of 2 kg/s on the belt.

Velocity of the belt,u = 10 m/s

∴ to maintain the same velocity, force required,

\(F=\frac{m_2-m_1}{t} \times u=\frac{2}{1} \times 10=20 \mathrm{~N}\)

Question 2. Starting from rest, a body of mass m attains a velocity V at a distance of x by application of a force F for time t. Show that, \(t=\frac{m v}{F} \text { and } x=\frac{m v^2}{2 F}\)

Asnwer:

Acceleration of the body, a = \(\frac{F}{m}\)

Velocity of the body after time t,

v = at or, v = \(\frac{Ft}{m}\)

∴ t = \(

Again, v² = lax or, v² = [latex]\frac{2F}{m}\)x

∴ \(x=\frac{m v^2}{2 F}\)

Question 3. A car of mass 100 kg is running with a velocity of 2 m/s. What amount of forte is to be applied to stop the car within a distance of 10 m?

Answer:

The car is running with a velocity of u = 2 m/s.

A force is applied on the car to stop it within s = 10 m .

Suppose, deceleration of the car = a.

Final velocity of the car, v = 0

So, the equation v² = u² – 2as gives 0 = u² – 2as or, 2as = u²

∴ \(a=\frac{u^2}{2 s}\)

Now, mass of the car, m = 100 kg

∴ applied force

\(F=m a=\frac{m u^2}{2 s}=\frac{100 \times 2^2}{2 \times 10}=20 \mathrm{~N}\)

Question 4. A body of mass 2 kg is moving with a velocity of 10 m/s. It Is stopped after a time of 10s. What amount of force is applied to the body?

Answer:

Initial velocity of the body, u = 10 m/s When the body is stopped after t = 10 s, its final velocity, v = 0

If deceleration of the body is a, then from the equation v = u – a t, we get

0 = 10 – a x 10 or, l0a = 10

∴ a = \(\frac{10}{10}\) = 1 m/s²

Now, mass of the body, m = 2 kg

∴ Force applied against the motion of the body,

F = ma = 2 kg x 1 m/s² = 2 N

Question 5. A tennis ball of mass 100 g comes running with a velocity of 10 m/s. It is sent back in the opposite’direction by a racket with a velocity of 15 m/s. If the duration of collision is 0.01s, what is the average force applied by the racket?

Answer:

If the direction of return of the ball is considered as positive, initial velocity of the ball, u = -10 m/s and final velocity, v = 15 m/s .

Duration of collision, t = 0.01s and mass of the ball, m = 100 g = 0.1 kg

∴ Average force applied on the tennis ball,

\(F=\frac{m(v-u)}{t}=\frac{0.1\{15-(-10)\}}{0.01}=250 \mathrm{~N}\)

Question 6. A force acts on a body of mass 100 g for 2 s and then stops acting. In the next 2 s, the body moves 100 cm. What amount of force is applied on the body?

Answer: After the force stops acting, the body moves a distance, s = 100 cm in t = 2 s.

If the velocity of the body is v, then from equation s = vt , we get

\(v=\frac{s}{t}=\frac{100 \mathrm{~cm}}{2 \mathrm{~s}}=50 \mathrm{~cm} / \mathrm{s}\)

The force acted on the body for t₁ = 2 s. If the acceleration of the body during that time is a, then

\(v=a t_1\)

∴ \(a=\frac{v}{t_1}=\frac{50 \mathrm{~cm} \cdot \mathrm{s}^{-1}}{2 \mathrm{~s}}=25 \mathrm{~cm} \cdot \mathrm{s}^{-2}\)

Now, mass of the body, m = 100 g applied force,

F = ma = 100 g x 25 cm/s² = 2500 dyn

Question 7. A force of 20 N acts on a mass of 10 kg for 4 s. What is the change in velocity of the body?

Answer:

Mass of the body, m = 10 kg; force applied, F = 20 N; force acted for a time t = 4s; initial velocity of the body = u and final velocity = v.

Change in velocity =(v-u) = V (say).

So, \(F=\frac{m(v-u)}{t}=\frac{m V}{t}\)

∴ V = \(\frac{F \cdot t}{m}=\frac{20 \times 4}{10}=8 \mathrm{~m} / \mathrm{s}\)

Question 8. A man of mass 50 kg is standing on a lift. Write down the reaction force on the man by the lift (or apparent weight of the man) in each of the following cases:

1. When the lift is at rest;
2. The lift is moving up with an acceleration of 30 cm/s² [acceleration due to gravity, g = 980 cm/s²]

Answer:

1. The weight of the man when the lift is at rest = 50 kg x g cm/s²

= 50 kg x \(\frac{980}{100}\) m/s2 = 50 x 9.8 N = 490 N

2. The effective acceleration of the man when the lift moves upwards with an acceleration of 30 cm/s2

= (g + 30) cm/s²

= (980 + 30) cm/s² = 1010 cm/s²

= 10.1 m/s²

∴ The apparent weight of the body

= 50 x 10.1 N = 505 N

Question 9. A bullet of mass 50 g moving with a velocity of 400 m/s penetrates 20 cm in a wall and comes to rest. What is the average resistance of the wall?

Answer:

In this case, initial velocity of the bullet, u = 400 m/s; final velocity of the bullet, v = 0;

mass of the bullet, m = 50 g = \(\frac{50}{1000}\)kg

Distance traversed by the bullet inside the wall,

s = 20 cm = \(\frac{20}{100}\) m

Suppose, deceleration of the bullet = a

∴ according to the formula, v² = u² – 2as or, 2as = u² – v²

\(v^2=u^2-2 a s\) \(2 a s=u^2-v^2\)

∴ \(a=\frac{u^2-v^2}{2 s}=\frac{(400)^2-0}{2 \times \frac{20}{100}}\)

= \(\frac{400^2 \times 100}{2 \times 20}=100000 \mathrm{~m} / \mathrm{s}^2\)

So, the average resistance of the wall,

∴ \(F=m a=\frac{50}{1000} \times 100000=5000 \mathrm{~N}\)

Question 10. Two blocks A and B have masses 2 kg and 3 kg, respectively. According to the figure, two blocks are kept on a smooth horizontal table such that they are in contact with each other. A force of 10 N is horizontally applied on block A. How much force will be applied on B by A?

 

Answer:

A force of amount 10 N acts on a combined system of (2 + 3) or 5 kg mass and produces an acceleration of a = \(\frac{F}{m}\) = \(\frac{10}{5}\) =2 m/s².

This means that acceleration of 2 m/s² is produced on each of the two blocks A and B.

Therefore, force applied on block 6 by block A,

F= mass of block B x acceleration of the block = 3 x 2 = 6N

Question 11. Two bodies of masses 3 kg and 5 kg are placed in rest. If same amount of force is applied to them then find the ratio of time required to acquire Same velocity.

Asnwer:

Here, initial velocities of the two bodies are zero.

Let, F be the same force applied on both of the bodies through time t₁ a t₂ respectively to acquire same velocity v.

According to Newton’s second law, for the 1st body,

\(F=\frac{3(v-0)}{t_1}=\frac{3 v}{t_1}\)

And for the second body,

\(F=\frac{5(v-0)}{t_2}=\frac{5 v}{t_2}\)

∴ The required ratio \(t_1: t_2=\frac{3 v}{F}: \frac{5 v}{F}=3: 5\).

Question 12. Two forces of magnitude 10 dyn each acts at an angle 120° with each other. Find resultant of this two forces using the law of parallelogram of vectors.

Answer:

ABCD is a parallelogram, whose AB and AD sides represent 10 dyn force both in direction and magnitude.

∠BAD = 120° and AB = 10 dyn, AD = 10 dyn .

The diagonal of the parallelogram, AC represents the resultant of the two forces.

Here, AB = BC= CD = AD

Thus ABCD is a rhombus and AC is the bisector of the angle ∠BAD.

∴ \(\angle B A C=\frac{120^{\circ}}{2}=60^{\circ}\)

AB = BC

∴ ∠BAC = ∠ACB = 60°

or ∠ABC = 180° – 60° – 60° = 60°

∴ ΔABC is a equilateral triangle.

So, AC – AB = 10 dyn

∴ The magnitude of the resultant of the two forces is 10 dyn and it inclined at an angle of 60° with any of the two forces.

Question 13. A hammer of mass 2 kg falling from a height 5 m, hit a nail partially fixed on a surface, and stops in 1/10 s. Find the average force exerted on the nail, [g = 10 m • s^-2]

Answer:

Mass of the hammer m = 2kg, distance covered by the hammer in 1/10 s is h = 5 m and its velocity,

v = \(\sqrt{2 g h}=\sqrt{2 \times 10 \times 5}=10 \mathrm{~m} \cdot \mathrm{s}^{-1}\)

∴ Total force exerted on the nail = Impulsive force exerted by the hammer + weight of the hammer

= \(\left(\frac{m v-m u}{t}\right)+m g=\frac{2 \times 10}{\frac{1}{10}}+2 \times 10\)

= 200 + 20 = 220 N

[If a large force acts on a body for a very short interval of time, it is called an impulsive force.

When a nail is hammered the force acts for a very short period of time. Hence it is impulsive force.

\(\text { Impulsive force } \left.=\frac{\text { change of momentum }}{\text { time }}\right]\)

Question 14. A force produces an acceleration of 2 m/s² in a block. Four such block are tied together and the same force is applied on the combination. What will be the acceleration of the combination?

Answer:

Let, mass of a single block = m kg and the force is = F N

∴ Acceleration produced in a single block, a = \(\frac{F}{m}\) = 2 m • s^-2

∴ The acceleration of the combination

= \(\frac{F}{4 m}=\frac{2}{4}\) = 0.5 m • s^-2