Physics Force And Laws Of Motion Notes

Force And Laws Of Motion

In the previous chapter, we learned about the unidirectional motion of an object in terms of speed, velocity, distance and acceleration in a given time interval without discussing about the cause of that motion. In this chapter, we will study about that cause and the dynamics of the motion.

We will learn how the system varies with time and what causes are responsible for that variation. To understand the dynamics we will  also learn about the laws which govern motion in our everyday life.

Physics Force And Laws Of Motion

 

Force And Laws Of Motion Force in Nature

Consider some situations which occurs in our day-to-day life, like a boy kicking a ball or a girl squeezing a lemon or a mother kneading a flour dough . In all these situations, we see that the objects interact with each other either by pushing or by pulling which further changes the position, speed or shape of the object.

Through these examples, we can come up with a definition of force. Force is the interaction of objects by pushing or pulling  which tends to change the shape or the state of motion of that object.

Following effects can be bring out when a force is applied on an object:

  • A stationery object will start moving once after a force is applied.
    Example: paddling the bicycle.
  • Force is required to stop the moving object.
    Example: apply brake to a running car.
  • Speed of a moving body can be changed.
    Example: Press accelerator to increase the speed of a car and apply brake to decrease or stop its speed.
  • Direction of moving object can be changed after applying force.
    Example: Change the direction of steering a car.
  • Among all these force can even change the shape and size of an object.
    Example: Hammering a stone will break it into several pieces. Force is a vector quantity i.e., it can be completely defined by both direction and magnitude.

Force And Laws Of Motion Balanced and Unbalanced Forces

Force can be defined as an agent, which can produce acceleration in a body on which it acts, or produce a change in its size or shape, or both.

  • Balanced Forces: When more than one force is applied on an object in such a way that the forces does not bring any change in its state of rest or motion, then the forces are said to be balanced forces. Balanced forces may bring out the change in the shape on the object. If the resultant of applied forces is equal to zero, it is called balanced force. Balanced forces do not cause any change of state of an object. Balanced forces are equal in magnitude and opposite in direction. Balanced forces can change the shape and size of the object.
  • Unbalanced Forces: When more than one force is applied on an object in such a way that the forces bring out a change in its state of rest or motion, then the forces are said to be unbalanced forces. If the resultant of applied forces are greater than zero, the forces are called unbalanced forces (Fig. 2.2). An object in rest can be moved because of applying balanced forces. Forces can be broadly classified into two types:
    • Contact Force
    • Field Force or Non-contact force

Contact Force

The force exerted by an object over the other object which is in contact with the former can be defined as contact force.

Example: Pulling a cart

Some important contact forces are:

  1. Normal force: If contact forces between the bodies are perpendicular to the ­surfaces in contact, the forces are known as contact forces.
    For example: A wooden block kept on table is in equilibrium. Block applies a downward force on a table due to its weight and table pushes the block upwards. Thus both table and block apply a normal force on each other.
  2. Frictional force: When we try to slide the heavy box on a rough floor by pushing it, a force acts parallel to the surface in contact with the floor and opposes the pushing force. This parallel force is called the frictional force.

Physics Force And Laws Of Motion Balanced forces

 

Example:

In the above example, the truck is being pushed from opposite sides in such a way that it does not move i.e., equal force is applied from both the sides. Such forces are called balanced forces where net force is zero.

Laws Of Motion Balanced forces

 

Non-contact Force

The force exerted by an object over the other object when they are not physically in contact with each other is called non-contact force or action at a distance.

Examples:

  • Gravitational pull: It is the downward force acting on an object due to gravitational pull of earth. If the object is not in contact with the earth the earth pulls it. If we release a stone or ball or apple from height it will fall on the ground.
  • Magnetic force or Electrostatic force: It is known that a magnet can pull an iron piece from distance. Also when we rub a comb with hair and bring it near bits of paper, it attracts them from distance.

In the above example, the truck moves in the direction of the force applied because there is no equal opposite force acting on it.

The outcome of such unbalanced forces are as following:

  • Stop a moving object suddenly
  • Increase the speed of a moving object
  • Decrease the speed of a moving object
  • Move a static object
  • Change the shape and size of an object

Force And Laws Of Motion Galileo’s Experiments

Galileo after a series of experiments concluded that a body continues to move with uniform velocity if no force acts on it. He observes when a marble rolls down a smooth incline plane, its velocity increases. Its velocity decreases when it climbs up.

He also said that if a marble, kept on one side of the smooth plane inclined at both sides, released from rest, it will roll down the slope and go up on the opposite side to the same height above bottom line, from which it is released.

Force And Laws Of Motion Galileo's experiments

 

If the angle of inclination of the right side plane were gradually decreased the marble would travel further distances till it reach the original height. If the right side of the plane is made horizontal the marble will continue to travel forever trying to reach the same height that it was released from.

Thus, it suggests that an unbalanced force (gravity) is required to change the motion of the marble, but not net force is needed to sustain the uniform motion of marble.

Newton further studied the Galileo’s experiment and postulated the three fundamental laws of motion which laid the foundation for classical mechanics. They describe the relationship between a body and the forces acting upon it, and its motion in response to those forces. They are summarized as following:

  1. First law: When viewed in an inertial reference frame, an object either remains at rest or continues to move at a constant velocity, unless acted upon by a net force.
  2. Second law: In an inertial reference frame, the vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object: F = ma.
  3. Third law: When one body exerts a force on a second body, the second body simultaneously exerts a force equal in magnitude and opposite in direction on the first body.

According to Galileo, there is no need of force to continue the motion of an object. If an object is moving, it will be in the same state until and unless an external force is applied to stop it.

Example: A boy riding a cycle applies brake. The cycle does not stop immediately In fact, speed gets reduced slowly and steadily.

As per the above discussion, we form the Galileo’s law of inertia which states that

  • If a body is at rest, it remain continues to at rest unless a force is applied to it.
  • If a body is moving, it will continue to move with the same speed in the same direction unless a force is applied to it.

We come across certain experiences while traveling in a car which can be explained on the basis of the law of inertia. For example, we tend to remain at rest with respect to the seat until the drives applies a brakes to stop the car. With the application of braking force, the car slows down but our body tends to remain in the same state of motion because of its inertia.

Force And Laws Of Motion Newton’s First Law of Motion

‘If an object is in a state of rest, it will remain in the state of rest and if it is in the state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.’

Newton's first law of motion

 

Force And Laws Of Motion Master Your Test

Question 1. Describe Newton’s First law.
Answer: If an object is in a state of rest, it will remain in the state of rest and if it is in state of motion, it will remain moving in the same direction with the same speed unless an external force is applied on it.

Question 2. What did Galileo conclude through his series of experiments?
Answer: Galileo concluded that the natural state of the body is to oppose change in its state of rest or motion.

Question 3. Why a groove is provided in the saucer to hold the cup?
Answer: A groove is provided in a saucer for placing the tea cup because it prevents the cup from toppling over in case of any sudden jerks.

Force And Laws Of Motion Track Your Learning Question And Answers

Question 1. Choose the correct statement.

  1. The object will move at constant speed when the net force is zero.
  2. Law of inertia is Newton’s second law of motion.
  3. All objects do not resist a change in their state of motion.
  4. All object maintains its state of motion under the continuous application of an unbalanced force.

Answer. 1. The object will move at constant speed when the net force is zero.

Question 2. Newton’s laws of motion laid the foundation for ___________.
Answer. classical mechanics

Question 3. Newton gave five laws of motion. (True/False)
Answer. False

Question 4. Seatbelts are required in the cars because of law of inertia. (True/False)
Answer. True

Force And Laws Of Motion Inertia and Mass

Inertia is derived from the Latin word inert which means unchangeable. Inertia is defined as the inability to an object to change its current state. The object with heavy weight has more inertia than the lighter one. Therefore it can be said that ‘The larger the mass, larger is the inertia, and vice versa’. In other words if the mass of the body is more, it is difficult to move the body from rest or if it is moving then it is difficult to stop it.

Therefore, mass is the measure of inertia. The S.I. unit is Kilogram (kg).

Example 1:

Passengers travelling in a bus are running with the same speed as that of the bus and also in the same direction. When driver applies brake, lower part of the body stops but upper part of body doesn’t, hence tends to fall forward, when brakes are applied to a running bus.

 

Example 2: It is easy to hit an empty box than a box filled with stones. The force ­applied while hitting both boxes is same. Empty box will move from its original place while box filled with stone doesn’t move from its place, in fact can cause injury to foot because inertia of box filled with stone is greater than inertia of empty box.

Types of Inertia

Inertia of Rest

A body being at rest will remain at rest until and unless an external force is applied to change its state of rest. This property of body is called inertia of rest.

Example: Passengers sitting in a bus fall backward as soon as the driver starts the bus. This is because the lower part of the passenger is in close contact with the bus. The lower part moves as the bus starts but the upper part of the body doesn’t move immediately, therefore it experiences jerk. This is due to the inertia of rest.

Inertia of Motion

An object which is in a state of motion and continues to be in a state of motion with the same speed in the same direction in a straight line unless an external force is applied to it to change its state. This property is known as inertia of motion.

When a boy jumps from the moving train, he falls down. This is because when the boy was inside the train, his whole body was in a state of motion with the running train. When he jumps from the train, on to a platform, lower part comes to rest while the upper part is still in the state of motion.

This results to the boy falling in the direction of motion of the train. To avoid falling down, the boy should continue to run on the platform in the same direction of motion of the train for some distance.

Example: An athlete runs for some distance before taking a long jump. The reason behind doing this is that running changes the state of body from rest to motion which makes it easier for him to do long jump.

Inertia of Direction

An object moving in a straight line will continue to move in the same direction until and unless some external force compels it to change the direction of motion.

Momentum: The momentum of a body is defined as the product of its mass and velocity.

Thus, momentum = mass × velocity

Or, p = m × v

Where, p = momentum

m = mass of the body

v = velocity of the body

The SI unit of momentum is kilograms meter per second (kg.m/s)

Note: The force required to stop a moving body is directly proportional to its ass and velocity.

Change in momentum: It is defined as the difference between final momentum and initial momentum. Suppose initial momentum is mu, and final momentum is mv, then

Change in momentum = mv – mu

Rate of change of momentum: The rate at which momentum is changing is known as rate of change of momentum.

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

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

Force And Laws Of Motion Conservation of Momentum

Conservation of momentum states that the momentum of a system is constant if there are no external forces acting on the system (Fig. 2.11). It is embodied in Newton’s first law.

Suppose we have two interacting objects 1 and 2 of different masses. The forces between them are equal and opposite. According to Newton’s second law, force is the time rate of change of the momentum, therefore the rate of change of momentum P1 of object 1 is equal to minus the rate of change of momentum P2 of an object 2,

\(\frac{d P_1}{d t}=-\frac{d P_2}{d t}\) (1)

Now, if the rate of change is always equal and opposite, it follows that the total change in the momentum of an object 1 is equal and opposite of the total change in the momentum of object 2. It means that if we sum up the two momenta, the result is zero,

\(\frac{d\left(P_1+P_2\right)}{d t}=0\) (2)

However, the statement that the rate of change of this sum is zero is equivalent to stating that the quantity P1 + P2 is constant. This sum is called the total momentum of a system, and in general it is the sum of individual momenta of each particle in the system.

Example: As shown in the above figure, consider two balls of different masses m1 and m2 that are moving towards each other. Velocity of both balls are v1 and v2. After collision both balls exert some force. Let t be the time taken for collision to last. Ball with mass m1 exerts force F12 on second ball. Ball with mass m2 will exert a force F21 on first ball. ­

Velocity of both balls after collision will be va1 and vb1, Momentum before ­collision = m1 v1

Rate of change of Momentum during collision for ball A F vab = m1 (va1 – v1)/t.

Rate of change of momentum during collision for Ball B F vba = m2 (vb1 – v2)/t.

According to Newton’s third law, Fab = −Fba

m1(va1 – v1)/t = –m2 (vb1 – v2)/t

m1 v1 + m2 v2 = m1 va1 + m2 vb1

Total momentum before collision = m1 v1 + m2 v2

Total momentum after collision = m1 va1 + m2 vb1

Force And Laws Of Motion Track Your Learning Question And Answers

Question 1. For every action there is an equal and opposite ______.
Answer. Reaction

Question 2. The product of mass and velocity of a moving body is known as ______.
Answer. Momentum

Question 3. Conservation of momentum states that the momentum of a system is constant if there are no ______ acting on the system.
Answer. external forces

Question 4. Rate of change of momentum is equal to

  1. ma
  2. mv
  3. pv
  4. ut

Answer. 1. ma

Force And Laws Of Motion Fill in the Blanks

Question 1. For a body of 2 kg moving under gravity, force is ______.
Answer. Zero

Question 2. When a stationary car starts suddenly, the passengers fall ______.
Answer. Backwards

Question 3. A water tanker filled up to half of its tank capacity is running with uniform speed when the brakes are applied suddenly. The water of its tank will move ______.
Answer. Forward

Question 4. Aforce of ______ is needed to stop a car of mass 1200 kg moving with an acceleration of 2 ms–2 .
Answer. 2400N

Question 5. Momentum of a system of particles does not change until ______ is applied.
Answer. Force

Question 6. When bomb explodes into fragments, these fragments must move in a ______ direction.
Answer. Opposite

Question 7. For the measurement of small forces, we can use a ______.
Answer. Force gauge

Question 8. For a body undergoing uniform motion, force is ______.
Answer. Zero

Question 9. A jet plane works on the principle of ______.
Answer. Third law of motion

Question 10. ______ law of motion is called universal law of motion.
Answer. First

Force And Laws Of Motion Match the Columns

Question 1. Choose the correct unit.

NEET Foundation Physics Force And Laws Of Motion Correct Option 1

  1. A-1, B-2, C-3, D-4
  2. A-4, B-3, C-2, D-1
  3. A-2, B-1, C-3, D-4
  4. A-2, B-3, C-1, D-4

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

Question 2. Choose the correct equation.

NEET Foundation Physics Force And Laws Of Motion Correct Option 2

  1. A-1, B-2, C-3, D-4
  2. A-4, B-3, C-2, D-1
  3. A-2, B-1, C-3, D-4
  4. A-2, B-3, C-1, D-4

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

Question 3. Choose the correct code related to Newton’s law of motion.

NEET Foundation Physics Force And Laws Of Motion Correct Option 3

  1. A-1, B-2, C-3, D-4
  2. A-4, B-3, C-2, D-1
  3. A-2, B-1, C-3, D-4
  4. A-3, B-2, C-1, D-4

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

Question 4. Choose the correct code.

NEET Foundation Physics Force And Laws Of Motion Correct Option 4

  1. A-1, B-2, C-3, D-4
  2. A-4, B-3, C-2, D-1
  3. A-2, B-1, C-3, D-4
  4. A-2, B-4, C-3, D-1

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

Force And Laws Of Motion Assertion Reasoning

Direction: For the following questions the options will remain the following:

  • Both A and R are correct and R is the correct explanation of A.
  • Both A and R are correct but R is not a logical explanation of A.
  • A is correct but R is incorrect.
  • R is correct but A is incorrect.
  1. Assertion: Force and acceleration graph is a straight line when force is needed to produce a given acceleration in an object is directly proportional to the mass of the object.
    Reason: As force increases, acceleration also increases.
  2. Assertion: On shaking the branch of a tree, fruits fall down.
    Reason: Branch comes in motion while fruits remain in the state of rest due to inertia.
  3. Assertion: When a passenger jumps from the running train, he falls down.
    Reason: Upper part of body comes to rest while lower part is still in motion.

Force And Laws Of Motion Comprehension Passage

Galileo Galilei was born on 15 February, 1564 in Italy. He had interest in mathematics and ­natural philosophy. In the year 1586, he wrote his first scientific book named as ‘The little Balance’. This book has the description about Archimedes method of finding the relative densities of substances using a balance.

In 1589, in his series of essays ‘De Motu’ he wrote about falling objects using an inclined plane to slow down the rate of descent.

In 1592, he got appointed as a professor of mathematics at the university of Padua in the Republic of venice. Here, he did research on motion. With the study of inclined planes and pendulum, he derived the law for uniformly accelerated objects which states that distance travelled by the object is directly proportional to the square of its time taken.

Question 1. Galileo was born in the year ______.

  1. 1560
  2. 1562
  3. 1564
  4. 1566

Answer. 3. 1564

Question 2. What was the name of his first scientific book?

  1. The secret
  2. The little Balance
  3. The magic
  4. De Motu

Answer. 2. The little Balance

Question 3. In which year Galileo became a professor in University of Padua?

  1. 1562
  2. 1592
  3. 1597
  4. 1599

Answer. 2. 1592

Question 4. What research he did in the university of Padua?

  1. Motion
  2. Speed
  3. Velocity
  4. Acceleration

Answer. 1. Motion

Question 5. Where was Galileo born?

  1. India
  2. America
  3. Europe
  4. Italy

Answer. 4. Italy

Force And Laws Of Motion Integer Type Questions

Question 1. A boy pushes a wall in the north direction with a force of 15 N. What force is exerted by the wall on the boy?
Answer. 15 N

Question 2. How much force is required to produce an acceleration of 2 m/s in an object of mass 0.8 kg?
Answer. 1.6 N

Question 3. A cricket ball of mass 100 g moving with a speed of 30 m/s is brought to rest by a player in 0.03 s. Find the average force.
Answer. 1500 N

Question 4. An object of mass 20 kg moves with an acceleration of 2 m/s2. Find the change in momentum.
Answer. 40 N

Question 5. Which object will require more force? One object of 2 kg mass is moving with acceleration of 5 m/s2. Another object has acceleration of 2 m/s2 and the weight is 4 kg.
Answer. The object of 2 kg mass moving with acceleration 5 m/s2 will require more force.

 Force And Laws Of Motion Long Answer Question And Answers

 Force And Laws Of Motion Long Answer Type Question And Answers

Question 1. A cricket ball weighing 200 g moving with a speed of 90 kmhr-1 is caught by a player. While doing so, he moves his hands back by 10 cm. How much force is exerted by

  1. the player on the ball?
  2. the ball on the player?

Answer.

Given m = 200 g = 0.2 kg

u = 90 kmhr−1 = \(90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\)

v = 0

S = 10 cm = 0.1 m

F = ?

We know that

v2 = u2 + 2aS

⇒ 0 = (25)2 + 2 × a × 0.1

−0.2 a = 625

⇒ a = \(-\frac{625}{0.2}\)

⇒ a = −3125 ms-2

According to Newton’s II law of motion,

F = ma

⇒ F = 0.2 × −3125

⇒ F = −625 N

Negative sign of force implies that it is a retarding force and acts in a direction opposite to the motion of the ball.

∴ Force exerted by the player on the ball = 625 N

According to Newton’s III law of motion, the ball exerts an equal and opposite force on the players hands

∴ F = 625 N

Question 2. When a car weighing 800 kg was moving on a horizontal road with a speed of 30 ms-1, its brakes stopped working. The car came to rest after travelling a distance of 150 m. Find

  1. the retardation of the car
  2. frictional force exerted by the ground on the car

Answer.

(1) Given u = 30 ms-1

v = 0

S = 150 m

a = ?

We know that

v2= u2 + 2aS

⇒ 0 = (30)2 + 2 × a × 150

∴ −300 a = 900

∴ a = −3 ms-2

(2) We know that

F = ma

= 800 × −3

F = −2400 N

Negative sign implies that it is a retarding force. This force which opposes the motion of a car on the ground is a frictional force.

∴ Frictional force exerted by ground on car = 2400 N

Question 3. A hockey player hits a ball (m = 100 g) lying on ground, the with his stick. It is found that the ball starts moving with a velocity of 40 ms−1.Find

  1. the impulse of the force exerted by the stick on the ball.
  2. If the stick was in contact with the ball for 1 ms, then find the magnitude of the force acting on the ball.

Answer.

(1) Impulse of a force = F × t

= ma × t

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

= m(v − u)

= 0.1 kg (40 ms-1 − 0)

= 4 kg ms-1

(2) I = F × t

4 kgms-1 = F × 1 × 10-3 s

∴ F = \(\frac{4 \mathrm{kgms}^{-1}}{1 \times 10^{-3} \mathrm{~s}}\)

= 4 × 103 kgms-2

F = 4 × 103 N

Question 4. Two trolleys m1 = 0.2 kg and m2 = 0.3 kg are held against a compressed spring on a smooth horizontal table, as shown in the figure below.

 

Force And Laws Of Motion Long Answer Question

When the spring is cut, m1 is found to move towards right with an initial acceleration of 6 ms-2. Find the initial acceleration of m2.

Answer.

Force acting on m2 is equal and opposite to that force acting on m1.

∴ F2 = − F1

i.e., m2a2 = − m1a1

0.3 kg × a2 = − 0.2 kg × 6 ms-2

∴ a2 = \(-\frac{0.2 \times 6}{0.3}\)

∴ a2 = −4 ms-2

Negative sign implies that it is in opposite direction of that of m1 i.e., a1.

∴ Acceleration of m2 = 4 ms-2 towards left.

Question 5. A toy rocket weighing 500 g is standing vertically on ground. How much force should act on it so that it starts ascending with a uniform acceleration of 5  ms-2? (Take g = 10 ms-2)
Answer.

To go up, the upward force acting on the rocket must be greater than its weight ‘mg’.

Net force acting on rocket

Fnet = Fup − mg (1)

By Newton’s II law,

Fnet = ma (2)

From equations (1) and (2),

Fup − mg = ma

∴ Fup = mg + ma

= m (g + a)

= 0.5 kg (10 + 5)ms-2

= 0.5 × 15 kg ms-2

= 7.5 N

Question 6. Name the physical quantity, which is equal to the rate of change of momentum and derive a relation between them. A car weighing 2000kg moving with a velocity of 90 km/h retards uniformly and comes to rest in 20 s. Calculate the force required to stop the car.
Answer.

Force is the physical quantity, which is equal to rate of change of momentum. Consider an object of mass m moving with an initial velocity u. When a force is acting on the object for a certain time t its velocity changes to v. According to Newton’s second law of motion,

F = ma (1)

Where a is the acceleration produced on the object due to the application of force. Acceleration = Change in velocity/Time

a = v – u/t

Substituting the value of a in equation (1) we get,

F = m (v – u/t)

F = (mv – mu)/t

Since v as is the final velocity, mv will be the final momentum and u is the initial velocity, mu will be the initial momentum.

F = Final momentum – Initial momentum/Time

Force = Rate of change of momentum

Mass of the car = 2000 kg

Initial velocity (u) = 90 km/h

= \(90 \times \frac{5}{18}\)

Initial velocity (u) = 25 m/s

Final velocity (v) = 0

Time (t) = 20 s

Force required to stop the car = m (v – u)/t

= 2000 (0 – 25)/20

= \(-2000 \times \frac{25}{20}\)

= –100 × 25

Force required to stop the car = –2500 N.

Question 7. A research balloon of total mass M is descending vertically with downward acceleration a as shown in the below figure. How much ballast must be thrown from the car to give the ­balloon an upward acceleration a, assuming that the upward lift of the air on the balloon does not change?
Answer.

Initially, let us consider the mass of the system (balloon) as M.

So the force ( f ) acting on the system having downward acceleration, a will be,

f = –Ma (1)

And the weight of the system (W) will be,

W = Mg (2)

Where g is the free fall acceleration of the system.

Therefore, the upward force acting on the system (F) will be,

F = W + f = Mg + (–Ma)

F = Mg – Ma (3)

Again let us consider m as the mass of single ballast which is thrown from the balloon.

Now mass of the system is, M – m.

So the force ( f1) acting on the system having an upward acceleration a and mass M – m will be,

f1 = (M – m) a (4)

And the weight of the system (W1) having mass M – m will be,

W1 = (M – m) g (5)

Where, g is the free fall acceleration of the system.

To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f1) acting on the system having upward acceleration a and mass, M-m and the weight of the system (W1) having mass M – m.

So the equation will be,

f1 + W1 = F (6)

(M – m) a + (M – m) g = Mg – Ma

Ma – ma + Mg – mg = Mg – Ma

ma + mg = Ma + Mg – Mg + Ma

m (a + g) = 2Ma

m = 2Ma/(a + g) (7)

From equation (7) we observed that, 2Ma/(a + g) mass of ballast must be thrown from the carto give the balloon an upward acceleration a.

Question 8. A meteor of mass 0.25 kg is falling vertically through Earth’s atmosphere with an acceleration of 9.2 m/s2. In addition to gravity, a vertical retarding force (due to frictional drag of the atmosphere) acts on the meteor as shown in the below figure. What is the magnitude of this retarding force?Answer.

Mass of the meteor, m = 0.25 kg

Acceleration of the meteor, a = 9.2 m/s2

The net force exerted (Fnet) on the meteor will be,

Fnet = ma

= (0.25 kg) (9.2 m/s2)

= (2.30 kg. m/s2) (1 N/1 kg. m/s2)

= 2.30 N (1)

If g (g = 9.80 m/s2) is the free fall acceleration of meteor, then the weight of the meteor (W) will be,

W = mg = (0.25 kg) (9.80 m/s2)

= (2.45 kg. m/s2) (1 N/ 1 kg. m/s2)

= 2.45 N (2)

The vertical retarding force would be equal to the net force exerted on the meteor (Fnet) minus weight of the meteor (W).

So, vertical retarding force = Fnet – W (3)

Substituting the value of Fnet and W in equation (3), the vertical retarding force will be,

Vertical retarding force = Fnet – W

= 2.30 N – 2.45 N = –0.15 N (4)

From equation (4) we observed that, magnitude of the vertical retarding force would be, –0.15N.

Question 9. Workers are loading equipment into a freight elevator at the top floor of a building. However, they overloaded the elevator and the worn cable snaps. The mass of the loaded elevator at the time of the accident is 1600 kg. As the elevator falls, the guide rails exert a constant retarding force of 3700 N on the elevator. At what speed does the elevator hit the bottom of the shaft 72 m below?
Answer.

To obtain the weight of the elevator W, substitute 1600 kg for mass of the elevator m and 9.81 m/s2 for free fall acceleration g in the equation W = (m) (g).

W = (m) (g)

= (1600 kg) (9.81 m/s2)

= 15680 kg, m/s2

= (15680 kg, m/s2)(1 N/1 kg ⋅ m/s2)

= 15680 N

The magnitude of the net force F will be,

F = W – R

To obtain the magnitude of the net force F, substitute 15680 N for W and 3700 N for retarding force R in the equation F = W – R,

F = W – R

= (15680 N) – (3700 N)

= 11980 N

Rounding off to two significant figures, the magnitude of the net force F will be 12000 N.

To obtain the acceleration a, substitute 12000 N for F and 1600 kg for m in the equation a = F/m, we get,

a = F/m

= 12000 N/1600 kg

= (7.5 N/kg) (1 kg. m/s2/1 N)

= 7.5 m/s2

To obtain the time t to fall, substitute –72 m for y and –7.5 m/s2 for a in the equation t = \(\sqrt{2} y / a\)

t = \(\sqrt{2} y / a\)

= \(\sqrt{2}(-72 \mathrm{~m}) /\left(-7.5 \mathrm{~m} / \mathrm{s}^2\right)\)

= 14 s

To obtain the final speed v at which the elevator hits the bottom of the shaft 72 m below, substitute 7.5 m/s2 (only magnitude of a) for a and 4.4 s for t in the equation v = at, we get,

v = at

= (7.5 m/s2)(14 s)

= 105 m/s

From the above observation we conclude that, the speed at which the elevator hits the bottom of the shaft 72 m below would be 105 m/s.

Question 10. A force produces an acceleration of 2.0 m s-2 in a body A and 5.0 m s-2 in another body B. Find the ratio of the masses of A to the mass of B.
Answer.

Let the mass of the body A be mAand that of the body B be mB. We have

F = (mA) (2.0 m s-2)

F = (mB) (5.0 m s-2)

Thus

(mA)(2.0 m s-2) = (mB) (5.0 m s-2)

Or \(\frac{m_A}{m_B}=\frac{5.0}{2.0}=2.5\)

Question 11. A man weighing 60 kg runs along the rails with the velocity of 18 km hr-1 and jumps into a car of mass 1 quintal standing on the rails. Calculate the velocity with which the car will start travelling along the rails.
Answer.

Here, mass of man, m1 = 60 kg.

Initial velocity of man,

u1= 18 km/1 h = 18 × 1000 m/60 × 60 s = 5 m s-1.

Mass of car, m2 = 1 quintal = 100 kg.

Initial velocity of car, u2 = 0.

After a man jumps into a car, let their common velocity be v

Applying the principle of conservation of momentum

Total momentum after jump = Total momentum before jump

(m1+ m2) v = m1u1 + m2u2

(60 + 100) v = 60 × 5 + 100 × 0

Or v = \(\frac{300}{160}=1.88 \mathrm{~m} \mathrm{~s}^{-1}\)

Question 12. A77-kg person is parachuting and experiencing a downward acceleration of 2.5  m/s1 shortly after opening the parachute. The mass of the parachute is 5.2 kg. (1)  Find the upward force exerted on the parachute by the air. (2) Calculate the downward force exerted by the person on the parachute.
Answer.

(1) The net force Fnet on the system is equal to the sum of force exerted by the person and force exerted by the parachute.

So, Fnet = (mpe+ mpa) (a)

Here, mpe is the mass of person, mpa is the mass of parachute and a is the downward acceleration.

To obtain the net force Fnet on the system, substitute 77 kg for mpe , 5.2 kg for mpa and –2.5 m/s2 for a in the equation Fnet = (mpe +mpa) (a),

Fnet = (mpe+ mpa) (a)

= (77 kg + 5.2 kg) (–2.5 m/s2)

= (–210 kg,m/s2) (1 N/1 kg,m/s2)

= –210 N

The weight W of the system will be,

W = (mpe+ mpa) (g)

To obtain the weight W of the system, substitute 77 kg for mpe, 5.2 kg for mpa and 9.81 m/s2 for free fall acceleration g in the equation W = (mpe+ mpa) (g),

W = (mpe+ mpa) (g)

= (77 kg + 5.2 kg) (9.81 m/s2)

= (810 kg, m/s2) (1 N/1 kg, m/s2)

= 810 N

If P is the upward force of the air on the system (parachute) then,

P = Fnet+ W = (–210 N) + (810 N) = 600 N

From the above observation we conclude that, the upward force exerted on the parachute by the air would be 600 N.

(2) The net force Fnet on the parachute will be

Fnet = mpa a

Here, mpa is the mass of parachute and a is the downward acceleration.

To obtain the net force Fnet on the parachute, substitute 5.2 kg for mpa and –2.5 m/s2 for a in the equation Fnet = mpa a,

Fnet = mpa a = (5.2 kg) (–2.5 m/s2)

= (–13 kg m/s2) (1 N/1 kg m/s2) = –13 N

The weight W of the parachute will be,

W = (mpa) (g)

To obtain the weight W of the system, substitute 5.2 kg for mpa and 9.81 m/s2 for free fall acceleration g in the equation W = (mpa) (g),

W = (mpa)(g)

= (5.2 kg) (9.81 m/s2)

= (51 kg, m/s2)(1 N/1 kg, m/s2)

= 51 N

If D is the downward force of the person on the parachute then,

D = –Fnet – W + P

= –(–13 N) – (51 N) + (600 N) = 560 N

Question 13. Are search balloon of total mass Misdescending vertically with downward acceleration a as shown in below figure. How much ballast must be thrown from the car to give the balloon an upward acceleration a, assuming that the upward lift of the air on the balloon does not change?
Answer.

Let us consider initially mass of the system (balloon) be, M.

So the force ( f ) acting on the system having downward acceleration, a will be,

f = –Ma (1)

And the weight of the system (W) will be,

W = Mg (2)

Where, g is the free fall acceleration of the system.

Therefore the upward force acting on the system (F) will be,

F = W + f = Mg + (–Ma)

F = Mg – Ma (3)

Again let us consider m as the mass of single ballast which is thrown from the balloon.

Now mass of the system is, M-m.

So the force ( f1) acting on the system having upward acceleration, a and mass, M-m will be,

f1 = (M – m) a (4)

And the weight of the system (W1) having mass M-m will be,

W1 = (M-m) g (5)

Where, g is the free fall acceleration of the system.

To give the balloon an upward acceleration a, the upward force (F) must be equal to the addition of the force (f1) acting on the system having upward acceleration, a and mass, M-m and the weight of the system (W1) having mass M-m.

So the equation will be,

f1 + W1 = F (6)

(M – m) a + (M – m) g = Mg – Ma

Ma – ma +Mg – mg = Mg – Ma

ma + mg = Ma + Mg – Mg + Ma

m (a + g) = 2Ma

m = 2Ma/(a + g) (7)

From equation (7) we observed that, 2Ma/(a + g) mass of ballast must be thrown from the carto give the balloon an upward acceleration a.

Question 14. A child’s toy consists of three cars that are pulled in tandem on small frictionless rollers as shown in below figure. The cars have masses m1 = 3.1 kg, m2 = 2.4 kg, and m3 = 1.2 kg. If they are pulled to the right with a horizontal force P = 6.5 N, find

  1. the acceleration of the system,
  2. the force exerted by the second car on the third car, and
  3. the force exerted by the first car on the second car.

Answer.

Given Data:

Mass of the first car, m1 = 3.1 kg

Mass of the second car, m2 = 2.4 kg

Mass of the third car, m3 = 1.2 kg

Horizontal force on the car, P = 6.5 N

Acceleration of a body is equal to the force exerted on the body divided by mass of the body.

(1) The acceleration of the system is equal to the total horizontal force acting on the system divided by total mass of the system.

So,

a = F/ (m1+ m2+ m2)

= (6.5 N)/(3.1 kg + 2.4 kg + 1.2 kg)

= (6.5 N)/(6.7 kg)

= (0.97 N/kg)(1 kg. m/s2/1N)

= 0.97 m/s2

Therefore, the acceleration of the system would be 0.97 m/s2.

(2) Force exerted by the second car on the third car would be,

F32 = (Mass of third car)(Total acceleration of the system)

= m3 a

= (1.2 kg)(0.97 m/s2)

= (1.164 kg. m/s2)(1 N/1 kg. m/s2)

= 1.164 N

Rounding off to two significant figures, the force exerted by the second car on the third car would be 1.2 N.

(3) Force exerted by the first car on the second car would be,

F32 = (Sum of mass of second car and third car)(Total acceleration of the system)

= (m2+ m3) a

= (2.4 kg +1.2 kg)(0.97 m/s2)

= (3.492 kg. m/s2)(1 N/1 kg. m/s2)

= 3.492 N

Rounding off to two significant figures, the force exerted by the first car on the second car would be 3.5 N.

Force And Laws Of Motion Short Answer Question And Answers

Force And Laws Of Motion Short Answer Type Question And Answers

Question 1. A Force of 8000 N acts on a truck, initially at rest for 10 s. Find

  1. velocity of the truck
  2. distance traveled by the truck
  3. Mass of the truck = 2000 kg

Answer.

∵ F = ma

8000 = 2000a

∴ a = \(\frac{8000}{2000}=4 \mathrm{~ms}^{-2}\)

(1) ∵ v = u + at

∴ v = 0 + 4 × 10 = 40 ms-1

(2) ∵ s = \(u t+\frac{1}{2} a t^2\)

= \(0+\frac{1}{2} \times 4 \times(10)^2\)

∴ S = 200 m.

Newtons Law Of Motion

Question 2. A ball weighing 500 g is lying on the floor. A player exerts a force of 250 N on the ball for a small time interval of 0.02 s. Find the velocity acquired by the ball.
Answer.

Given u = 0

m = 500 g = 0.5 kg

F = 250 N

t = 0.02 s

v = ?

We know that

F = ma = \(m\left(\frac{v-u}{t}\right)\)

⇒ 250 = \(0.5\left(\frac{v-0}{0.02}\right)\)

∴ v = \(\frac{250 \times 0.02}{0.5}=\frac{250 \times 2}{50}=10 \mathrm{~m} / \mathrm{s}\)

Question 3. A book weighing 800 g is kept on a horizontal table. How much force is exerted by the table on the book? (Take g = 10 ms−2)
Answer.

NEET Foundation Physics Force And Laws Of Motion Force Equal To Weight

The book exerts a force equal to its weight on the table.

This force is called ‘action’.

∴ Action = W = mg

= 0.8 × 10

= 8 N

According to Newton’s III law of motion, the table exerts an equal and opposite force on the book. It is called a reaction.

∴ R = Action

= 8 N

It acts vertically upwards.

Question 4. The figure below show a trolley of mass 300 g lying on a smooth horizontal surface. A boy pulls it with a force 1.2 N. If a constant frictional force of 0.3 N acts on it, how much distance it will travel in 0.8 s?

NEET Foundation Physics Force And Laws Of Motion Frictional Force 1

Answer.

Since the two forces are acting in opposite directions, the net force acting on the trolley

= F1 − F2

= 1.2 N − 0.3 N

= 0.9 N

We know that Newton’s II law of motion gives us

F = ma

0.9 N = 0.3 kg × a

∴ a = \(\frac{0.9 \mathrm{~N}}{0.3 \mathrm{~kg}}\)

= 3 ms-2

∵ S = \(u t+\frac{1}{2} a t^2\)

= \(0+\frac{1}{2} \times 3 \times(0.8)^2\)

= \(\frac{1}{2} \times 3 \times 0.8 \times 0.8\)

∴ S = 0.96 m

Question 5. In a cathode ray tube shown in the figure below, an electron starts from cathode c. What will be its speed on reaching the anode, if an electric force of N acts on it?

  1. (Mass of electron = 9 × 10-31 kg)
  2. (Distance between cathode C and anode A is cm)

Answer.

By Newton’s II law of motion,

F = ma

1.8 × 10 = (9 × 10-31 kg) a

⇒ 9 × 10-31 a = 1.8 × 10-15

∴ a = \(\frac{1.8 \times 10^{-15}}{\left(9 \times 10^{-31}\right)}\)

∴ a = 2 × 1015 ms-2

We know that,

v2 = u + 2as

0 = 2 × 2 × 1015 × 0.4

⇒ v2 = 1.6 × 1015

⇒ v2 = 16 × 1014

∴ v = 4 × 107 ms-1

Newton’s Third Law of Motion

Newton’s Third Law of Motion

Newton’s second law tells us the magnitude of acceleration which is produced by a force when applied on an object.

These two laws do not tell how the force acts on the object. To answer this question let us study Newton’s third law of motion, which states:

In other words, if an object exerts a force on another object, then in return it will exert a force of the same magnitude on it in the opposite direction (Fig. 2.9).

Example 1: A boy throws a ball on the ground. In this case ball exerts a force on the ground. As a result, the ground exerts an equal and opposite force on the ball which makes the ball bounce.

Example 2: Let us consider a day-to-day example. While walking on the floor we exert a force by our feet to push the ground backward, in return the ground exerts a force of the same magnitude on our feet forward which makes us move forward. Here our force is called action and the force by ground is called reaction.

The linear momentum of an object is the product of its mass and velocity. To understand it better, let us elaborate on it.

Newton's Third Law of Motion

Illustration

Let two objects with different masses move with the same velocity. When they are brought to a stop, the one with more mass will require more force to stop as compared to the lighter one.

Now consider two moving objects of the same masses but different velocities. When they are brought to a stop the one with higher velocity will require more force to stop as compared to the slower one.

This concludes that the force needed to stop an object at a particular time depends on both the product of the mass and velocity of a moving object. The product of mass and velocity of a moving body is known as linear momentum or momentum. It is represented by a symbol p.

Let the mass of an object be m. It is moving with velocity v, then linear momentum is

p = mv

Linear momentum is a vector quantity as it has the direction of the motion of the body.

Unit

As we know

Unit of momentum = unit of mass × unit of velocity = kg × m/s

S.I. unit of momentum is kg m/s

C.G.S. unit of momentum is g cm/s

Change in Momentum

The change in momentum equals the mass times the change in velocity.

∆p = m (∆v)

Where ∆ denotes a small change in the quantity.

The change in product mv can be due to the change in mass m change in velocity v or change in both m and v.

The symbol ∆ before mv denotes a small change in the product of m and v. If mass m does not change, then the product mv will change only due to the change in v and so m can be written before the symbol ∆. The quantity ∆v now represents a small change in velocity only.

The velocity of a moving object is changed when a force is applied to it. Thus, a change in velocity results in a change in momentum.

Equation

Let F = force applied on an object

m = Mass of the object

t = time

Suppose its velocity changes from u to v in time t

The initial momentum of an object = mu

Final momentum of an object = mv

Change in momentum in t second = mv − mu

= m (v − u)

\(\text { Rate of change of momentum }=\frac{\text { change in momentum }}{\text { time }}\)

= \(\frac{m(v-u)}{t}\) (1)

\(\text { Acceleration } a=\frac{\text { change in velocity }}{\text { time }}\)

i.e. a = \(\frac{v-u}{t}\)

Therefore, in equation (1) the omes rate of change of momentum = mass × acceleration = m.a

Newton’s Second Law of Motion

Newton’s Second Law of Motion

Newton’s first law of motion implies that force produces acceleration in a body. The ­second law of Newton relates force to measurable quantities like acceleration and mass. The rate of change of momentum of a body is directly proportional to the net force acting on it and takes place in the direction of the net force.

Illustration 1: Push a tennis ball. It will produce a small acceleration and require a small velocity in a certain time. If we push the same tennis ball a little hard then it will produce a large acceleration and also it acquires a large velocity in the same interval of time. Therefore, the magnitude of two forces can be compared by measuring the accelerations produced by them at the same time.

Graphical Representation

This experiment by Newton concludes that

If force F1 produces an acceleration of 10 m/s

And force F2 produces acceleration of 20 m/s on the same object,

Magnitude of force F2 = 2 magnitude of F1

Illustration 2: Take a tennis ball and a football which are initially at rest. Apply the same change of velocity in the same time to both balls, we need to apply more force on the football as compared to the tennis ball.

Therefore, the force needed to produce the same acceleration in both balls if masses are ­different. Force for football is more than the tennis ball.

Example: If force F is required to produce an acceleration of 5 m/s2 in an object of mass 2 kg, then to produce the same acceleration in another object of mass 4 kg, a force of ­magnitude 2F is required.

Graphical Representation

F ∝ m if acceleration remains same

After combining both the equations, the new relation is

F ∝ m a

F = Kma where K is a constant

The unit of force is chosen by taking K = 1 when m = 1 and a = 1. Thus, the amount of force that when applied on a body of unit mass produces a unit acceleration in the body is taken as one unit of force. With the above unit of force, the equation is as follows.

Newton's Second Law of Motion

Mathematical Expression of Newton’s Second Law of Motion

Force = Mass × acceleration

F = ma

Unit of Force

As we know,

F = ma.

Therefore,

F = kg m/s2.

S.I. Unit of Force = Newton

One newton is the force which when acts on an object of mass 1 kg, produces an ­acceleration of 1 m/s2

i.e., 1 Newton = 1 kg × 1 m/s2

Newton is represented as N.

C.G.S. Unit of Force = Dyne

One dyne is the force which when acts on an object of mass 1 g, produces an acceleration of 1 cm/s2

i.e., 1 Dyne = 1 g × 1 cm/s2.

Relationship Between Newton and Dyne

1 Newton = 1 kg × 1 m/s2

= 1000 g × 100 cm/s2

= 105 g cm/s2

= 105 Dyne

Newton’s Second Law of Motion in Terms of Rate of Change of Momentum

When a force is applied on a body, it produces an acceleration in the body, because of which, the velocity and thereafter the momentum of the body changes.

Therefore, \(\frac{\Delta p}{\Delta t}=m a\), if mass remains constant

Newton’s second law of motion states

F = m a

Therefore, force = rate of change of momentum

F = \(\frac{\Delta p}{\Delta t}\)

= \(\frac{m \Delta v}{\Delta t}=ma\), if mass remains constant

The rate of change of momentum of an object is directly proportional to the force applied on it and the change in momentum takes place in the direction in which the force is applied.

As we have studied F = \(\frac{\Delta p}{\Delta t}=\frac{\Delta(m v)}{\Delta t}\)

This equation shows that momentum changes because of the change in mass or change in velocity or because of the change in both mass and velocity.

It is seen that the mass of an object increases with the increase in velocity when the velocity v of an object is comparable with the speed of light c (3 × 108 m/s), but at velocities v << c, the change in mass is not perceptible.

As such velocities (v << c), mass m can be considered to be a constant.

Then, Newton’s second law of motion is described as

F = \(\frac{\Delta v}{\Delta t}=m a\)

Thus, relation F = \(m \frac{\Delta v}{\Delta t}\) = ma holds in two conditions:

  • When velocities are much smaller than the velocity of light.
  • When mass remains constant.

Relationship between First Law of Motion and Second Law of Motion

From Newton’s second law F = m a

If F = 0 then a = 0

It means when there is no force applied then acceleration will be zero i.e., if the object is at rest, it will continue to be at rest and if it is moving, it will remain moving in the same direction with the same speed. This is what Newton’s first law of motion states.

Example: A boy throws a bottle on the floor and it breaks, but when he throws the same bottle on the carpet it doesn’t break. The reason is when the bottle falls on the hard floor, it comes to rest in a very short time due to which the floor exerts a large force on the bottle, and as a result it breaks.

But when the bottle falls on a carpet the time duration in which the bottle comes to rest increases, therefore carpet exerts less force on the bottle and as a result, it doesn’t break.

Force And Laws Of Motion Very Short Answer Questions

Force And Laws Of Motion Master Your Test Question And Answers

Question 1. What are the effects of force?
Answer: A stationary object will start moving once force is applied. Force can stop a moving object and can change the speed of the moving object. It can also change the direction of the moving object. It can also change the shape and size of an object.

Question 2. Define contact forces.
Answer: Contact force represents the result of physical contact between two objects. Ex-Normal force, frictional force etc.

Question 3. What are field forces?
Answer: If an object attracts or repels another object from a distance without being in contact is called field forces. Example: Gravitational force, electrostatic force etc.

Question 4. Differentiate balanced and unbalanced force.
Answer: A balanced force is the one in which after applying force on an object from different sides, it is still at rest. Unbalanced force is that force which produces motion in the object. In this case, the object will move in the direction of greater force.

Question 5. Calculate the net force applied on the rope in a tug of war where rope does not move in any direction.
Answer: If the rope does not move in any direction, it means balanced force is applied on it i.e., equal and opposite forces are applied from both the direction.
Therefore, the net force in this scenario will be zero.

Question 7. Define inertia. How is inertia related to mass?
Answer: Inertia is the property of inability of an object to change its state of rest or motion. Inertia of an object is directly proportional to the mass of the object.

Question 8. What are the different kinds of inertia?
Answer: There are three different kinds of inertia:

  1. Inertia of rest.
  2. Inertia of motion.
  3. Inertia of direction.

Question 9. The mud from the tyres of the automobiles flies off tangentially. Why?
Answer: The mud from the tyres of the automobiles flies off tangentially due to inertia of direction. Because of the absence of external force on it, the mud, when leave the rotating tyre, follow a tangential linear motion. This is the reason why mudguards are placed in the vehicles.

Question 10. Give Reason: An athlete runs for some distance before taking a long jump.
Answer: This is because of inertia of motion. When the athlete jumps and leaves the ground, his body inclines to move with the same velocity which he had before leaving the ground. And therefore, the length of the jump will hinge on the velocity of the athlete at the time when he jumps.

Question 11. Define linear momentum. Write its SI unit.
Answer: Linear momentum of an object is equal to the product of mass and velocity. The SI unit of linear momentum is kgm/s.

Question 12. What is relation between force, mass and acceleration?
Answer: Force applied on a body is equal to the product of mass and acceleration produced in the body.

Question 13. What is the SI unit of force?
Answer: SI unit of force is Newton (N) or kgm/s2.

Question 14. What is relation between two units Newton and Dyne?
Answer: 1 Newton = 105 Dyne

Question 15. Calculate the force needed to speed up a car with a rate of 5 ms-2, if the mass of the car is 1000 kg.
Answer: Acceleration (a) = 5 m/s2 and

Mass (m) = 1000 kg,

Therefore, Force (F) =?

We know that,

F = m × a

= 1000 kg × 5 m/s2

= 5000 kg m/s2

Therefore, required Force = 5000 kg m/s2 or 5000 N

Question 16. An object requires the force of 100 N to achieve the acceleration ‘a’. If the mass of the object is 500 kg what will be the value of ‘a’?
Answer: According to the question,

Mass (m) = 500 kg,

Force (F) = 100 N,

Acceleration (a) =?

We know that,

Force=Mass×Acceleration

or F = m × a

Therefore,

100 N = 500 kg × a

a =100N/500kg

a =100kgms-2/500kg

a = 0.2 ms-2+

Thus acceleration of the vehicle = 0.2 ms-2

Question 17. Define force with respect to momentum.
Answer: Force is defined as the rate of change of momentum produced in an object.

Question 18. Define conservation of momentum.
Answer: Momentum of a system is constant if there are no external force acting on the system.

Question 19. A body of mass 15 kg moving with a velocity of 15 m/s is brought to rest in 10 seconds. Find the change in momentum and also the retarding force.
Answer: Mass of the body (m) = 15 kg

Initial Velocity (u) = 15 m/s

Final Velocity (v) = 0

Time (t) = 10 s

Change in momentum = mv – mu

= 15 × 0 – 15 × 15 = 0 – 225

= –225 kg m/s

Force applied is equal to rate of change in momentum.

F = mv – mu/t

= –225/10

= –22.5 N

Retarding Force = 22.5 N

Question 20. The speed of a car weighing 1000 kg increases from 36 km/h to 108 km/h. Calculate the change in momentum.
Answer: Mass of the Car (m) = 1000 kg

Initial Velocity (u) = 36 km/h (1 km/h = 5/18m/s)

= 36 × 5/18 m/s

= 10 m/s

Final Velocity (v) = 108 km/h

= 108 × 5/18 m/s

= 6 × 5

= 30 m/s

Change in momentum = mv – mu

= m(v – u)

= 1000 (30 – 10)

= 1000 × 20

Change in momentum = 20,000 N s

Force And Laws Of Motion Newtons law of motion

Force And Laws Of Motion Multiple Choice Questions

Force And Laws Of Motion Multiple Choice Questions

Direction: Choose the correct option for each question. There is only one correct response for each question.

Force And Laws Of Motion

Question 1. Choose the S.I. unit of force

  1. cm/s
  2. Newton
  3. m/s
  4. km/s

Answer. 2. Newton

Question 2. Which among the following are the types of inertia?

  1. Rest
  2. Motion
  3. Direction
  4. Volume

Answer. 4. Volume

Question 3. Which law of Newton states that ‘Every action has an equal and opposite reaction’?

  1. The first law of motion
  2. The second law of motion
  3. Third law of motion
  4. The fourth law of motion

Answer. 3. Third law of motion

Question 4. If a train starts suddenly in the forward direction, passengers standing on the train will fall ______.

  1. Backwards
  2. Forward
  3. Sideways
  4. Downwards

Answer. 1. Backwards

Question 5. The rate of change of momentum is equal to ______.

  1. Work done
  2. Acceleration
  3. Speed
  4. Force

Answer. 4. Force

Question 6. kg.m/s2 is the unit of _____.

  1. Force
  2.  Speed
  3. Velocity
  4. Acceleration

Answer. 1. Force

Question 7. A car is running with a force of 5 N. The weight of the car is 0.5 kg. What is the acceleration?

  1. 5
  2. 10
  3. 15
  4. 20

Answer. 2. 10

Question 8. The principle on which the jet engine works is _______.

  1. Conservation of mass
  2. Conservation of momentum
  3. Newton’s first law of motion
  4. Newton’s Second law of motion

Answer. 2. Conservation of momentum

Question 9. Inertia is a property of an object by which object is _______.

  1. Not able to change its state of rest or uniform motion in a straight line
  2. Not able to change its state of rest
  3. Not able to change its direction of motion
  4. Not able to change its state of uniform motion in a straight line

Answer. 1. Not able to change its state of rest or uniform motion in a straight line

Question 10. When a net force acts on an object, the object will be accelerated in the direction of force with acceleration proportional to ______ of the object.

  1. Speed
  2. Velocity
  3. Acceleration
  4. Force

Answer. 4. Force

Question 11. When an object undergoes ­acceleration, ___________.

  1. Speed decreases
  2. Speed increases
  3. Force is acting on it
  4. An object falls on the ground

Answer. 3. Force is acting on it

Question 12. Force can be defined from ______.

  1. Newton’s first law of motion
  2. Newton’s second law of motion
  3. Newton’s third law of motion
  4. Newton’s fourth law of motion

Answer. 1. Newton’s first law of motion

Question 13. If A and B are two objects of masses 10 kg and 20 kg respectively, then which among the following statements is correct?

  1. A has more inertia than B
  2. B has more inertia than A
  3. A and B have the same inertia
  4. None of the Above

Answer. 2. B has more inertia than A

Question 14. An object moves with constant acceleration if constant ______ is applied to the object.

  1. Speed
  2. Velocity
  3. Force
  4. Acceleration

Answer. 3. Force

Question 15. One Newton is the force equal to ______.

  1. Gravity on 1 kg object
  2. Mass of 1 kg and acceleration of 1 m/s2
  3. Mass of 1 g and acceleration of 1 cm/s2
  4. Gravity on 1 g object

Answer. 2. Mass of 1 kg and acceleration of 1 m/s2

Question 16. A man tosses a coin in a moving train and the coin falls behind him. It shows that the train is ______.

  1. Accelerated
  2. Retarded
  3. Uniform
  4. Moving in a circular track

Answer. 1. Accelerated

Question 17. A car of mass 20 kg is moving with an acceleration of 2 m/s2. What is the rate of change of momentum?

  1. 10
  2. 20
  3. 30
  4. 40

Answer. 4. 40

Question 18. An object is moving in a straight line in an acceleration motion. Choose the incorrect statement related to the statement.

  1. Velocity keeps changing
  2. Speed keeps changing
  3. The object goes away from the earth
  4. Force is always acting on an object

Answer. 3. Object goes away from the earth

Question 19. Rocket works on the principle of conservation of _______.

  1. Speed
  2. Velocity
  3. Energy
  4. Momentum

Answer. 4. Momentum

Question 20. As per the third law of motion, action, and reaction _______.

  1. Always act on different objects in the opposite direction
  2. Always act on the same body in the opposite direction
  3. Always act on the same body in the same direction
  4. Always act on the same body

Answer. 1. Always act on different objects in the opposite direction

Question 21. A goalkeeper holding the football pulls his hand backward because it helps the goalkeeper to ______.

  1. Increase rate of change of momentum
  2. Decrease rate of change of momentum
  3. Exert larger force on football
  4. Reduce the force exerted by the ball on the hand

Answer. 4. Reduce force exerted by the ball on the hand

Question 22. An athlete before doing a long jump, runs for some distance because ______.

  1. Helps in applying a large force
  2. Action and reaction force increases
  3. Gives him a larger inertia of motion
  4. Gains energy

Answer. 3. Gives him larger inertia of motion

Question 23. While riding a horse, the rider falls back when the horse starts because the ______.

  1. The rider is afraid of falling back
  2. The rider is taken back
  3. The inertia of rest keeps the upper part of the body at rest while the lower part of the body moves forward with the horse
  4. The inertia of rest keeps the lower part of the body at rest while the upper part of the body moves forward with the horse

Answer. 3. Inertia of rest keeps the upper part of the body at rest while the lower part of the body moves forward with the horse

Question 24. The property of inertia is more in ______.

  1. Toy car
  2. Car
  3. Truck
  4. Horse cart

Answer. 4. Horse cart

Question 25. A boy has a tennis ball and a football at rest. To move it, which among the following is correct

  1. The same force is required for both
  2. More force is required for a football ball
  3. More force is required for a tennis ball
  4. No force is required to move the balls

Answer. 2. More force is required for a football ball

Question 26. Which among the following is the equation for the second law of motion?

  1. F = ma
  2. F = mv
  3. F = av
  4. F = mav

Answer. 1. F = ma

Question 27. When a running car stops suddenly, the passenger tends to fall ______.

  1. Forward
  2. Backward
  3. Sideways
  4. Downside

Answer. 1. Forward

Question 28. ______ is an external cause that can move a still object to move and can stop a moving object.

  1. Force
  2. Speed
  3. Velocity
  4. Acceleration

Answer. 1. Force

Question 29. Force is a ______ quantity

  1. Scalar
  2. Vector
  3. Both
  4. None of the above

Answer. 2. Vector

Question 30. The inertia of an object depends on ______.

  1. Force
  2. Speed
  3. Velocity
  4. Mass

Answer. 4. Mass

Question 31. On applying equal and opposite forces on an object, the object will.

  1. Move left
  2. Move right
  3. Doesn’t move
  4. Move up

Answer. Doesn’t move

Question 32. The unit of Linear momentum is ______.

  1. Ns
  2. N/s
  3. kg/s
  4. g/s

Answer. 1. Ns

Question 33. The correct form of Newton’s second law of motion is ______.

  1. F = \(\frac{\Delta p}{\Delta t}\)
  2. F = \(m\left(\frac{\Delta v}{\Delta t}\right)\)
  3. \(F=v\left(\frac{\Delta m}{\Delta t}\right)\)
  4. F = mv

Answer. 1. F = \(\frac{\Delta p}{\Delta t}\)

Question 34. The C.G.S. unit of force is ______.

  1. Newton
  2. Dyne
  3. kg/m
  4. m/s

Answer. 2. Dyne

Question 35. Which law states that ’When an object is at rest it will remain at rest and if it is ­moving, it will continue to move in the same direction unless an external force is applied on it‘.

  1. Newton’s first law of motion
  2. Newton’s second law of motion
  3. Newton’s third law of motion
  4. Newton’s fourth law of motion

Answer. 1. Newton’s first law of motion

Question 36. Newton’s first law of motion is discussed through the definition of ______.

  1. Inertia
  2. Force
  3. Speed
  4. Acceleration

Answer. 

2. Force

3. Speed

Question 37. Which among the following are the units of Force?

  1. kg/m
  2. Newton
  3. Dyne
  4. m/s

Answer. 

3. Newton

4. Dyne

Question 38. Choose the correct statement:

  1. If an object is at rest it will remain at rest until an external force is applied.
  2. If an object is moving, it will continue to move until an external force is applied.
  3. If an object is at rest it will remain at rest even after applying external force.
  4. If an object is moving it will continue moving even after applying external force.

Answer.

1. If an object is at rest it will remain at rest until an external force is applied.

4. If an object is moving it will continue moving even after applying external force.

Question 39. Newton’s third law of motion states that ‘For every action, there is an equal and ______ reaction’.

  1. Equal
  2. less
  3. Greater
  4. Opposite

Answer.

1. Equal

4. Opposite

Question 40. Choose the incorrect equation:

  1. 1 N = 1 kg.m/s2
  2. F = m a
  3. a = \(\frac{v}{t}\)
  4. a = v – u

Answer. 

3. a = \(\frac{v}{t}\)

4. a = v – u

Question 41. Choose the correct statement:

  1. The inertia of an object tends to cause the object to resist any change in its state of motion.
  2. Inertia is a property of an object by the virtue of which an object is unable to change by itself the state of rest or uniform motion in a straight line.
  3. Inertia is a property of an object by the virtue of which an object can change by itself the state of rest or uniform motion in a straight line.
  4. The inertia of an object tends to cause the object to resist no change in its state of motion.

Answer.

  1. The inertia of an object tends to cause the object to resist any change in its state of motion.
  2. Inertia is a property of an object by the virtue of which an object is unable to change by itself the state of rest or uniform motion in a straight line.

Question 42. Choose the correct statement:

  1. When a train starts suddenly, passengers standing on the train tend to fall backward.
  2. When a train starts suddenly, passengers standing on the train tend to fall forward.
  3. When a train stops suddenly, passengers standing on the train tend to fall forward.
  4. When a train stops suddenly, passengers standing on the train tend to fall backward.

Answer. 

1. When a train starts suddenly, passengers standing on the train tend to fall backward.

3. When a train stops suddenly, passengers standing on the train tend to fall forward.

Question 43. Which of the following statements is correct for an object moving along a straight line in an accelerated motion?

  1. Speed keeps changing
  2. Velocity always change
  3. It goes away from earth
  4. Force is acting on it

Answer. 

1. Speed keeps changing

2. Velocity always changes

4. Force is acting on it

Question 44. Choose the incorrect statement. According to newton’s third law of motion, action, and reaction ______.

  1. Acts on the same object in the same direction
  2. Acts on different objects in the same direction
  3. Acts on different objects in the opposite direction
  4. Acts on the same object in a different direction

Answer. 

1. Acts on the same object in the same direction

2. Acts on different objects in the same direction

4. Acts on the same object in a different direction

Question 45. Choose the incorrect statement:

  1. Rocket works on the principle of conservation of momentum.
  2. Rocket works on the principle of conservation of mass.
  3. Rocket works on the principle of conservation of energy.
  4. Rocket works on the principle of conservation of velocity.

Answer. 

2. Rocket works on the principle of conservation of mass.

3. Rocket works on the principle of conservation of energy.

4. Rocket works on the principle of conservation of velocity.

Question 46. A body at rest breaks into two pieces of equal masses. The parts will move

  1. in the same direction
  2. along different lines
  3. in opposite directions with equal speeds
  4. in opposite directions with unequal speeds

Answer. 3. In opposite directions with equal speeds

Question 47. When a bus starts suddenly, the passengers standing inside lean backward. This is an example of

  1. Newton’s first law of motion
  2. Newton’s second law of motion
  3. Newton’s third law of motion
  4. Law of conservation of momentum

Answer. 1. Newton’s first law of motion

Question 48. A block of mass 2 kg is sliding with a constant velocity of 8 m/s on a frictionless horizontal surface. The force exerted on the horizontal surface is nearly

  1. 20 N
  2. 10 N
  3. 40 N
  4. 16 N

Answer. 1. 20 N

Question 49. A machine gun fires n bullets per second, each of mass m. If the speed of each bullet is u, then the force of recoil i

  1. Mng
  2. inv
  3. moving
  4. (Nov)/g

Answer. 2. inv

Question 50. A force of 5 N acts on a body of weight 9.8 N. What is the acceleration produced in m/s2

  1. 0.51
  2. 1.96
  3. 5.00
  4. 49.00

Answer. 3. 5.00

Question 51. In a game of tug of war, a condition of equilibrium exists. Both teams pull the rope with a force of 104 N. Tension in the rope is

  1. 104 N
  2. 108 N
  3. 2 × 104 N
  4. zero

Answer. 1. 104 N

Question 52. The action and reaction forces referred to in Newton’s third law of motion

  1. must act on the same body
  2. must act on different bodies
  3. need not be equal in magnitude but must have the same line of action
  4. must be equal in magnitude but need not have the same line of action

Answer. 2. must act on different bodies

Question 53. You are marooned on the frictionless horizontal plane and cannot exert any horizontal force by pushing against the surface. How can you get off?

  1. By running on the plane
  2. By jumping
  3. By rolling your body on the surface
  4. By sneezing or spitting

Answer. 3. By rolling your body on the surface

Question 54. Choose the wrong statement of the following

  1. 1 kg wt = 9.8 N
  2. Momentum is a vector quantity
  3. Force is always conserved
  4. Momentum is conserved in the absence of an external force

Answer. 3. Force is always conserved

Question 55. A long jumper runs before jumping because

  1. he covers a greater distance
  2. he maintains momentum conservation
  3. he gains energy by running
  4. he gains momentum

Answer. 4. he gains momentum

Question 56. A man is standing on a boat in still water. If he walks toward the shore the boat will

  1. move away from the shore
  2. remain stationary
  3. move towards the shore
  4. sink

Answer. 1. move away from the shore

Question 57. A body of mass 300 gm is at rest. What force in Newton will you have to apply to move it through 200 cm in 10 sec.?

  1. Zero
  2. 12 newton
  3. 1.2 newton
  4. 0.12 newton

Answer. 4. 0.12 Newton

Question 58. A boat of mass 3000 kg, initially at rest is pulled by a force of 1.8 × 104 Newton through a distance of 3 m. Assuming that the resistance due to water is negligible, the velocity of the boat is

  1. 6 m/s
  2. 8 m/s
  3. 9 m/s
  4. 11 m/s

Answer. 1. 6 m/s

Question 59. A certain force applied to mass m1 gives it an acceleration of 15 m/s2. The same force is applied on m2 to give it an acceleration of 10 m/s2, if the two masses are fixed together and the same force is applied to the combination, then the acceleration will be

  1. 6 m/s2
  2. 8 m/s2
  3. 9 m/s2
  4. 11 m/s2

Answer. 1. 6 m/s2

Question 60. A bullet of mass A and velocity B is fired into a block of wood of mass C. If loss of any mass and friction is neglected, then the velocity of the system must be

  1. \(\frac{A B}{A+C}\)
  2. \(\frac{A+C}{B+C}\)
  3. \(\frac{A C}{B+C}\)
  4. \(\frac{A+B}{A C}\)

Answer. 1. \(\frac{A B}{A+C}\)

Question 61. A driver accelerates his car first at the rate of 18 m/s2 and then at the rate of 12 m/s2. The ratio of the forces exerted by the engines respectively will be

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

Answer. 1. 2 : 3

Question 62. A body of mass 5  kg undergoes a change in speed from 30 to 40 m/s. Its momentum would increase by

  1. 50 kg m/s
  2. 75 kg m/s
  3. 150 kg m/s
  4. 350 kg m/s

Answer. 1. 50 kg m/s

Question 63. A body of mass 5 kg undergoes a change in speed from 20 to 0.20 m/s. The momentum of the body would

  1. increase by 99 kg m/s
  2. decrease by 99 kg m/s
  3. increase by 101 kg m/s
  4. decrease by 101 kg m/s

Answer. 2. decrease by 99 kg m/s

Question 64. A body of mass 100 g is moving with a velocity of 15 m/s. The momentum associated with that ball will be

  1. 0.5 kg m/s
  2. 1.5 kg m/s
  3. 2.5 kg m/s
  4. 3.2 kg m/s

Answer. 2. 1.5 kg m/s

Question 65. A stationary ball weighing 0.25 kg acquires a speed of 10 m/s when hit by a hockey stick. The impulse imparted to the ball is

  1. 2.5 N s
  2. 2.0 N s
  3. 1.5 N s
  4. 0.5 N s

Answer. 1. 2.5 N s

Question 66. Momentum is a measure of

  1. weight
  2. mass
  3. quantity of motion
  4. velocity

Answer. 3. Quantity of motion

Question 67. It is difficult to walk on ice because of

  1. absence of friction
  2. absence of inertia
  3. more inertia
  4. more friction

Answer. 1. absence of friction

Question 68. A and B are two objects with mass 6 kg and 34 kg respectively. Then

  1. A has more inertia than B
  2. B has more inertia than A
  3. A and B both have the same inertia
  4. None of these

Answer. 2. B has more inertia than A

Question 69. When a body is stationary

  1. there is no force acting on it
  2. the forces acting on it are not in contact with it
  3. the combination of forces acting on it balanced each other
  4. the body is in a vacuum

Answer. 3. The combination of forces acting on it balanced each other

Question 70. Which of the following represents the SI unit of force?

  1. dyne
  2. gram-weight
  3. newton
  4. kilogram-weight

Answer. 3. newton

Question 71. When a constant force is applied to a body, it moves with a uniform

  1. acceleration
  2. velocity
  3. speed
  4. momentum

Answer. 1. acceleration

Question 72. Passengers standing in a bus are thrown outwards when the bus takes a sudden turn, due to

  1. outward pull on them
  2. inertia
  3. change in momentum
  4. change in acceleration

Answer. 2. inertia

Question 73. A body of mass M collides against a wall with velocity V and rebounds with the same speed. Its change of momentum is

  1. zero
  2. MV
  3. 2 MV
  4. –MV

Answer. 3. 2 MV

Question 74. Tires are made circular because

  1. they can be inflated
  2. they require less material
  3. they look beautiful
  4. they face less friction

Answer. 4. they face less friction

Question 75. If a 1 kg mass v-t graph is as shown, then which of the following is true?

NEET Foundation Physics Force And Laws Of Motion Question 75

  1. force between A and B ≠ force between C and D
  2. force between A and B = force between C and D
  3. force between B and C = 1
  4. data insufficient to calculate the force

Answer. 1. force between A and B ≠ force between C and D

Question 76. Two forces of 6N and 3N are acting on the two blocks of 2kg and 1kg kept on a frictionless floor. What is the force exerted on a 2kg block by a 1 kg block?

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

Answer. 3. 4 N

Question 77. Three masses of 1 kg, 6 kg, and 3 kg are connected with threads and are placed on the table as shown in the figure, What is the acceleration with which the system is moving? Take g = 10 m s-2.

NEET Foundation Physics Force And Laws Of Motion Question 77

  1. Zero
  2. 1 m s-2
  3. 2 m s-2
  4. 3 m s-2

Answer. 3. 2 m s-2

Question 78. Two blocks of mass 4 kg and 2 kg are connected by a heavy string and placed on a rough horizontal plane. The 2 kg block is pulled with a constant force of F. The coefficient of friction between the blocks and the ground is 0.5. What is the value of F so that tension in the string is constant throughout during the motion of the blocks: (g = 10 m/s2)

NEET Foundation Physics Force And Laws Of Motion Question 78

  1. 40 N
  2. 30 N
  3. 50 N
  4. 60 N

Answer. 2. 30 N

Question 79. A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as shown in the figure. The force on the pulley by the clamp is given by

NEET Foundation Physics Force And Laws Of Motion Question 79

  1. \(\sqrt{2}\) Mg
  2. \(\sqrt{2}\) mg
  3. \(\left(\sqrt{(M+m)^2+m^2}\right) g\)
  4. \(\left(\sqrt{(M+m)^2+M^2}\right) g\)

Answer. 4. \(\left(\sqrt{(M+m)^2+M^2}\right) g\)

Force And Laws Of Motion Track Your Learning Questions and Answers

Question 1. If an object is still, it will remain still and if it is moving it will remain moving until an external force is applied. This is called ______.

  1. Newton’s first law of motion
  2. Newton’s second law of motion
  3. Newton’s third law of motion
  4. Newton’s fourth law of motion

Answer. 1. Newton’s first law of motion

Question 2. An object moving in a straight line will continue to move in the same direction until and unless some external force compels it to change the direction of motion. This is called ______.

  1. Inertia of motion
  2. Inertia of direction
  3. Inertia of speed
  4. Inertia of velocity

Answer. 2. Inertia of direction

Question 3. ______ is the external cause which tends to change the state of rest or the state of motion of an object.

  1. Velocity
  2. Speed
  3. Force
  4. Acceleration

Answer. 3. Force

Question 4. S.I. unit of inertia is ______.

  1. t
  2. kg
  3. g
  4. m

Answer. 2. kg

Question 5. _____ is the measure of inertia of a body at rest.

  1. Mass
  2. Time
  3. Acceleration
  4. Speed

Answer. 1. Mass

Force And Laws Of Motion Track Your Learning Questions and Answers

Question 1. C.G.S. unit of force is ______.

  1. Newton
  2. Dyne
  3. kg
  4. m/s

Answer. 2. Dyne

Question 2. Choose the correct equation

  1. F = m a
  2. F = m/a
  3. F = m t
  4. F = mv

Answer. 1. F = m a

Question 3. When a glass falls on the floor, it ______.

  1. Jumps
  2. Moves backward
  3. Moves forward
  4. Breaks

Answer. 4. Breaks

Question 4. S.I. unit of momentum is ______.

  1. kgm/s
  2. cm m/s
  3. m/s
  4. g cm/s

Answer. 1. kg m/s

Question 5. C.G.S. unit of momentum is ______.

  1. kg m/s
  2. cm m/s
  3. m/s
  4. g cm/s

Answer. 4. g cm/s

Force And Laws Of Motion Practice Exercises

Question 1. Which of the following statements is not correct for an object moving along a straight path in an accelerated motion?

  1. Its speed keeps changing
  2. Its velocity always changes
  3. It always goes away from the earth
  4. A force is always acting on it

Answer. 3. It always goes away from the earth

Question 2. According to the third law of motion, action and reaction

  1. always act on the same body
  2. always act on different bodies in opposite directions
  3. have the same magnitude and directions
  4. act on either body at normal to each other

Answer. 2. always act on different bodies in opposite directions

Question 3. A goalkeeper in a game of football pulls his hands backward after holding the ball shot at the goal. This enables the goalkeeper to

  1. exert larger force on the ball
  2. reduce the force exerted by the ball on the hands
  3. increase the rate of change of momentum
  4. decrease the rate of change of momentum

Answer. 2. reduce the force exerted by the ball on hands

Question 4. The inertia of an object tends to cause the object

  1. to increase its speed
  2. to decrease its speed
  3. to resist any change in its state of motion
  4. to decelerate due to friction

Answer. 3. To resist any change in its state of motion

Question 5. A passenger in a moving train tosses a coin which falls behind him. It means that the motion of the train is

  1. accelerated
  2. uniform
  3. retarded
  4. along circular tracks

Answer. 1. accelerated

Question 6. An object of mass 2 kg is sliding with a constant velocity of 4 ms-1 on a frictionless horizontal table. The force required to keep the object moving with the same velocity is

  1. 32 N
  2. 0 N
  3. 2 N
  4. 8 N

Answer. 2. 0 N

Question 7. Rocket works on the principle of conservation of

  1. mass
  2. energy
  3. momentum
  4. velocity

Answer. 3. momentum

Question 8. A water tanker filled up to 2/3 of its height is moving with a uniform speed. On sudden application of the brake, the water in the tank would

  1. move backward
  2. move forward
  3. be unaffected
  4. rise upwards

Answer. 2. move forward

Question 9. Which among the following is not an unbalanced force?

  1. Stop a moving object
  2. Two boys playing on a see-saw
  3. Arm-wrestling competition
  4. Pushing an object from both sides

Answer. 4. Pushing an object from both sides

Question 10. ________ is the force acting parallel to the surface of the contact.

  1. Inertia
  2. Balanced Force
  3. Unbalanced Force
  4. Frictional force

Answer. 2. Balanced Force

Question 11. If one object applies more force than the other one it is called _____ force.

  1. Inertia
  2. Balanced
  3. Unbalanced
  4. Linear

Answer. 3. Unbalanced