A car of mass 1000 kg is moving with a velocity of 20 m/s. It applies brakes and comes to a stop in 5 seconds. What is the average force exerted by the brakes?

Correct Option: A. Solution: The change in momentum $\Delta p$ is $1000 \times 20 = 20000 \text{ kg m/s}$. The average force $F$ is given by $F = \frac{\Delta p}{\Delta t} = \frac{20000}{5} = 4000 \text{ N}$.

Laws of Motion

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

Understand the concept of inertia and its implications in motion.
Explain Newton's three laws of motion and their applications.
Calculate momentum and understand its significance in mechanics.
Apply Newton's second law to solve problems involving force and acceleration.
Analyze the concept of impulse and its relationship to momentum.
Differentiate between static and kinetic friction and their applications in real-world scenarios.
Utilize free-body diagrams to visualize forces acting on a system.
Solve problems involving tension in strings and forces in circular motion.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Laws of Motion

4.1 Introduction

4.2 Aristotle's Fallacy

Aristotle's view that a force is necessary to keep a body in uniform motion is incorrect. A force is necessary to counteract friction.

4.3 The Law of Inertia

Galileo's observations led to the law of inertia, which states that a body remains at rest or in uniform motion unless acted upon by an external force.
Newton's First Law of Motion: If the external force on a body is zero, its acceleration is zero.

4.4 Newton's First Law of Motion

Rephrased: Everybody continues in its state of rest or uniform motion unless compelled by an external force.

4.5 Newton's Second Law of Motion

Definition: The rate of change of momentum is proportional to the applied force and occurs in the direction of the force.
Formula: F = ma, where F is the net external force and a is acceleration.
SI Unit of Force: 1 N = 1 kg m s⁻².
Key Points:
- Consistent with the First Law (F = 0 implies a = 0).
- Applicable to particles and systems of particles.
- Local law: acceleration at a point does not depend on the history of motion.

4.6 Newton's Third Law of Motion

Statement: To every action, there is always an equal and opposite reaction.
Important Notes:
- Action and reaction forces act on different bodies.
- They occur simultaneously, not sequentially.

4.7 Conservation of Momentum

Momentum (p) is defined as the product of mass (m) and velocity (v): p = mv.

4.8 Equilibrium of a Particle

In equilibrium, the sum of forces acting on a particle is zero.

4.9 Common Forces in Mechanics

Types of forces include:
- Gravitational force
- Tension
- Friction
- Normal force

4.10 Circular Motion

Centripetal force is not a separate force but the name for the net force causing circular motion.

4.11 Solving Problems in Mechanics

Drawing free-body diagrams is essential for visualizing forces acting on a system.

Summary

Aristotle's view on force and motion is incorrect.
Galileo's law of inertia is foundational to Newton's First Law.
Momentum is defined as p = mv.
Newton's Second Law relates force, mass, and acceleration.
Impulse is the product of force and time, equating to change in momentum.
Newton's Third Law emphasizes action-reaction pairs.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding Newton's Laws: Many students confuse the terms 'action' and 'reaction' in Newton's third law, thinking that action precedes reaction. Remember, they act simultaneously on different bodies.
Static Friction Misapplication: Do not assume that static friction always equals µₛ N. It is a self-adjusting force and only reaches its maximum value when necessary.
Forces in Equilibrium: The equation mg = R is only valid when the body is in equilibrium. In non-equilibrium situations, these forces can differ.
Centripetal Force Confusion: Centripetal force is not a separate force; it is the net force causing circular motion, such as tension or gravity.
Impulse Calculation Errors: When calculating impulse, ensure you understand that it is the change in momentum and can be calculated even if the forces during contact are unknown.

Exam Tips

Draw Free-Body Diagrams: This practice helps clarify the forces acting on a system and is essential for solving mechanics problems.
Understand the Context of Forces: Familiarize yourself with different types of forces (e.g., tension, friction, normal force) and their contexts to avoid confusion.
Practice with Real-World Examples: Relate problems to everyday situations to better grasp concepts like inertia, momentum, and forces.
Review Key Definitions and Formulas: Make sure you can define and apply key terms like impulse, momentum, and Newton's laws accurately.
Check Units Consistency: Always ensure that your units are consistent, especially when applying formulas in mechanics.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

The body moves in the direction of the force.

An equal and opposite force acts on a different body.

The body accelerates indefinitely.

The force is absorbed by the body.

Correct Answer: B

Solution:

Newton's third law states that for every action, there is an equal and opposite reaction, meaning forces act in pairs on different bodies.

Static friction is always greater than kinetic friction.

Kinetic friction is always greater than static friction.

Static friction and kinetic friction are equal.

There is no relationship between static and kinetic friction.

Correct Answer: A

Solution:

Experimentally, it is found that the coefficient of static friction is greater than that of kinetic friction, making static friction greater than kinetic friction.

Force is inversely proportional to momentum.

Force is equal to the rate of change of momentum.

Force is directly proportional to momentum.

Force is independent of momentum.

Correct Answer: B

Solution:

Newton's second law states that force is equal to the rate of change of momentum.

0 kg m/s

2 kg m/s

4 kg m/s

6 kg m/s

Correct Answer: C

Solution:

The initial momentum of the ball is

p_i = mv = 0.2 \times 10 = 2 \text{ kg m/s}

. After rebounding, the velocity is

-10 \text{ m/s}

, so the final momentum is

p_f = 0.2 \times (-10) = -2 \text{ kg m/s}

. The change in momentum is

\Delta p = p_f - p_i = -2 - 2 = -4 \text{ kg m/s}

. Since we are interested in the magnitude, the change is

4 \text{ kg m/s}

14 m/s

20 m/s

10 m/s

9.8 m/s

Correct Answer: A

Solution:

Using the equation of motion

v^2 = u^2 + 2gh

, where

u = 0

g = 9.8 \text{ m/s}^2

, and

h = 10 \text{ m}

. Thus,

v^2 = 0 + 2 \times 9.8 \times 10 = 196

. Therefore,

v = \sqrt{196} = 14 \text{ m/s}

9.5 m

11.5 m

12.5 m

14.5 m

Correct Answer: B

Solution:

Using the equation

v^2 = u^2 - 2gh

, where

v = 0

at the maximum height,

u = 15 \text{ m/s}

, and

g = 9.8 \text{ m/s}^2

, we solve for

h

0 = 15^2 - 2 \cdot 9.8 \cdot h \Rightarrow h = \frac{225}{19.6} \approx 11.5 \text{ m}

1.8 Ns

3.6 Ns

0 Ns

6.0 Ns

Correct Answer: B

Solution:

The impulse is calculated as the change in momentum:

0.15 \times 12 - (-0.15 \times 12) = 3.6 \text{ Ns}

For every action, there is an equal and opposite reaction.

A force is necessary to keep a body in uniform motion.

The rate of change of momentum is proportional to the applied force.

Momentum is the product of mass and velocity.

Correct Answer: A

Solution:

Newton's third law states that for every action, there is an equal and opposite reaction.

A body at rest will remain at rest unless acted upon by an external force.

The acceleration of a body is always zero.

A body in motion will eventually stop without any external force.

A force is necessary to keep a body in uniform motion.

Correct Answer: A

Solution:

Newton's first law states that a body at rest will remain at rest, and a body in motion will remain in motion unless acted upon by an external force.

It requires a force to maintain its motion.

It continues in motion without any force.

It comes to rest without a force.

It accelerates without a force.

Correct Answer: A

Solution:

Aristotle's fallacy states that a force is necessary to keep a body in uniform motion, which is incorrect as per modern physics.

25 N

39.5 N

50 N

75 N

Correct Answer: B

Solution:

The tension in the string is calculated using the formula for centripetal force:

T = m \cdot \omega^2 \cdot r

, where

\omega = \frac{2\pi \cdot 40}{60}

rad/s.

Force is inversely proportional to acceleration.

Force is directly proportional to acceleration.

Force is independent of acceleration.

Force is equal to mass times velocity.

Correct Answer: B

Solution:

Newton's second law states that the rate of change of momentum of a body is proportional to the applied force, which is directly proportional to the acceleration.

5 Ns

10 Ns

15 Ns

20 Ns

Correct Answer: B

Solution:

The impulse imparted is equal to the change in momentum. Initial momentum =

0.5 \times 10 = 5 \text{ kg m/s}

. Final momentum =

0.5 \times (-10) = -5 \text{ kg m/s}

. Change in momentum =

|-5 - 5| = 10 \text{ Ns}

5 m/s²

10 m/s²

2 m/s²

20 m/s²

Correct Answer: A

Solution:

According to Newton's second law, acceleration is given by

a = \frac{F}{m}

. Therefore,

a = \frac{10\, \text{N}}{2\, \text{kg}} = 5\, \text{m/s}^2

It states that force is the product of mass and velocity.

It states that the rate of change of momentum is proportional to the applied force.

It states that every action has an equal and opposite reaction.

It states that a body at rest will remain at rest unless acted upon by an external force.

Correct Answer: B

Solution:

Newton's second law of motion states that the rate of change of momentum of a body is proportional to the applied force and occurs in the direction of the force.

1 m/s

2 m/s

3 m/s

4 m/s

Correct Answer: C

Solution:

In an elastic collision, both momentum and kinetic energy are conserved. Using conservation of momentum:

0.1 \times 5 + 0.2 \times 0 = 0.1 \times v_1 + 0.2 \times v_2

. Using conservation of kinetic energy:

\frac{1}{2} \times 0.1 \times 5^2 = \frac{1}{2} \times 0.1 \times v_1^2 + \frac{1}{2} \times 0.2 \times v_2^2

. Solving these equations gives

v_1 = 3 \text{ m/s}

1 m/s²

2 m/s²

3 m/s²

4 m/s²

Correct Answer: C

Solution:

The total mass of the system is 15 kg. Using Newton's second law,

F = ma

, the acceleration

a = \frac{F}{m} = \frac{30}{15} = 2 \text{ m/s}^2

0.5 m/s

1 m/s

2.5 m/s

0 m/s

Correct Answer: A

Solution:

Using the conservation of momentum:

m_1v_1 + m_2v_2 = (m_1 + m_2)v_f

. Here,

5 \times 2 + 5 \times (-3) = 10v_f

. Thus,

10 - 15 = 10v_f \Rightarrow v_f = -0.5 \text{ m/s}

(the negative sign indicates the direction).

0.3 Ns

0.6 Ns

0.9 Ns

1.2 Ns

Correct Answer: B

Solution:

The impulse imparted to each ball is calculated as the change in momentum, which is

0.05 \times 6 - (-0.05 \times 6) = 0.6 \text{ Ns}

The stone moves radially outwards.

The stone flies off tangentially.

The stone flies off at an angle with the tangent.

The stone remains in circular motion.

Correct Answer: B

Solution:

When the string breaks, the stone flies off tangentially due to inertia.

The 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.

To every action, there is always an equal and opposite reaction.

A body continues in its state of rest or uniform motion unless acted upon by an external force.

The total momentum of an isolated system is conserved.

Correct Answer: A

Solution:

Newton's second law states that the rate of change of momentum of a body is proportional to the applied force and occurs in the direction of the force.

Force

Impulse

Momentum

Acceleration

Correct Answer: C

Solution:

Momentum is defined as the product of mass and velocity of a body, represented as

p = mv

10 N

15 N

20 N

25 N

Correct Answer: B

Solution:

The total mass of the system is

5 + 10 = 15 \text{ kg}

. The acceleration

a

of the system is

\frac{30}{15} = 2 \text{ m/s}^2

. The tension

T

in the string is the force required to accelerate the 5 kg block, so

T = 5 \times 2 = 10 \text{ N}

4000 N

5000 N

6000 N

8000 N

Correct Answer: A

Solution:

The change in momentum

\Delta p

1000 \times 20 = 20000 \text{ kg m/s}

. The average force

F

is given by

F = \frac{\Delta p}{\Delta t} = \frac{20000}{5} = 4000 \text{ N}

1 m/s²

2 m/s²

3 m/s²

4 m/s²

Correct Answer: B

Solution:

The total mass of the system is

m = 5 + 10 = 15 \, \text{kg}

. The force applied is

F = 30 \, \text{N}

. The acceleration

a

is given by

a = \frac{F}{m} = \frac{30}{15} = 2 \, \text{m/s}^2

Centripetal force

Centrifugal force

Gravitational force

Magnetic force

Correct Answer: A

Solution:

The centripetal force is the force that provides the inward radial acceleration to a body in circular motion.

It remains at rest or moves with constant velocity.

It accelerates.

It changes direction.

It comes to a stop.

Correct Answer: A

Solution:

Newton's first law states that a body remains at rest or in uniform motion unless acted upon by an external force.

Gravitational force

Tension in the satellite

Frictional force

Electromagnetic force

Correct Answer: A

Solution:

The centripetal force required for circular motion is provided by the gravitational force between the Earth and the satellite. This is the force that acts towards the center of the Earth, keeping the satellite in its orbit.

26 m/s

20 m/s

18 m/s

16 m/s

Correct Answer: A

Solution:

Using the equation of motion,

v = u + at

, where initial velocity

u = 10 \text{ m/s}

, acceleration

a = \frac{F}{m} = \frac{15}{5} = 3 \text{ m/s}^2

, and time

t = 4 \text{ s}

. Thus, final velocity

v = 10 + 3 \times 4 = 22 \text{ m/s}

A body at rest will remain at rest unless acted upon by an external force.

The rate of change of momentum is proportional to the applied force.

For every action, there is an equal and opposite reaction.

Momentum is the product of mass and velocity.

Correct Answer: A

Solution:

Newton's first law of motion states that a body at rest will remain at rest, and a body in motion will continue in its motion unless acted upon by an external force.

19.8 m/s

20.0 m/s

22.1 m/s

24.5 m/s

Correct Answer: C

Solution:

Using the equation of motion,

v^2 = u^2 + 2gh

, where

u = 0

g = 9.8 \text{ m/s}^2

, and

h = 20 \text{ m}

. Thus,

v^2 = 0 + 2 \times 9.8 \times 20 = 392

, giving

v = \sqrt{392} = 22.1 \text{ m/s}

1 m/s

2 m/s

-1 m/s

-2 m/s

Correct Answer: D

Solution:

Using the conservation of momentum and the fact that kinetic energy is conserved in elastic collisions, we have the equations:

m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2

and

\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2

. Solving these with

m_1 = 0.2 \text{ kg}

u_1 = 3 \text{ m/s}

m_2 = 0.3 \text{ kg}

, and

u_2 = 0

, we find

v_1 = -2 \text{ m/s}

20 m/s

5 m/s

10 m/s

15 m/s

Correct Answer: A

Solution:

Using Newton's second law, we have

F = ma

. Here,

F = 50 \text{ N}

and

m = 10 \text{ kg}

, so

a = \frac{F}{m} = \frac{50}{10} = 5 \text{ m/s}^2

. The final velocity

v

can be calculated using

v = u + at

, where

u = 0

a = 5 \text{ m/s}^2

, and

t = 4 \text{ s}

. Thus,

v = 0 + 5 \times 4 = 20 \text{ m/s}

50 kg m/s

5 kg m/s

10 kg m/s

15 kg m/s

Correct Answer: A

Solution:

Momentum is calculated as the product of mass and velocity. Therefore, momentum = 10 kg * 5 m/s = 50 kg m/s.

10 N

0 N

9.8 N

98 N

Correct Answer: B

Solution:

According to Newton's first law of motion, if no external force is applied, the net force is zero.

The stone moves radially outwards.

The stone flies off tangentially from the instant the string breaks.

The stone flies off at an angle with the tangent.

The stone moves in a spiral path.

Correct Answer: B

Solution:

When the string breaks, the stone will continue to move in the direction it was moving at the instant of the break, which is tangentially to the circular path.

7 Ns

9.8 Ns

14 Ns

19.6 Ns

Correct Answer: D

Solution:

The speed just before impact is

v = \sqrt{2gh} = \sqrt{2 \cdot 9.8 \cdot 10} = 14 \text{ m/s}

. The change in velocity is

\Delta v = 14 - (-14) = 28 \text{ m/s}

. Impulse

J = m \cdot \Delta v = 0.5 \cdot 28 = 14 \text{ Ns}

Static friction is always greater than kinetic friction.

Kinetic friction is always greater than static friction.

Static friction does not depend on the area of contact.

Kinetic friction depends on the speed of the moving object.

Correct Answer: A

Solution:

Static friction is generally greater than kinetic friction because it needs to overcome the initial interlocking between surfaces.

5000 N

6000 N

7000 N

8000 N

Correct Answer: B

Solution:

Using the equation

F = ma

, where

m = 1200 \text{ kg}

and

a = \frac{\Delta v}{\Delta t} = \frac{0 - 25}{6} = -4.17 \text{ m/s}^2

. Therefore,

F = 1200 \times (-4.17) = -5000 \text{ N}

. The magnitude of the force is

6000 \text{ N}

Newton

Joule

Watt

Pascal

Correct Answer: A

Solution:

The SI unit of force is the newton, which is equivalent to 1 kg m s⁻².

1.96 m/s²

2.45 m/s²

3.27 m/s²

4.90 m/s²

Correct Answer: B

Solution:

The acceleration

a

can be found using the formula

a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{(12 - 8) \times 9.8}{12 + 8} = 2.45 \text{ m/s}^2

10 m/s

5 m/s

15 m/s

20 m/s

Correct Answer: A

Solution:

Using Newton's second law, the acceleration

a

is given by

F = ma \Rightarrow a = \frac{F}{m} = \frac{20}{10} = 2 \text{ m/s}^2

. The final velocity

v

can be calculated using

v = u + at

, where

u = 0

a = 2 \text{ m/s}^2

, and

t = 5 \text{ s}

. Thus,

v = 0 + 2 \times 5 = 10 \text{ m/s}

5 m/s²

8.66 m/s²

4.33 m/s²

10 m/s²

Correct Answer: C

Solution:

The horizontal component of the force is

F_x = 10 \cos(30°) = 10 \times \frac{\sqrt{3}}{2} = 8.66 \text{ N}

. The acceleration is given by

a = \frac{F_x}{m} = \frac{8.66}{2} = 4.33 \text{ m/s}^2

0.016 m/s

0.02 m/s

0.04 m/s

0.08 m/s

Correct Answer: A

Solution:

Using conservation of momentum: (mass of shell × speed of shell) = (mass of gun × recoil speed of gun). Thus, 0.020 kg × 80 m/s = 100 kg × recoil speed, giving a recoil speed of 0.016 m/s.

Impulse is the rate of change of momentum.

Impulse is the product of momentum and time.

Impulse is equal to the change in momentum.

Impulse is the force required to maintain momentum.

Correct Answer: C

Solution:

Impulse is defined as the product of force and time, which equals the change in momentum.

1000 N

2000 N

500 N

250 N

Correct Answer: A

Solution:

The deceleration

a

can be calculated using

v = u + at

, where

v = 0

u = 20 \text{ m/s}

, and

t = 10 \text{ s}

. Thus,

0 = 20 + a \times 10

, giving

a = -2 \text{ m/s}^2

. The braking force

F

F = ma = 500 \times (-2) = -1000 \text{ N}

. The magnitude is

1000 \text{ N}

2000 N

2500 N

3000 N

3200 N

Correct Answer: C

Solution:

The average force can be calculated using the formula

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

, where

\Delta p

is the change in momentum and

\Delta t

is the time. Initial momentum

p_i = mv = 800 \times 25 = 20000\, \text{kg m/s}

. Final momentum

p_f = 0

. Thus,

\Delta p = 20000\, \text{kg m/s}

F = \frac{20000}{10} = 2000\, \text{N}

The wall exerts an equal and opposite force against you.

The wall moves slightly.

The wall exerts no force.

The wall exerts a force greater than your push.

Correct Answer: A

Solution:

Newton's third law states that for every action, there is an equal and opposite reaction. When you push against a wall, it pushes back with an equal force.

20 m/s

30 m/s

40 m/s

50 m/s

Correct Answer: C

Solution:

The initial momentum of the object is

p_i = mv = 15 \times 10 = 150 \, \text{kg m/s}

. The force applied is

F = 30 \, \text{N}

, and the time duration is

t = 5 \, \text{s}

. The change in momentum (impulse) is

\Delta p = Ft = 30 \times 5 = 150 \, \text{kg m/s}

. The final momentum is

p_f = p_i + \Delta p = 150 + 150 = 300 \, \text{kg m/s}

. Therefore, the final velocity is

v_f = \frac{p_f}{m} = \frac{300}{15} = 20 \, \text{m/s}

2 Ns

0 Ns

-2 Ns

-1 Ns

Correct Answer: C

Solution:

Impulse is the change in momentum. The impulse due to gravity is

\Delta p = m \cdot \Delta v = 0.2 \cdot (-10) = -2 \text{ Ns}

5 m/s²

8.66 m/s²

9.8 m/s²

4.9 m/s²

Correct Answer: D

Solution:

The acceleration of the block down the incline can be calculated using the component of gravitational force along the incline:

a = g \sin \theta = 9.8 \times \sin 30° = 4.9 \text{ m/s}^2

5 N

10 N

20 N

25 N

Correct Answer: B

Solution:

The centripetal force

F_c

is given by

F_c = \frac{mv^2}{r}

. Substituting the values,

F_c = \frac{0.2 \times 5^2}{0.5} = 10 \text{ N}

2 m/s

1 m/s

-2 m/s

-1 m/s

Correct Answer: C

Solution:

By conservation of momentum, the total momentum before and after must be equal. Initially, both skaters are at rest, so total momentum is zero. After they push off,

m_1 v_1 + m_2 v_2 = 0

. Given

m_1 = m_2 = 60\, \text{kg}

and

v_1 = 2\, \text{m/s}

, we find

v_2 = -\frac{m_1 v_1}{m_2} = -\frac{60 \times 2}{60} = -2\, \text{m/s}

5 seconds

10 seconds

15 seconds

20 seconds

Correct Answer: B

Solution:

Using Newton's second law,

a = \frac{F}{m} = \frac{1000}{500} = 2 \text{ m/s}^2

. Using

v = u + at

, where

u = 0

v = 20

, we solve for

t

20 = 0 + 2t \Rightarrow t = 10 \text{ seconds}

8 N

7.84 N

9.8 N

10 N

Correct Answer: B

Solution:

The maximum static friction force

f_s

is given by

f_s = \mu_s N

, where

N = mg = 2 \times 9.8 = 19.6 \text{ N}

. Thus,

f_s = 0.4 \times 19.6 = 7.84 \text{ N}

. This is the minimum force required to overcome static friction and start moving the block.

The force of gravity acting on a falling apple.

The force of a bat hitting a ball and the ball hitting the bat.

The force of friction slowing down a sliding box.

The tension in a rope holding a hanging weight.

Correct Answer: B

Solution:

According to Newton's third law, the force of a bat on a ball and the force of the ball on the bat are equal and opposite, forming an action-reaction pair.

4 N

8 N

6 N

16 N

Correct Answer: B

Solution:

The centripetal force

F_c

is given by

F_c = \frac{mv^2}{r}

, where

m = 0.5 \text{ kg}

v = 4 \text{ m/s}

, and

r = 2 \text{ m}

. Thus,

F_c = \frac{0.5 \times 4^2}{2} = \frac{8}{2} = 4 \text{ N}

The products move in the same direction.

The products move in opposite directions.

The products remain stationary.

The products move randomly.

Correct Answer: B

Solution:

According to the conservation of momentum, if a nucleus disintegrates at rest, the two smaller nuclei must move in opposite directions to conserve momentum.

12 N

24 N

6 N

36 N

Correct Answer: B

Solution:

The centripetal force is given by

F_c = \frac{mv^2}{r} = \frac{3 \times 4^2}{2} = \frac{48}{2} = 24 \text{ N}

14 Ns

18 Ns

22 Ns

26 Ns

Correct Answer: C

Solution:

Impulse is the integral of force over time:

I = \int_0^2 (3t^2 + 2t + 1) \, dt = [t^3 + t^2 + t]_0^2 = (8 + 4 + 2) - (0) = 14 \text{ Ns}

0 kg m/s

1 kg m/s

2 kg m/s

4 kg m/s

Correct Answer: C

Solution:

The total momentum before the collision is

0.5 \times 4 + 0 = 2 \text{ kg m/s}

. By conservation of momentum, the total momentum after the collision is also 2 kg m/s.

1.5 m/s

2 m/s

3 m/s

4 m/s

Correct Answer: C

Solution:

By conservation of momentum, the total momentum before and after must be zero. If one skater moves with 3 m/s, the other must move with 3 m/s in the opposite direction to conserve momentum:

50 \times 3 + 50 \times v = 0

, thus

v = -3 \text{ m/s}

A new type of force specific to circular motion.

The force that acts perpendicular to the direction of motion.

Any material force like tension or gravity that acts towards the center.

The force that acts in the direction of motion.

Correct Answer: C

Solution:

Centripetal force is not a new type of force; it is provided by material forces like tension, gravity, etc., that act towards the center of the circular path.

1 m/s

2 m/s

3 m/s

0 m/s

Correct Answer: B

Solution:

In a perfectly elastic collision, both momentum and kinetic energy are conserved. Using the conservation of momentum:

0.2 \times 3 + 0.3 \times 0 = 0.2 \times v_1 + 0.3 \times v_2

. Using conservation of kinetic energy:

\frac{1}{2} \times 0.2 \times 3^2 = \frac{1}{2} \times 0.2 \times v_1^2 + \frac{1}{2} \times 0.3 \times v_2^2

. Solving these equations,

v_1 = 2 \text{ m/s}

It opposes impending relative motion.

It is always greater than kinetic friction.

It is independent of the area of contact.

It acts only when there is motion.

Correct Answer: D

Solution:

Static friction acts to prevent motion and only opposes impending motion, not actual motion.

0.3 Ns

0.6 Ns

0.1 Ns

0.5 Ns

Correct Answer: B

Solution:

Impulse is the change in momentum. The change in velocity is 12 m/s (from 6 m/s to -6 m/s), so impulse = mass × change in velocity = 0.05 kg × 12 m/s = 0.6 Ns.

Tension in the string

Gravitational force

Frictional force

Air resistance

Correct Answer: A

Solution:

In circular motion, the tension in the string provides the necessary centripetal force to keep the stone moving in a circle.

2 m/s²

3 m/s²

4 m/s²

5 m/s²

Correct Answer: A

Solution:

The total mass of the system is

4 + 6 = 10 \text{ kg}

. Using Newton's second law,

F = ma

, where

F = 20 \text{ N}

and

m = 10 \text{ kg}

. Therefore,

a = \frac{F}{m} = \frac{20}{10} = 2 \text{ m/s}^2

2 m/s²

4 m/s²

5 m/s²

10 m/s²

Correct Answer: B

Solution:

According to Newton's second law,

F = ma

. Here,

F = 20\, \text{N}

and

m = 5\, \text{kg}

. Therefore,

a = \frac{F}{m} = \frac{20}{5} = 4\, \text{m/s}^2

20 N

15 N

25 N

10 N

Correct Answer: A

Solution:

The maximum static frictional force is given by

f_s = \mu_s N

, where

\mu_s = 0.4

and

N = mg = 5 \times 9.8 = 49 \text{ N}

. Thus,

f_s = 0.4 \times 49 = 19.6 \text{ N}

. The maximum horizontal force that can be applied without moving the block is approximately

20 \text{ N}

Centripetal force is not a distinct type of force; it is a name given to the force that provides inward radial acceleration to a body in circular motion, such as tension, gravitational force, or friction.

Correct Answer: True

Solution:

It is experimentally found that the coefficient of static friction

\mu_s

is greater than the coefficient of kinetic friction

Solution:

Centripetal force is not a new type of force; it is the name given to the force that provides the necessary inward radial acceleration for circular motion, such as tension or gravitational force.

Correct Answer: True

Solution:

Momentum is defined as the product of mass and velocity, expressed as

p = mv

Correct Answer: True

Solution:

According to Newton's third law of motion, forces always occur in pairs and the mutual forces between two bodies are equal and opposite.

Correct Answer: False

Solution:

The law of inertia, also known as Newton's first law of motion, states that a body remains in its state of rest or uniform motion in a straight line unless compelled by an external force to change that state.

Laws of Motion

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Learning Objectives

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter Notes on Laws of Motion

4.1 Introduction

4.2 Aristotle's Fallacy

4.3 The Law of Inertia

4.4 Newton's First Law of Motion

4.5 Newton's Second Law of Motion

4.6 Newton's Third Law of Motion

4.7 Conservation of Momentum

4.8 Equilibrium of a Particle

4.9 Common Forces in Mechanics

4.10 Circular Motion

4.11 Solving Problems in Mechanics

Summary

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips

Common Pitfalls

Exam Tips

Multiple Choice Questions

Q1.According to Newton's third law, what happens when a force is applied to a body?

Correct Answer: B

Q2.What is the relationship between static friction and kinetic friction?

Correct Answer: A

Q3.What is the relationship between force and momentum according to Newton's second law?

Correct Answer: B

Q4.A 0.2 kg ball is moving with a velocity of 10 m/s. It collides with a wall and rebounds with the same speed. What is the change in momentum of the ball?

Correct Answer: C

Q5.A 0.5 kg ball is dropped from a height of 10 m. What is the speed of the ball just before it hits the ground? (Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2g=9.8 m/s2 and air resistance is negligible.)

Correct Answer: A

Q6.A 0.2 kg ball is thrown vertically upward with a speed of 15 m/s. What is the maximum height reached by the ball? (Assume g=9.8 m/s2g = 9.8 \text{ m/s}^2g=9.8 m/s2)

Correct Answer: B

Q7.What is the impulse imparted to a 0.15 kg ball hit back straight by a batsman with an initial speed of 12 m/s?

Correct Answer: B

Q8.Which of the following statements correctly describes Newton's third law of motion?

Correct Answer: A

Q9.Which of the following statements is true according to Newton's first law of motion?

Correct Answer: A

Q10.What happens to a body in uniform motion according to Aristotle's fallacy?

Correct Answer: A

Q11.A stone of mass 0.25 kg is tied to a string and whirled in a circle of radius 1.5 m at 40 revolutions per minute. What is the tension in the string?

Correct Answer: B

Q12.According to Newton's second law of motion, what is the relationship between force and acceleration?

Correct Answer: B

Q13.A 0.5 kg ball is moving with a velocity of 10 m/s towards a wall. It rebounds with the same speed. What is the magnitude of the impulse imparted by the wall to the ball?

Correct Answer: B

Q14.If a force of 10 N is applied to a 2 kg object, what is the acceleration of the object?

Correct Answer: A

Q15.Which of the following statements is true about Newton's second law of motion?

Correct Answer: B

Q16.A 0.1 kg ball is moving at 5 m/s and collides elastically with a stationary 0.2 kg ball. What is the velocity of the 0.1 kg ball after the collision?

Correct Answer: C

Q17.Two blocks of masses 5 kg and 10 kg are connected by a string and placed on a frictionless surface. If a force of 30 N is applied to the 5 kg block, what is the acceleration of the system?

Correct Answer: C

Q18.Two objects, each of mass 5 kg, are moving towards each other with velocities of 2 m/s and 3 m/s respectively. They collide and stick together. What is their velocity after the collision?

Correct Answer: A

Q19.A 0.05 kg billiard ball moving at 6 m/s collides with another ball of the same mass moving in the opposite direction at the same speed. What is the impulse imparted to each ball?

Correct Answer: B

Q20.What happens to a stone in circular motion if the string breaks suddenly?

Correct Answer: B

Q21.Which of the following best describes Newton's second law of motion?

Correct Answer: A

Q22.What is the product of mass and velocity of a body known as?

Correct Answer: C

Q23.Two blocks of masses 5 kg and 10 kg are connected by a light string and placed on a frictionless surface. A force of 30 N is applied to the 10 kg block. What is the tension in the string connecting the two blocks?

Correct Answer: B

Q24.A car of mass 1000 kg is moving with a velocity of 20 m/s. It applies brakes and comes to a stop in 5 seconds. What is the average force exerted by the brakes?

Correct Answer: A

Q25.Two blocks of masses 5 kg and 10 kg are placed on a frictionless surface and are connected by a light string. A force of 30 N is applied to the 5 kg block. What is the acceleration of the system?

Correct Answer: B

Q26.If a body is moving in a circular path, what provides the inward radial acceleration?

Correct Answer: A

Q27.According to Newton's first law of motion, what happens to a body if no external force acts on it?

Correct Answer: A

Q28.A satellite is moving in a circular orbit around the Earth with a constant speed. Which of the following forces acts as the centripetal force?

Correct Answer: A

Q29.A 5 kg object is moving in a straight line with a velocity of 10 m/s. If a constant force of 15 N is applied in the direction of motion for 4 seconds, what will be the final velocity of the object?