9th Class Physics Notes 2026 – Chapter 3
Class 9 Physics Notes – Chapter 3: Dynamics
Punjab Board | 2026
A. Multiple Choice Questions
3.1. When we kick a stone, we get hurt. This is due to:
(a) Inertia
(b) Velocity
(c) Momentum
(d) Reaction
Correct Answer: (d) Reaction
Explanation: According to Newton’s third law of motion, every action has an equal and opposite reaction. When you exert force on the stone, the stone exerts an equal and opposite force on your foot, causing pain.
3.2. An object will continue its motion with constant acceleration until:
(a) The resultant force on it begins to decrease
(b) The resultant force on it is zero
(c) The resultant force on it begins to increase
(d) The resultant force is at right angle to its tangential velocity
Correct Answer: (b) The resultant force on it is zero
Explanation: By Newton’s second law, F = ma, acceleration is directly proportional to the resultant force. Constant acceleration implies a constant resultant force acting on the object.
3.3. Which of the following is a non-contact force?
(a) Friction
(b) Air resistance
(c) Electrostatic force
(d) Tension in the string
Correct Answer: (c) Electrostatic force
Explanation: Electrostatic, gravitational, and magnetic forces act at a distance without physical contact.
3.4. A ball with initial momentum p hits a solid wall and bounces back with the same velocity. Its momentum p′ after collision will be:
(a) p′ = p
(b) p′ = −p
(c) p′ = 2p
(d) p′ = −2p
Correct Answer: (b) p′ = −p
Explanation: Momentum is a vector quantity. When the ball bounces back in the opposite direction with the same speed, its momentum vector reverses sign.
3.5. A particle of mass m moving with a velocity v collides with another particle of the same mass at rest. The velocity of the first particle after collision is:
(a) v
(b) −v
(c) 0
(d) −½v
Correct Answer: (c) 0
Explanation: In a perfectly elastic head-on collision between two equal masses, where one is at rest, the moving particle transfers all its velocity and momentum to the stationary particle and comes to rest.
3.6. Conservation of linear momentum is equivalent to:
(a) Newton’s first law of motion
(b) Newton’s second law of motion
(c) Newton’s third law of motion
(d) None of these
Correct Answer: (c) Newton’s third law of motion
Explanation: The law of conservation of momentum is a direct consequence of Newton’s third law of motion. Action and reaction forces during a collision are equal and opposite.
3.7. An object with a mass of 5 kg moves at a constant velocity of 10 m s⁻¹. A constant force then acts for 5 seconds on the object and gives it a velocity of 2 m s⁻¹ in the opposite direction. The force acting on the object is:
(a) 5 N
(b) −10 N
(c) −12 N
(d) −15 N
Correct Answer: (c) −12 N
Given:
Mass, m = 5 kg
Initial velocity, vᵢ = +10 m/s
Final velocity, vᶠ = −2 m/s
Time, t = 5 s
Using:
F = m(vᶠ − vᵢ)/t
F = 5(−2 − 10)/5
F = −60/5
F = −12 N
3.8. A large force acts on an object for a very short interval of time. In this case, it is easy to determine:
(a) Magnitude of force
(b) Time interval
(c) Product of force and time
(d) None of these
Correct Answer: (c) Product of force and time
Explanation: The product of force and time is called impulse.
J = FΔt = Δp
Impulse is equal to the change in momentum.
3.9. A lubricant is usually introduced between two surfaces to decrease friction. The lubricant:
(a) Decreases temperature
(b) Acts as ball bearings
(c) Prevents direct contact of the surfaces
(d) Provides rolling friction
Correct Answer: (c) Prevents direct contact of the surfaces
Explanation: Lubricants form a thin fluid film between moving surfaces, separating them and replacing solid-to-solid friction with fluid friction.
B. Short Answer Questions
3.1. What kind of changes in motion may be produced by a force?
Answer: A force can cause an object at rest to move, speed up or slow down a moving object, change the direction of motion, or change the shape of an object.
3.2. Give five examples of contact forces.
Answer:
Frictional force
Normal force
Tension force
Air resistance (drag)
Applied pushing or pulling force
3.3. An object moves with constant velocity in free space. How long will the object continue to move with this velocity?
Answer: According to Newton’s first law of motion (inertia), the object will continue to move with constant velocity indefinitely unless acted upon by an external unbalanced force.
3.4. Define impulse of force.
Answer: Impulse is the product of a force acting on an object and the time interval for which it acts. It is mathematically equal to the change in momentum.
J = FΔt = Δp
3.5. Why has Newton’s first law not been proved on the Earth?
Answer: Newton’s first law describes an ideal situation where no external forces act. On Earth, it is impossible to completely eliminate external opposing forces such as friction and air resistance, so a moving body may eventually slow down.
3.6. When sitting in a car which suddenly accelerates from rest, you are pushed back into the seat. Why?
Answer: This is due to inertia (Newton’s first law). Your body tends to remain at rest while the car moves forward, causing you to be pressed against the seat.
3.7. The force expressed in Newton’s second law is a net force. Why is it so?
Answer: Acceleration is produced by the resultant (net) unbalanced force acting on a body, taking into account all forces acting on it, including opposing forces such as friction.
3.8. How can you show that rolling friction is less than sliding friction?
Answer: Pull a heavy block or suitcase on wheels and compare the force required with that needed to drag it across the floor. The force required when it rolls is significantly smaller than the force required to slide it.
3.9. Define terminal velocity of an object.
Answer: Terminal velocity is the constant maximum velocity reached by a falling object when the downward force of gravity equals the upward drag (air resistance) force, resulting in zero acceleration.
3.10. An astronaut walking in space wants to return to his spaceship by firing a hand rocket. In what direction does he fire the rocket?
Answer: He fires the rocket in the direction opposite to where he wants to go. By Newton’s third law, the expelled gases produce an opposite reaction that pushes him toward the spaceship.
C. Constructed Response Questions
3.1. Two ice skaters weighing 60 kg and 80 kg push off against each other on a frictionless ice track. The 60 kg skater gains a velocity of 4 m s⁻¹. Considering all the relevant calculations involved, explain how Newton’s third law applies to this situation.
Answer: According to Newton’s third law, the force exerted by skater 1 on skater 2 is equal and opposite to the force exerted by skater 2 on skater 1.
F₁₂ = −F₂₁
Since the impulses and changes in momentum are equal and opposite:
m₁v₁ = −m₂v₂
For the 60 kg skater:
60 × 4 = 240 kg·m/s
For the 80 kg skater:
80v₂ = −240
v₂ = −240/80 = −3 m/s
Therefore, the 80 kg skater moves in the opposite direction at a speed of 3 m/s.
3.2. Inflatable airbags are installed in vehicles as safety equipment. In terms of momentum, what is the advantage of airbags over seatbelts?
Answer: Airbags increase the impact time, Δt, during a collision. Since:
F = Δp/Δt
increasing the time interval reduces the magnitude of the impact force experienced by the passenger, helping to prevent severe injury.
3.3. A horse refuses to pull a cart. The horse argues, “According to Newton’s third law, whatever force I exert on the cart, the cart will exert an equal and opposite force on me. Since the net force will be zero, therefore, I have no chance of accelerating (pulling) the cart.” What is wrong with this reasoning?
Answer: The horse considers the action and reaction forces as if they act on the same body. However, action and reaction forces always act on two different bodies.
The horse pulls the cart forward, and the cart pulls the horse backward. The horse also pushes backward against the ground with its feet. The ground pushes the horse forward through friction, enabling the horse and cart to accelerate when the forward force is greater than the opposing resistance.
3.4. When a cricket ball is hit high, a fielder tries to catch it. While holding the ball, he or she draws their hands backward. Why?
Answer: Drawing the hands backward increases the time interval, Δt, over which the ball’s momentum is brought to zero. Since:
F = Δp/Δt
increasing the time reduces the impact force on the hands and helps prevent injury.
3.5. When someone jumps from a small boat onto the river bank, why does the jumper often fall into the water? Explain.
Answer: When jumping forward, the person exerts a backward force on the boat. By Newton’s third law, the boat moves backward away from the bank, increasing the effective distance the jumper must cover and potentially causing the person to fall short into the water.
3.6. Imagine that friction suddenly vanishes from everything. What could be the scenario of daily life activities?
Answer: Without friction:
We would not be able to walk or run because our feet would slip.
Cars could not accelerate, brake, or turn properly.
Nails would slip out of walls.
Knots would come undone.
Writing with a pencil on paper would become impossible.
D. Comprehensive Questions
3.1. Explain the concept of force with practical examples.
Answer: A force is a push or pull that changes or tends to change the state of rest or uniform motion of a body, its direction, or its shape.
Examples include pushing a cart, kicking a football, and pulling a drawer.
3.2. Describe Newton’s laws of motion.
Answer:
First Law (Law of Inertia): Every body continues in its state of rest or uniform motion in a straight line unless acted upon by an external unbalanced force.
Second Law: The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
F = ma
Third Law: To every action, there is always an equal and opposite reaction.
3.3. Define momentum and express Newton’s second law of motion in terms of change in momentum.
Answer: Momentum (p) is the quantity of motion possessed by a body. It is defined as the product of mass and velocity.
p = mv
Newton’s second law states that the rate of change of momentum of a body is directly proportional to the applied net force.
F = Δp/Δt
3.4. State and explain the principle of conservation of momentum.
Answer: The total linear momentum of an isolated system of two or more interacting bodies remains constant, provided no external resultant force acts on the system.
For two bodies:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Here, u₁ and u₂ are the initial velocities, while v₁ and v₂ are the final velocities.
3.5. Describe the motion of a block on a table, taking into account the friction between the two surfaces. What are static friction and kinetic friction?
Answer: When a pulling force is applied to a block on a table, static friction opposes the impending motion up to a maximum limit called limiting friction. Once the block starts moving, kinetic friction opposes its sliding motion and is generally slightly less than static friction.
3.6. Explain the effect of friction on the motion of vehicles in the context of tyre surfaces and braking force.
Answer: Friction between tyres and the road provides the necessary centripetal force for turning and traction for acceleration. During braking, friction helps slow down the vehicle and bring it to a halt.
E. Numerical Problems
3.1. A 10 kg block is placed on a smooth horizontal surface. A horizontal force of 5 N is applied to the block. Find: (a) the acceleration produced in the block, and (b) the velocity of the block after 5 seconds.
Answer:
(a) Calculate acceleration
Given:
Mass, m = 10 kg
Force, F = 5 N
Using:
a = F/m
a = 5/10 = 0.5 m/s²
(b) Calculate final velocity
Given:
Initial velocity, vᵢ = 0
Time, t = 5 s
Acceleration, a = 0.5 m/s²
Using:
vᶠ = vᵢ + at
vᶠ = 0 + (0.5 × 5) = 2.5 m/s
3.2. The mass of a person is 80 kg. What will be his weight on the Earth? What will be his weight on the Moon? The value of acceleration due to gravity on the Moon is 1.6 m s⁻².
Answer:
Given:
Mass, m = 80 kg
Earth’s gravitational acceleration, gₑₐᵣₜₕ = 10 m/s²
Moon’s gravitational acceleration, gₘₒₒₙ = 1.6 m/s²
(a) Weight on Earth
Wₑₐᵣₜₕ = mgₑₐᵣₜₕ
Wₑₐᵣₜₕ = 80 × 10 = 800 N
(b) Weight on the Moon
Wₘₒₒₙ = mgₘₒₒₙ
Wₘₒₒₙ = 80 × 1.6 = 128 N
3.3. What force is required to increase the velocity of an 800 kg car from 10 m s⁻¹ to 30 m s⁻¹ in 10 seconds?
Answer:
Given:
Mass, m = 800 kg
Initial velocity, vᵢ = 10 m/s
Final velocity, vᶠ = 30 m/s
Time, t = 10 s
Using:
F = m(vᶠ − vᵢ)/t
F = 800(30 − 10)/10
F = 800 × 20/10
F = 1600 N
3.4. A 5 g bullet is fired by a gun. The bullet moves with a velocity of 300 m s⁻¹. If the mass of the gun is 10 kg, find the recoil speed of the gun.
Answer:
Given:
Mass of bullet, m₁ = 5 g = 0.005 kg
Velocity of bullet, v₁ = 300 m/s
Mass of gun, m₂ = 10 kg
Recoil velocity of gun, v₂ = ?
By conservation of momentum:
m₁v₁ + m₂v₂ = 0
(0.005 × 300) + (10v₂) = 0
1.5 + 10v₂ = 0
v₂ = −1.5/10
v₂ = −0.15 m/s
The recoil speed of the gun is 0.15 m/s, opposite to the direction of the bullet.
3.5. An astronaut weighs 70 kg. He throws a wrench of mass 300 g at a speed of 3.5 m s⁻¹. Determine: (a) the speed of the astronaut as he recoils away from the wrench, and (b) the distance covered by the astronaut in 30 minutes.
Answer:
Given:
Mass of astronaut, m₁ = 70 kg
Mass of wrench, m₂ = 300 g = 0.3 kg
Velocity of wrench, v₂ = 3.5 m/s
(a) Calculate recoil speed
By conservation of momentum:
m₁v₁ + m₂v₂ = 0
v₁ = −m₂v₂/m₁
v₁ = −(0.3 × 3.5)/70
v₁ = −1.05/70
v₁ = −0.015 m/s
The astronaut recoils at a speed of 0.015 m/s, opposite to the wrench.
(b) Calculate distance travelled
Time:
t = 30 × 60 = 1800 s
Using S = vt:
S = 0.015 × 1800
S = 27 m
3.6. A 6.5 × 10³ kg bogie of a goods train is moving with a velocity of 0.8 m s⁻¹. Another bogie of mass 9.2 × 10³ kg coming from behind with a velocity of 1.2 m s⁻¹ collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.
Answer:
Given:
Mass of first bogie, m₁ = 6.5 × 10³ kg
Velocity of first bogie, v₁ = 0.8 m/s
Mass of second bogie, m₂ = 9.2 × 10³ kg
Velocity of second bogie, v₂ = 1.2 m/s
By conservation of momentum:
m₁v₁ + m₂v₂ = (m₁ + m₂)vᶠ
Substituting values:
(6.5 × 10³ × 0.8) + (9.2 × 10³ × 1.2) = (6.5 × 10³ + 9.2 × 10³)vᶠ
5200 + 11040 = 15700vᶠ
16240 = 15700vᶠ
vᶠ = 16240/15700
vᶠ ≈ 1.03 m/s
3.7. A cyclist weighing 55 kg rides a bicycle of mass 5 kg. He starts from rest and applies a force of 90 N for 8 seconds. Then he continues at a constant speed for another 8 seconds. Calculate the total distance travelled by the cyclist.
Answer:
Given:
Mass of cyclist = 55 kg
Mass of bicycle = 5 kg
Applied force, F = 90 N
First time interval, t₁ = 8 s
Second time interval, t₂ = 8 s
Total mass:
m = 55 + 5 = 60 kg
Acceleration:
a = F/m = 90/60 = 1.5 m/s²
Distance during the first 8 seconds
Since the cyclist starts from rest, vᵢ = 0.
S₁ = vᵢt₁ + ½at₁²
S₁ = 0 + ½ × 1.5 × 8²
S₁ = 0.75 × 64 = 48 m
Final velocity after 8 seconds:
vᶠ = vᵢ + at
vᶠ = 0 + (1.5 × 8) = 12 m/s
Distance during the next 8 seconds
The cyclist continues at a constant speed of 12 m/s.
S₂ = vt₂
S₂ = 12 × 8 = 96 m
Total distance:
S = S₁ + S₂
S = 48 + 96 = 144 m
3.8. A ball of mass 0.4 kg is dropped on the floor from a height of 1.8 m. The ball rebounds straight upward to a height of 0.8 m. What is the magnitude and direction of the impulse applied to the ball by the floor?
Answer:
Given:
Mass, m = 0.4 kg
Initial height, h₁ = 1.8 m
Rebound height, h₂ = 0.8 m
Gravitational acceleration, g = 10 m/s²
Velocity just before hitting the floor
v₁ = √(2gh₁)
v₁ = √(2 × 10 × 1.8)
v₁ = √36 = 6 m/s
Taking upward as positive, the velocity before impact is −6 m/s.
Velocity just after hitting the floor
v₂ = √(2gh₂)
v₂ = √(2 × 10 × 0.8)
v₂ = √16 = 4 m/s
The velocity after impact is +4 m/s.
Calculate impulse
J = Δp = m(v₂ − v₁)
J = 0.4[4 − (−6)]
J = 0.4 × 10
J = 4 N·s
The impulse has a magnitude of 4 N·s and acts upward.
3.9. Two balls of masses 0.2 kg and 0.4 kg are moving towards each other with velocities 20 m s⁻¹ and 5 m s⁻¹, respectively. After collision, the velocity of the 0.2 kg ball becomes 6 m s⁻¹. What will be the velocity of the 0.4 kg ball?
Answer:
Given:
Mass of first ball, m₁ = 0.2 kg
Initial velocity of first ball, u₁ = +20 m/s
Mass of second ball, m₂ = 0.4 kg
Initial velocity of second ball, u₂ = −5 m/s
Final velocity of first ball, v₁ = +6 m/s, using the supplied solution’s assumption that it rebounds forward
Final velocity of second ball, v₂ = ?
By conservation of momentum:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
(0.2 × 20) + (0.4 × −5) = (0.2 × 6) + (0.4v₂)
4 − 2 = 1.2 + 0.4v₂
2 = 1.2 + 0.4v₂
0.8 = 0.4v₂
v₂ = 0.8/0.4
v₂ = +2 m/s
The positive sign indicates that the 0.4 kg ball moves in the direction chosen as positive.
Notes by Sir Dilawar
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