9th Class Physics
Chapter-3: Dynamics
A. Multiple Choice Questions
Tick (✓) the correct option
3.1 When we kick a stone, we get hurt. This is due to:
Explanation: According to Newton's third law, for every action there is an equal and opposite reaction. When your foot exerts a force on the stone, the stone exerts an equal and opposite force back on your foot, which you feel as pain.
3.2 An object will continue its motion with constant acceleration until:
Explanation: Newton's second law states F = ma. If the net (resultant) force F is constant and non-zero, the acceleration a remains constant. Only when F becomes zero does acceleration cease (a = 0), causing the object to move with constant velocity.
3.3 Which of the following is a non-contact force?
Explanation: Electrostatic force acts between charged objects without physical contact. Friction, air resistance, and tension all require direct contact between surfaces or objects.
3.4 A ball with initial momentum p hits a solid wall and bounces back with the same speed. Its momentum p′ after collision will be:
Explanation: Momentum is a vector. Rebounding with the same speed reverses the velocity direction, hence the momentum changes sign: p′ = –p.
3.5 A particle of mass m moving with velocity v collides with another particle of the same mass at rest. The velocity of the first particle after collision is:
Explanation: In an elastic head-on collision between equal masses where one is initially at rest, the moving mass stops and transfers all its velocity to the initially stationary mass (conserving both momentum and kinetic energy).
3.6 Conservation of linear momentum is equivalent to:
Explanation: The law of conservation of linear momentum arises directly from Newton's third law: internal action–reaction pairs cancel, so the total momentum of an isolated system remains constant.
3.7 An object of mass 5 kg moves at a constant velocity of 10 m s⁻¹. A constant force then acts for 5 s on the object and gives it a velocity of 2 m s⁻¹ in the opposite direction. The force acting on the object is:
Solution:
Take the original direction as positive.
Initial velocity: v₀ = +10 m s⁻¹
Final velocity: v = –2 m s⁻¹ (opposite direction)
Time interval: Δt = 5 s
Acceleration:
a = (v – v₀)/Δt = (–2 – 10)/5 = –12/5 = –2.4 m s⁻²
Force:
F = m a = 5 kg × (–2.4 m s⁻²) = –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:
Explanation: When the interaction time is extremely short, measuring the exact force or duration individually is hard. However, their product—the impulse (F Δt)—equals the measurable change in momentum (Δp), so it is the quantity that can be reliably determined.
3.9 A lubricant is usually introduced between two surfaces to decrease friction. The lubricant:
Explanation: By forming a thin fluid film, the lubricant keeps the solid surfaces apart, thereby replacing high-friction solid–solid contact with lower-friction fluid shear.
Newton's three laws of motion form the foundation of dynamics
B. Short Answer Questions
3.1 What kind of changes in motion may be produced by a force?
Answer: A force can change the speed (magnitude of velocity), the direction of motion, or both—i.e., it produces acceleration.
3.2 Give 5 examples of contact forces.
Answer:
- Friction between sliding surfaces
- Normal force from a table on a book
- Tension in a rope pulling a cart
- Air resistance on a moving cyclist
- Applied push/pull when you open a door
3.3 An object moves with constant velocity in free space. How long will the object continue to move with this velocity?
Answer: Forever; with no net external force, Newton's 1st law guarantees the velocity stays constant indefinitely.
3.4 Define impulse of force.
Answer: Impulse = F Δt (average force × time interval)
It equals the change in momentum of the object: J = Δp.
3.5 Why has Newton's first law not been proved on the Earth?
Answer: Earth-bound objects always experience at least one external force (gravity, air resistance, friction), so we can never achieve a perfectly isolated system to demonstrate the law directly.
3.6 When sitting in a car that suddenly accelerates from rest, you are pushed back into the seat—why?
Answer: Your body tends to stay at rest due to inertia; the seat accelerates forward beneath you, so its back presses forward on you, creating the sensation of being "pushed" backward.
3.7 The force expressed in Newton's second law is a net force. Why is it so?
Answer: F_net = m a; only the vector sum of all forces determines the actual acceleration. Individual forces may cancel, leaving zero net force and no acceleration.
3.8 How can you show that rolling friction is less than sliding friction?
Answer: Place identical masses on two identical inclined planes: one slides, one rolls. The rolling object accelerates more and reaches the bottom sooner, indicating a smaller retarding force.
3.9 Define terminal velocity of an object.
Answer: The maximum constant speed reached when the drag force equals the gravitational force, producing zero net force and hence 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 away from the spaceship (action). By Newton's 3rd law, the reaction thrust pushes him toward the spaceship.
Astronaut using rocket thrust to return to spaceship - an application of Newton's third law
C. Constructed Response Questions
3.1 Two ice skaters (60 kg and 80 kg) push off on frictionless ice; the 60 kg skater moves away at 4 m s⁻¹. Show how Newton's 3rd law applies.
Answer:
Action–reaction pair: Each skater exerts an equal-magnitude, opposite-direction force on the other for the same time Δt.
Impulse–momentum:
– Impulse on 60 kg skater: J = +60 kg × 4 m s⁻¹ = 240 N s.
– Therefore impulse on 80 kg skater must be −240 N s.
– Velocity of 80 kg skater: v = −240 N s / 80 kg = −3 m s⁻¹.
Conclusion: Equal and opposite impulses produce equal and opposite momentum changes—exactly Newton's 3rd law in action.
3.2 Air-bags vs. seat-belts: momentum advantage.
Answer:
Same Δp must be delivered to the passenger during a crash.
Air-bag increases Δt (time to stop) five- to ten-fold compared with a belt alone.
F_avg = Δp / Δt, so force on chest drops dramatically, reducing injury.
Seat-belt alone gives a short Δt (high F); air-bag spreads impulse, cutting peak force.
3.3 "Horse can't pull cart because action equals reaction, so net force is zero." What is wrong?
Answer:
The two forces (horse on cart and cart on horse) act on different bodies; they never cancel for either body.
Net force on the cart = pull from horse − friction; if that net is > 0, cart accelerates.
Net force on the horse = push from ground − pull from cart; if that net is > 0, horse accelerates.
Ground's external horizontal force is what allows the system to accelerate.
3.4 Fielder draws hands backward while catching a fast ball—why?
Answer:
Impulse needed: J = Δp = m Δv (fixed by ball's mass and speed).
J = F_avg Δt; by increasing Δt (pulling hands back), fielder reduces F_avg on hands/palms.
Lower force → less sting and lower chance of the ball popping out.
3.5 Jumping from a small boat to river-bank, why does the jumper often fall backward into water?
Answer:
To jump, the person pushes backward on the boat (action).
By Newton's 3rd law, the boat pushes the person forward (reaction).
Light boat accelerates backward significantly; person's forward velocity relative to shore is therefore smaller than expected.
Result: horizontal reach is insufficient—center of gravity lands short, so the jumper topples backward into the water.
3.6 Imagine friction suddenly vanishes everywhere—describe daily-life consequences.
Answer:
- Walking: No grip; every step makes feet slide backward, people fall.
- Transport: Cars, bikes, trains spin wheels uselessly; brakes don't work—collisions everywhere.
- Objects: Anything on an incline slides; furniture, utensils, books glide to the floor.
- Knots & fasteners: Knots untie, screws/nails slip out, buildings may collapse.
- Life processes: Even muscle–tendon friction needed for joint movement would fail—biomechanical chaos.
Net result: Human society, machinery, and biology rely on friction; its disappearance would bring immediate paralysis and danger.
Consequences of a world without friction
9th Class Physics
Chapter 3: Dynamics
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.
(b) the velocity of block after 5 seconds.
Given:
m = 10 kg
F = 5 N (horizontal, smooth surface ⇒ no friction)
(a) Acceleration
Newton's 2nd law:
a = F / m = 5 N / 10 kg = 0.5 m s⁻²
(b) Velocity after 5 s
Starts from rest (u = 0):
v = u + a t = 0 + (0.5 m s⁻²)(5 s) = 2.5 m s⁻¹
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 of Moon is 1.6 ms².
Given:
m = 80 kg
gearth = 9.8 m s⁻²
gmoon = 1.6 m s⁻²
Weight
Earth: W = m g = 80 kg × 9.8 m s⁻² = 784 N
Moon: W = 80 kg × 1.6 m s⁻² = 128 N
3.3 What force is required to increase the velocity of 800 kg car from 10 ms¹ to 30 m s¹ in 10 seconds?
Given:
m = 800 kg
u = 10 m s⁻¹
v = 30 m s⁻¹
t = 10 s
Required force
a = (v – u)/t = (30 – 10)/10 = 2 m s⁻²
F = m a = 800 kg × 2 m s⁻² = 1,600 N
3.4 A 5 g bullet is fired by a gun. The bullet moves with a velocity of 300 m s. If the mass of the gun is 10 kg, find the recoil speed of the gun.
Given:
mbullet = 5 g = 0.005 kg
vbullet = 300 m s⁻¹
mgun = 10 kg
By conservation of momentum (initial total = 0):
0 = mbullet vbullet + mgun vgun
⇒ vgun = – (mbullet vbullet)/mgun
= – (0.005 kg × 300 m s⁻¹)/10 kg
= – 0.15 m s⁻¹ (negative sign indicates opposite direction)
3.5 An astronaut weighs 70 kg. He throws a wrench of mass 300 g at a speed of 3.5 m s. Determine:
(a) the speed of astronaut as he recoils away from the wrench.
(b) the distance covered by the astronaut in 30 minutes.
Given:
mast = 70 kg
mwrench = 300 g = 0.3 kg
vwrench = 3.5 m s⁻¹ (thrown away from astronaut)
(a) Recoil speed of astronaut
Momentum conservation (initial total = 0):
0 = mast vast + mwrench vwrench
⇒ vast = – (mwrenchvwrench)/mast
= – (0.3 kg × 3.5 m s⁻¹)/70 kg
= – 0.015 m s⁻¹
= – 1.5 × 10⁻² m s⁻¹
(b) Distance in 30 min
t = 30 min = 1,800 s
d = |vast| t = 0.015 m s⁻¹ × 1,800 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. Another bogie of mass 9.2 × 10³ kg coming from behind with a velocity of 1.2 m s collides with the first one and couples to it. Find the common velocity of the two bogies after they become coupled.
Given:
m₁ = 6.5 × 10³ kg v₁ = 0.8 m s⁻¹
m₂ = 9.2 × 10³ kg v₂ = 1.2 m s⁻¹ (same direction)
Perfectly inelastic collision (coupled after impact):
Momentum conservation:
m₁ v₁ + m₂ v₂ = (m₁ + m₂) vcommon
vcommon = (m₁ v₁ + m₂ v₂)/(m₁ + m₂)
= [(6.5 × 0.8) + (9.2 × 1.2)] / (6.5 + 9.2)
= (5.2 + 11.04) / 15.7
= 16.24 / 15.7
≈ 1.03 m s⁻¹
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.
Given:
total mass m = 55 kg + 5 kg = 60 kg
driving force F = 90 N (assumed horizontal, no friction)
phase 1: 8 s under force, starting from rest
phase 2: next 8 s with force removed (constant speed)
Phase 1 – accelerated motion
a = F / m = 90 N / 60 kg = 1.5 m s⁻²
Distance while force acts:
s₁ = ½ a t₁² = ½ (1.5 m s⁻²)(8 s)² = 0.75 × 64 = 48 m
Speed reached at end of phase 1:
v = a t₁ = 1.5 m s⁻² × 8 s = 12 m s⁻¹
Phase 2 – constant speed
t₂ = 8 s speed = 12 m s⁻¹
s₂ = v t₂ = 12 m s⁻¹ × 8 s = 96 m
Total distance
stotal = s₁ + s₂ = 48 m + 96 m = 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?
Given:
m = 0.4 kg
drop height h₁ = 1.8 m
rebound height h₂ = 0.8 m
g = 9.8 m s⁻²
Impulse = change of momentum.
Take upward as positive.
Speed just before impact (downward):
v₁ = √(2 g h₁) = √(2·9.8·1.8) = √35.28 ≈ –5.94 m s⁻¹ (negative because down)
Speed just after impact (upward):
v₂ = √(2 g h₂) = √(2·9.8·0.8) = √15.68 ≈ +3.96 m s⁻¹
Impulse delivered by floor
J = m(v₂ – v₁)
= 0.4 kg [3.96 – (–5.94)] m s⁻¹
= 0.4 × 9.90
≈ 3.96 kg m s⁻¹
Magnitude: 4.0 N s (2-significant-figure rounding)
Direction: vertically upward (that is the direction of the force the floor exerted on the ball).
3.9 Two balls of masses 0.2 kg and 0.4 kg are moving towards each other with velocities 20 m s and 5 m s respectively. After collision, the velocity of 0.2 kg ball becomes 6 ms¹. What will be the velocity of 0.4 kg ball?
Given:
Take the original direction of the 0.2 kg ball as positive.
m₁ = 0.2 kg, u₁ = +20 m s⁻¹
m₂ = 0.4 kg, u₂ = –5 m s⁻¹ (moving the other way)
After collision: v₁ = +6 m s⁻¹ (given)
Conservation of momentum (no external forces):
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
0.2(20) + 0.4(–5) = 0.2(6) + 0.4 v₂
4 – 2 = 1.2 + 0.4 v₂
2 = 1.2 + 0.4 v₂
0.4 v₂ = 0.8
v₂ = 2 m s⁻¹
The 0.4 kg ball moves at 2 m s⁻¹ in the same original direction the 0.2 kg ball had (i.e. the positive direction).
Step-by-step approach to solving numerical problems in physics