RowQ
The Vault
RowQ
The Vault
CBSE Class 9 Physics · 12 questions · 29 marks
Why does a passenger lurch forward when a bus brakes, and why does a rifle kick back when fired? Newton answered both with three short laws, and this chapter turns those laws into tools you can calculate with. Once you can write F = ma confidently and track momentum before and after a collision, most numericals in this unit solve themselves in three lines.
A wooden crate is pushed with a steady horizontal force but slides across the floor at a constant velocity. What can you conclude?
Answer
The applied force and friction are balanced, so the net force is zero — constant velocity means zero acceleration, and by F = ma a zero acceleration requires a zero net force. Friction must therefore be acting backwards with exactly the same magnitude as the applied push.
A trolley of mass 5 kg is acted on by an unbalanced force of 20 N. What acceleration does it acquire?
Answer
4 m/s² — from Newton's second law F = ma, so a = F/m = 20/5 = 4 m/s². The trolley's velocity increases by 4 metres per second every second while the force acts.
Two bodies have the same momentum. Body X has a larger mass than body Y. Which statement is true?
Answer
Y is moving faster than X — since p = mv is the same for both, mass and velocity are inversely related. The lighter body Y must have the greater velocity to carry the same momentum as the heavier body X.
A swimmer pushes the wall of the pool backwards with her feet and moves forward. Which law best explains this and what is the reaction force?
Answer
Third law; the wall pushes her forward with an equal and opposite force — the action is the backward push of her feet on the wall, and the reaction is the equal forward push of the wall on her feet. The two forces act on different bodies, which is why the swimmer actually moves.
Assertion (A): Action and reaction forces are equal in magnitude and opposite in direction, yet a football still accelerates when kicked. Reason (R): The action and reaction forces of a Newton's third law pair act on two different bodies and so cannot cancel each other.
Answer
Both A and R are true and R is the correct explanation of A — forces cancel only when they act on the same body. Here the foot's force acts on the ball and the ball's equal, opposite force acts on the foot, so the ball experiences an unbalanced force and accelerates away.
Explain, using inertia, why the dust flies out of a carpet when it is beaten with a stick.
Answer
When the carpet is struck, the fibres of the carpet are suddenly set into motion, but the dust particles resting in and on them tend to stay where they are because of their inertia of rest. The carpet moves away from under them while the dust stays behind, so the dust separates from the fabric and then falls out under gravity.
A 0.15 kg cricket ball moving at 30 m/s towards a fielder is caught, and she brings it to rest in 0.5 s. Calculate the change in momentum and the average force she must apply.
Answer
Taking the ball's initial direction as positive, u = 30 m/s and v = 0, with m = 0.15 kg. Initial momentum = mu = 0.15 × 30 = 4.5 kg·m/s. Final momentum = mv = 0. Change in momentum = 0 - 4.5 = -4.5 kg·m/s, i.e. a change of 4.5 kg·m/s directed opposite to the ball's motion. Average force F = change in momentum / time = -4.5/0.5 = -9 N. So she must apply an average force of 9 N against the ball's motion. Drawing her hands back to stretch the stopping time would reduce this force.
State Newton's second law of motion and use it to define the newton.
Answer
Newton's second law states that the rate of change of momentum of a body is directly proportional to the unbalanced force applied on it, and the change takes place in the direction of that force. For constant mass this gives F = ma. Using this relation, one newton is defined as the force which, acting on a body of mass 1 kg, produces in it an acceleration of 1 m/s². Hence 1 N = 1 kg·m/s².
A 4 kg trolley moving at 6 m/s collides head-on with a stationary 2 kg trolley, and the two lock together. Find their common velocity after the collision.
Answer
Let the direction of the first trolley be positive. Before the collision: m₁ = 4 kg, u₁ = 6 m/s; m₂ = 2 kg, u₂ = 0. Total momentum before = (4 × 6) + (2 × 0) = 24 kg·m/s. After the collision the two move together with a common velocity v, so total mass = 4 + 2 = 6 kg and total momentum after = 6v. By conservation of momentum, 6v = 24, so v = 4 m/s. The joined trolleys move off at 4 m/s in the original direction of the 4 kg trolley.
(a) State the law of conservation of momentum and explain how it follows from Newton's third law. (b) A 3 kg air-gun fires a 0.02 kg pellet at 150 m/s. Calculate the recoil velocity of the gun and comment on why the recoil feels far gentler than the pellet's speed suggests.
Answer
(a) The law of conservation of momentum states that when two or more bodies interact and no external unbalanced force acts on the system, the total momentum of the system before the interaction equals the total momentum after it. It follows from Newton's third law because during the interaction each body exerts an equal and opposite force on the other for exactly the same duration. Equal and opposite forces acting for the same time produce equal and opposite changes of momentum, so the two changes cancel and the total momentum of the system stays unchanged. (b) Before firing, both gun and pellet are at rest, so the total momentum of the system is zero. After firing, momentum of pellet = 0.02 × 150 = 3 kg·m/s forward. Let the recoil velocity of the gun be v. Total momentum after = 3v + 3 = 0 (taking forward as positive), so 3v = -3 and v = -1 m/s. The gun recoils at 1 m/s, opposite to the pellet. The recoil speed is small because the gun's mass is 150 times that of the pellet, and the same momentum shared by a much larger mass gives a much smaller velocity — which is why the shooter feels only a mild backward push.
A 50 kg sledge is pulled along level snow by a horizontal rope with a force of 90 N, while friction opposes the motion with a force of 40 N. (a) Draw in words the forces acting and find the net force. (b) Find the acceleration of the sledge. (c) If it starts from rest, find its velocity and the distance covered after 6 s. (d) State what would happen if the pulling force were reduced to exactly 40 N.
Answer
(a) Four forces act on the sledge: the weight acting vertically downward, the normal reaction of the snow acting vertically upward (these two balance each other), the applied pull of 90 N forward, and friction of 40 N backward. Net horizontal force = 90 - 40 = 50 N in the direction of the pull. (b) Using F = ma, a = F/m = 50/50 = 1 m/s². (c) Starting from rest, u = 0, a = 1 m/s², t = 6 s. Velocity: v = u + at = 0 + 1 × 6 = 6 m/s. Distance: s = ut + ½at² = 0 + ½ × 1 × (6)² = 18 m. (d) If the pull were reduced to 40 N it would exactly balance friction, making the net force zero. The sledge would then stop accelerating; if already moving it would continue at whatever constant velocity it had at that moment, and if at rest it would stay at rest.
Read the following and answer the questions that follow: A road-safety team demonstrates seat belts using a test trolley carrying a 40 kg dummy. The trolley runs at 15 m/s into a barrier. In the first run the dummy is unbelted and is thrown forward, stopping in 0.05 s against the dashboard. In the second run a belt stretches slightly and stops the same dummy in 0.4 s. (a) Which law of motion explains why the unbelted dummy keeps moving forward after the trolley stops? (b) Calculate the average stopping force on the dummy in the first run. (c) Calculate the average stopping force in the second run. (d) Using your two answers, explain how seat belts and crumple zones reduce injury.
Answer
(a) Newton's first law, the law of inertia. The barrier stops the trolley, but no force acts immediately on the dummy, so its inertia of motion keeps it travelling forward at 15 m/s until something stops it. (b) Change in momentum = m(v - u) = 40 × (0 - 15) = -600 kg·m/s. Force = change in momentum / time = -600/0.05 = -12000 N, i.e. an average stopping force of 12000 N. (c) With the belt, force = -600/0.4 = -1500 N, i.e. an average stopping force of 1500 N. (d) The change in momentum is identical in both runs, so the only way to reduce the force is to increase the time over which the momentum changes. Stretching the stop from 0.05 s to 0.4 s cuts the force from 12000 N to 1500 N — eight times smaller. Seat belts stretch, airbags inflate, and crumple zones deform for exactly this reason: they lengthen the collision time and so lower the force on the body.
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