A ball of 0·1 kg mass is dropped on a hard floor from a height of 0·45 m and rises to a height of 0·20 m. If it was in touch with the floor for 0·1 s, the net force it applied on the floor while bouncing is: (take the gravitational acceleration g = 10 m s⁻²)
- (a)1·0 N
- (b)6·0 N
- (c)3·0 N
- (d)5·0 N
Correct — D, 5·0 N (our pick, on a disputed item). Just before impact the ball's speed is sqrt(2 x 10 x 0.45) = 3 m/s downward; just after it is sqrt(2 x 10 x 0.20) = 2 m/s upward. The change in momentum is Delta_p = 0.1 x (2 - (-3)) = 0.5 kg m/s, so the average net force during the 0.1 s of contact is Delta_p/t = 0.5/0.1 = 5 N. Reading 'net force' as this rate of change of momentum gives 5 N (option d). Be honest that the item is split: if instead you want the actual contact (normal) force the ball pushes on the floor with, you must add the weight, giving 5 + 1 = 6 N (option b), which some keys mark. We choose (d) because the question says 'net force', but the 6 N reading is defensible.
- (a)1·0 N — This is just the ball's weight, mg = 0.1 x 10 = 1 N. It ignores the change in momentum on bouncing entirely, so it cannot be the force involved in the impact.
- (b)6·0 N — This is the strong alternative, not a careless error: it is the full normal (contact) force = Delta_p/t + mg = 5 + 1 = 6 N, i.e. the force the ball actually presses on the floor by Newton's third law. Because the question's wording ('net force ... applied on the floor') can be read either way, this item is disputed; we mark (d) only because of the word 'net'.
- (c)3·0 N — This comes from using only the downward speed (m x v_down / t = 0.1 x 3 / 0.1 = 3 N) and forgetting that the rebound reverses the momentum. The correct change adds the upward and downward momenta, giving 5 N.
By the impulse-momentum theorem, the average force on a body during a collision equals its change in momentum divided by the contact time: F = Delta_p / t. Momentum is a vector, so a bounce reverses its sign and the change is the sum of the incoming and outgoing magnitudes. The heights before and after the bounce fix the speeds through v = sqrt(2gh).
The whole difficulty is the phrase 'net force it applied on the floor'. During contact the ball feels two forces — the upward push from the floor (the normal force N) and its own weight downward (mg). The NET force on the ball equals Delta_p/t = 5 N, and this net force is what accelerates it. The actual contact force is N = Delta_p/t + mg = 6 N, and by Newton's third law the ball presses on the floor with that same 6 N. So 'net force' points to 5 N while 'force on the floor' points to 6 N — which is why examiners and coaching keys disagree.
- Speed just before impact: v = sqrt(2 x 10 x 0.45) = 3 m/s (downward).
- Speed just after impact: v = sqrt(2 x 10 x 0.20) = 2 m/s (upward).
- Change in momentum: Delta_p = 0.1 x (2 - (-3)) = 0.5 kg m/s; net force = Delta_p/t = 5 N.
- Full contact/normal force = net force + weight = 5 + 1 = 6 N (the competing answer).
- Forgetting that momentum reverses on the bounce, so the change adds the up and down speeds (5, not 1 or 3).
- Mixing up the net force (Delta_p/t) with the actual contact force (Delta_p/t + mg) — the source of the 5 N vs 6 N dispute.
Asked as a bouncing-ball or bat-and-ball impact: find the impulse, the average force, or the contact force from the drop and rebound heights.
Assertion (A): A man standing on a completely frictionless surface can propel himself by whistling. Reason (R): If no external force acts on a system, its momentum cannot change.
- (a) Both A and R are true, and R is the correct explanation of A
- (b) Both A and R are true, but R is not a correct explanation of A
- (c) A is true, but R is false
- (d) A is false, but R is true
Answer(a) Both A and R are true, and R is the correct explanation of A
Same concept — change of momentum and force. UPSC's reason states the momentum principle (force relates to change of momentum) that underlies the impulse calculation in this NDA bouncing-ball problem.
- practice — not a real PYQ
A ball dropped from height h rebounds to a smaller height. The speed just before impact is found from:
- (a)v = gh
- (b)v = sqrt(2gh)
- (c)v = 2gh
- (d)v = sqrt(gh)
Answer(b) v = sqrt(2gh) — free-fall kinematics relate the drop height to the impact speed.
- practice — not a real PYQ
The average force during a collision equals:
- (a)change in momentum x contact time
- (b)change in momentum / contact time
- (c)mass x height
- (d)weight x contact time
Answer(b) change in momentum / contact time — this is the impulse-momentum theorem, F = Delta_p / t.