Momentum & impulse — Physical Sciences | Learn on Stanmorephysics

5 exam-style question sets on Momentum & impulse (Paper 1), each with a hint and a fully worked answer. The app holds 25 questions on this section in total, including variants of every set below, and lets you mark yourself part by part.

Spring separation & energy on a rampMomentum–kinetic energy relationshipImpulse & change in momentumMomentum vs contact time graph (hockey)Momentum ratios (p = mv)

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Question 1

Spring separation & energy on a rampApplication12 marksPaper 1

Two gliders, N (mass 1,2 kg) and P (mass 1,8 kg), are held at rest on a frictionless horizontal air track with a compressed light spring between them. When they are released, the spring expands to its natural length and then drops away.

Glider N moves to the left. Glider P moves to the right and then runs up a frictionless curved ramp at the end of the track, rising to a maximum vertical height of 0,8 m.

Ignore air resistance.

aState the principle of conservation of mechanical energy in words. (2)
bCalculate the speed of glider P at the bottom of the ramp. (4)
cCalculate the change in momentum of glider P while the spring expands. (3)
dWrite down the change in momentum of glider N over the same time interval. (1)
eCalculate the speed of glider N just after the spring drops away. (2)

Hint

Use energy conservation on the ramp to get P's speed; the spring gives the two gliders equal and opposite changes in momentum.

Worked answer

a) In an isolated system (no friction or air resistance), the total mechanical energy — the sum of the gravitational potential energy and the kinetic energy — remains constant. ✓✓

b) On the frictionless ramp: $(E_{k} + E_{p})_{b o tt o m} = (E_{k} + E_{p})_{t o p}$ ✓
$\frac{1}{2} (1, 8) v^{2} + 0 = 0 + (1, 8) (9, 8) (0, 8)$ ✓✓
$v = 2 (9, 8) (0, 8) = 3, 96$ m·s⁻¹ ✓

c) $Δ p = m v_{f} - m v_{i}$ ✓ $= 1, 8 (3, 96) - 0$ ✓
$Δ p = 7, 13$ kg·m·s⁻¹ to the right ✓

d) 7,13 kg·m·s⁻¹ to the left ✓ (equal in magnitude, opposite in direction — total momentum of the system stays zero).

e) $Δ p_{N} = m_{N} v_{N}$ : $7, 13 = 1, 2 v_{N}$ ✓ → $v_{N} = 5, 94$ m·s⁻¹ ✓

Modelled on Nov 2024 Paper 1, Question 4 — same skills, new scenario.

Question 2

Momentum–kinetic energy relationshipApplication2 marksPaper 1

A laboratory trolley moves in a straight line with momentum of magnitude p and kinetic energy K. A constant net force acts on the trolley in its direction of motion until the magnitude of its momentum is 2p. The mass of the trolley remains constant.

The kinetic energy of the trolley is now …

Hint

Use K = p²/2m with the mass constant.

Worked answer

C. $K = \frac{p ^{2}}{2 m}$ , so at constant mass $K \propto p^{2}$ . Doubling $p$ multiplies $K$ by $2^{2} = 4$ , giving $4 K$ . A assumes K is directly proportional to p. B has the energy decreasing although the trolley speeds up. D applies the square twice ( $2^{4}$ ).

Modelled on Nov 2024 Paper 1, Question 1.3 — same skill, new scenario.

Question 3

Impulse & change in momentumComprehension2 marksPaper 1

A tennis ball is dropped from rest and strikes a concrete floor at speed w. It rebounds vertically and reaches a maximum height LOWER than its release height. Air resistance is ignored.

Which combination is CORRECT for the direction of the impulse of the floor on the ball, and the ball's speed just after leaving the floor?

Hint

Impulse points in the direction of the change in momentum; the rebound height fixes the rebound speed.

Worked answer

B. The momentum changes from downward to upward, so the change in momentum — and therefore the impulse on the ball — points upwards. Since it rises to a lower height than it fell from, its launch speed must be less than w (using $v^{2} = 2 g h$ ). A would require the bounce to add energy. C and D have the impulse downwards, which would push the ball into the floor instead of reversing its motion.

Modelled on Nov 2024 Paper 1, Question 1.5 — same skill, new scenario.

Question 4

Momentum vs contact time graph (hockey)Application11 marksPaper 1

Hockey players at a sports academy investigate how the final momentum of a ball depends on the contact time with the stick, for a constant average net force.

A hockey ball of mass 160 g is fed from a ball machine so that it approaches the striker at the same speed in every trial. The striker hits the ball straight back in the opposite direction. The average net force on the ball is horizontal and is kept the same in every trial; only the contact time $Δ t$ between stick and ball is varied.

The sketch graph of the results is a straight line that cuts the $Δ t$ -axis at 0,02 s and passes through a final momentum $p_{f}$ of 3,6 kg·m·s⁻¹ at $Δ t$ = 0,05 s. Take the direction away from the striker (after the hit) as positive.

aDefine the term impulse. (2)
bUse the graph to calculate the magnitude of the average net force acting on the ball. (3)
cCalculate the magnitude of the initial velocity of the ball. (4)
dRedraw the sketch graph and label it A. On the same set of axes, sketch the graph that will be obtained if a ball of SMALLER mass is used, with the initial speed and average net force unchanged. Label this graph B. (2)

This question has a diagram, shown in the app.

Hint

From F_netΔt = p_f − p_i the graph of p_f against Δt has gradient F_net and cuts the Δt-axis where F_netΔt equals the magnitude of the initial momentum.

Worked answer

a) Impulse is the product of the net force acting on an object and the time for which the net force acts. ✓✓

b) From the impulse–momentum theorem: $p_{f} = F_{n e t} Δ t + p_{i}$ , so the gradient of the line equals $F_{n e t}$ : ✓
$F_{n e t} = \frac{3 , 6 - 0}{0 , 05 - 0 , 02}$ ✓ $= 120$ N ✓

c) At the $Δ t$ -intercept, $p_{f} = 0$ :
$F_{n e t} Δ t = p_{f} - p_{i}$ ✓
$120 \times 0, 02 = 0 - p_{i} \Rightarrow p_{i} = - 2, 4$ kg·m·s⁻¹ ✓
$v_{i} = \frac{p _{i}}{m} = \frac{- 2 , 4}{0 , 16}$ ✓ $= - 15$ m·s⁻¹, so the magnitude is 15 m·s⁻¹ ✓

d) Graph A as given. Graph B: a straight line PARALLEL to A (same gradient, since $F_{n e t}$ is unchanged) ✓ but with a smaller (less negative) $p_{f}$ -intercept, so it cuts the $Δ t$ -axis at a smaller contact time (B lies above A) ✓ — the smaller mass has a smaller magnitude of initial momentum $p_{i} = - m v_{i}$ .

Modelled on Nov 2025 Paper 1, Question 4 — same skills, new scenario.

Question 5

Momentum ratios (p = mv)Comprehension2 marksPaper 1

Two trolleys, R of mass 3m and T of mass m, move along the same straight line with EQUAL momentum.

The velocity of R is ...

Hint

Set $3 m \cdot v_{R} = m \cdot v_{T}$ and solve for $v_{R}$ .

Worked answer

A. Equal momentum means $3 m v_{R} = m v_{T}$ , so $v_{R} = \frac{1}{3} v_{T}$ — the heavier trolley must move more slowly. B inverts the ratio. C would give R three times the momentum of T. D squares the mass ratio, which has no basis in $p = m v$ . Modelled on Nov 2025 Paper 1, Question 1.4 — same skills, new scenario.

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Grade 12 Physical Sciences: Momentum & impulse

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