Answers · Momentum and collisions

Impulse and safety — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. CQ16 · Practice

Model answer. Impulse is defined as force × the time for which the force acts, F∆t. It equals the change in momentum, ∆(mv).

A full answer includes

I chose force × time the force acts for.
I did not choose force ÷ time the force acts for.

Also accept

force × time the force acts for

Watch for

“mass × velocity of the object” — That is momentum. Impulse is F∆t, which equals the change in momentum.
“force ÷ time the force acts for” — Multiply the force by the time; do not divide. A force that acts for longer gives a bigger impulse.
“force × distance the force moves” — Force × distance is the work done, in joules. Impulse uses the time, not the distance.

2. CQ17 · Practice

Model answer. 250 ms = 0.25 s. Impulse = F∆t = 20 N × 0.25 s = 5.0 N s.

Checked automatically

Impulse: 5.0 N s (2 significant figures)
- Common wrong answer 5000 : Change 250 ms to 0.25 s first.
- Common wrong answer 80 : Multiply the force by the time; do not divide.

A full answer includes

I changed 250 ms to 0.25 s.
I wrote impulse = 20 N × 0.25 s.
I wrote 5.0 N s.
I did not multiply by 250.

Also accept

5.0 N s

Watch for

“Impulse = 20 × 250 = 5000 N s.” — Change 250 ms to 0.25 s first.
“Impulse = 20 ÷ 0.25 = 80 N s.” — Multiply the force by the time.

3. CQ18 · Practice

Model answer. Momentum = mass × velocity, so its unit is kg × m/s = kg m/s. Impulse = force × time, so its unit is N × s = N s. They are the same unit: 1 N s = 1 kg m/s. Row A.

A full answer includes

I chose row A.
I chose kg m/s for momentum.
I chose N s for impulse.
I did not choose N/s for impulse.

Also accept

Row A

Watch for

“Row B.” — kg m/s² is mass × acceleration: that is the newton, the unit of force, not of momentum.
“Row C.” — N/s means newtons divided by seconds. Impulse is force multiplied by time: N s.
“Row D.” — The newton is the unit of force and the joule is the unit of energy.

4. CQ19 · Practice

Model answer. The impulse equals the change in momentum. The ball starts at rest, so its momentum after is 2.3 kg m/s. 46 g = 0.046 kg, so v = p ÷ m = 2.3 kg m/s ÷ 0.046 kg = 50 m/s.

Checked automatically

Speed: 50 m/s (2 significant figures)
- Common wrong answer 0.050 : Change 46 g to 0.046 kg before you divide.
- Common wrong answer 0.11 : Divide the impulse by the mass; do not multiply.

A full answer includes

I used impulse = change in momentum.
I changed 46 g to 0.046 kg.
I wrote v = 2.3 ÷ 0.046.
I wrote 50 m/s.
I did not divide by 46.

Also accept

50 m/s

Watch for

“You need the force and the time to find the speed.” — Not here: the impulse is the change in momentum, so divide it by the mass.
“v = 2.3 ÷ 46 = 0.050 m/s.” — Change 46 g to 0.046 kg before you divide.

5. CQ20 · Practice

Model answer. Take towards the buffer as positive. Momentum before = 0.40 kg × 1.5 m/s = +0.60 kg m/s; after = 0.40 kg × (−1.0 m/s) = −0.40 kg m/s. Change = −0.40 − 0.60 = −1.0 kg m/s: 1.0 N s, away from the buffer.

Checked automatically

Change in momentum: 1.0 N s (2 significant figures)
- Common wrong answer 0.20 : The direction reverses, so add the sizes: 0.60 + 0.40.
- Common wrong answer 0.60 : That only stops the trolley. Add the momentum it has after the rebound.
- Common wrong answer -1.0 : That is the change with its sign, towards the buffer positive. The question asks for its size, so give it without the minus sign.

A full answer includes

I wrote the momentum before as 0.40 × 1.5 = +0.60 kg m/s.
I wrote the momentum after as 0.40 × (−1.0) = −0.40 kg m/s.
I wrote the change as −0.40 − 0.60 = −1.0 kg m/s.
I wrote the size as 1.0 N s.
I did not write 0.20 N s.

Also accept

1.0 N s

Watch for

“The change is 0.60 − 0.40 = 0.20 kg m/s.” — The trolley reverses, so its momentum changes from +0.60 to −0.40 kg m/s: a change of 1.0 kg m/s.
“The change is 0.60 kg m/s, the momentum it had.” — That is only the change needed to stop it. The buffer also sends it back at 1.0 m/s.

6. CQ21 · Practice

Model answer. Impulse = area under the force–time graph. The area is a triangle: ½ × base × height = ½ × 0.10 s × 40 N = 2.0 N s.

Checked automatically

Impulse: 2.0 N s (2 significant figures)
- Common wrong answer 4.0 : The area is a triangle, not a rectangle.
- Common wrong answer 400 : Impulse is the area: multiply, do not divide.

A full answer includes

I used impulse = area under the force–time graph.
I wrote ½ × 0.10 s × 40 N.
I wrote 2.0 N s.
I did not use the whole rectangle, 40 N × 0.10 s.

Also accept

2.0 N s

Watch for

“Impulse = 40 N × 0.10 s = 4.0 N s.” — That is the rectangle. The force is 40 N only at the peak, so the area is a triangle.
“Impulse = 40 ÷ 0.10 = 400 N s.” — Impulse is an area: multiply, do not divide.

7. CQ22 · Practice

Model answer. The area under a force–time graph is the impulse, which equals the change in momentum, so the impulses are equal. The foam spreads the same change in momentum over a longer time, so its average force, F = ∆p ÷ ∆t, is smaller.

A full answer includes

I used: the area under the graph is the impulse.
I chose: the impulses are equal, and the foam gives the smaller average force.
I did not choose that a taller peak means a bigger impulse.

Also accept

The impulses are equal, and the foam gives the smaller average force.

Watch for

“The impulses are equal, and the wall gives the smaller average force.” — The wall stops the car in a shorter time. The same ∆p in a shorter time needs a bigger force.
“The wall gives the bigger impulse, because its peak force is taller.” — The impulse is the area, not the height of the peak. The areas are equal.
“The foam gives the bigger impulse, because it presses for longer.” — It presses for longer but with a smaller force. The areas are equal, so the impulses are equal.

8. CQ23 · Practice

Model answer. The resultant force on an object is its change in momentum per unit time: F = ∆p ÷ ∆t. Its unit is (kg m/s) ÷ s = kg m/s², the newton.

A full answer includes

I chose change in momentum ÷ time taken.
I did not choose change in momentum × time taken.

Also accept

change in momentum ÷ time taken

Watch for

“momentum × time taken” — Its unit would be kg m, not a newton. Force is the change in momentum divided by the time.
“change in momentum × time taken” — Divide by the time; do not multiply. Force × time is the impulse.
“momentum ÷ mass of the object” — Momentum ÷ mass is the velocity, in m/s.

9. CQ24 · Practice

Model answer. F = ∆p ÷ ∆t = 12 kg m/s ÷ 0.40 s = 30 N.

Checked automatically

Resultant force: 30 N (2 significant figures)
- Common wrong answer 4.8 : Divide the change in momentum by the time.
- Common wrong answer 0.033 : Divide the change in momentum by the time, not the time by the change in momentum.

A full answer includes

I wrote F = 12 kg m/s ÷ 0.40 s.
I wrote 30 N.
I did not multiply 12 by 0.40.

Also accept

30 N

Watch for

“F = 12 × 0.40 = 4.8 N.” — Force is the change in momentum divided by the time.

10. CQ25 · Practice

Model answer. (a) ∆p = momentum after − momentum before = 1500 kg × 20 m/s − 1500 kg × 8.0 m/s = 30 000 − 12 000 = 18 000 kg m/s. (b) F = ∆p ÷ ∆t = 18 000 kg m/s ÷ 4.0 s = 4500 N.

Checked automatically

(a) Change in momentum: 1.8×10^4 kg m/s (2 significant figures)
- Common wrong answer 30000 : That is the momentum after. Take away the 12 000 kg m/s the car had before.
- Common wrong answer 42000 : Subtract the momentum before; do not add it.
(b) Average resultant force: 4.5×10^3 N (2 significant figures)
- Common wrong answer 7500 : Use the change in momentum, 18 000 kg m/s, not the momentum after.
- Common wrong answer 72000 : Divide by the time; do not multiply.

A full answer includes

(a) I wrote ∆p = 1500 kg × (20 − 8.0) m/s.
(a) I wrote 18 000 kg m/s.
(a) I did not use 20 m/s on its own.
(b) I wrote F = 18 000 kg m/s ÷ 4.0 s.
(b) I wrote 4500 N.

Also accept

(a) 18 000 kg m/s (b) 4500 N

Watch for

“∆p = 1500 × 20 = 30 000 kg m/s.” — The car already had momentum at 8.0 m/s. Take the momentum before away from the momentum after.
“F = 18 000 × 4.0 N.” — Divide the change in momentum by the time.

11. CQ26 · Practice

Model answer. The change in momentum is 0.15 kg × 4.0 m/s = 0.60 kg m/s both times. (a) 3.0 ms = 0.0030 s, so F = 0.60 ÷ 0.0030 = 200 N. (b) 40 ms = 0.040 s, so F = 0.60 ÷ 0.040 = 15 N. The carpet makes the stop about 13 times longer, so the force is about 13 times smaller.

Checked automatically

(a) Force on the tiles: 2.0×10^2 N (2 significant figures)
- Common wrong answer 0.20 : Change 3.0 ms to 0.0030 s before you divide.
- Common wrong answer 0.0018 : Divide the change in momentum by the time; do not multiply.
(b) Force on the carpet: 15 N (2 significant figures)
- Common wrong answer 0.015 : Change 40 ms to 0.040 s before you divide.
- Common wrong answer 0.024 : Divide the change in momentum by the time; do not multiply.

A full answer includes

I wrote ∆p = 0.15 × 4.0 = 0.60 kg m/s for both.
(a) I changed 3.0 ms to 0.0030 s.
(a) I wrote 0.60 ÷ 0.0030 = 200 N.
(b) I changed 40 ms to 0.040 s.
(b) I wrote 0.60 ÷ 0.040 = 15 N.
I did not give the carpet a smaller change in momentum.

Also accept

(a) 200 N (b) 15 N

Watch for

“The carpet gives a smaller change in momentum.” — The phone stops from the same speed both times, so ∆p is the same. Only the time changes.
“F = 0.60 ÷ 3.0 = 0.20 N on the tiles.” — Change 3.0 ms to 0.0030 s before you divide.

12. CQ27 · Practice

Model answer. Take right as positive. The momentum goes from positive to a bigger negative value, so the change, after − before, is negative: to the left. F = ∆p ÷ ∆t, so the force is in the same direction as the change in momentum: to the left. Row A.

A full answer includes

I chose row A.
I wrote that the change in momentum is to the left.
I wrote that the force is in the same direction as the change in momentum.
I did not give the force the direction the ball moved before.

Also accept

Row A

Watch for

“Row B.” — The ball ends up moving left, so its momentum changes towards the left, not the right.
“Row C.” — F = ∆p ÷ ∆t, so the force is always in the same direction as the change in momentum.
“Row D.” — The momentum after is to the left, and bigger, so the change is to the left. The force is in the same direction as the change.

13. CQ28 · Practice

Model answer. Both heads have the same mass and stop from the same speed, so their change in momentum is the same. The steel head hardly squashes, so it stops in a shorter time. F = ∆p ÷ ∆t, so the same ∆p in a shorter time gives a bigger force on the nail. That is why a steel hammer drives a nail in. A crumple zone, an airbag and a seat belt use the same idea the other way round: they make the stop take longer, so the force is smaller. Row A.

A full answer includes

I chose row A.
I wrote that the change in momentum is the same.
I wrote that the steel head stops in a shorter time.
I wrote that the force on the nail is bigger.
I did not choose a bigger change in momentum for the steel head.

Also accept

Row A

Watch for

“Row B.” — Rubber squashes, so it is the rubber head that stops over a longer time, not the steel one.
“Row C.” — Both heads have the same mass and speed and both stop, so their change in momentum is the same.
“Row D.” — The same change in momentum in a shorter time means a bigger force, not a smaller one.

14. CQ29 · Practice

Model answer. (a) ∆p = F∆t = 4.0 N × 3.0 s = 12 N s = 12 kg m/s. (b) It started at rest, so its momentum is now 12 kg m/s and v = 12 kg m/s ÷ 2.0 kg = 6.0 m/s.

Checked automatically

(a) Change in momentum: 12 kg m/s (2 significant figures)
- Common wrong answer 1.3 : Multiply the force by the time.
(b) Velocity: 6.0 m/s (2 significant figures)
- Common wrong answer 24 : Divide the momentum by the mass.

A full answer includes

(a) I wrote ∆p = F∆t = 4.0 N × 3.0 s.
(a) I wrote 12 kg m/s.
(b) I wrote v = 12 kg m/s ÷ 2.0 kg.
(b) I wrote 6.0 m/s.
(b) I did not multiply by the mass.

Also accept

(a) 12 kg m/s (b) 6.0 m/s

Watch for

“∆p = 4.0 ÷ 3.0 = 1.3 kg m/s.” — Change in momentum = force × time.
“v = 12 × 2.0 = 24 m/s.” — p = mv, so v = p ÷ m.

15. CQ30 · Practice

Model answer. Change in momentum = 900 kg × 12 m/s = 10 800 kg m/s. ∆t = ∆p ÷ F = 10 800 kg m/s ÷ 5000 N = 2.16 s = 2.2 s to 2 significant figures.

Checked automatically

Stopping time: 2.2 s (2 significant figures)
- Common wrong answer 0.46 : Divide the change in momentum by the force.

A full answer includes

I wrote ∆p = 900 × 12 = 10 800 kg m/s.
I wrote ∆t = 10 800 ÷ 5000.
I wrote 2.2 s.
I did not divide 5000 by 10 800.

Also accept

2.2 s

Watch for

“∆t = 5000 ÷ 10 800 = 0.46 s.” — Time = change in momentum ÷ force.

16. CQ31 · Practice

Model answer. Take towards the board as positive. 160 g = 0.16 kg and 20 ms = 0.020 s. The velocity changes from +9.0 m/s to −6.0 m/s, a change of 15 m/s, so ∆p = 0.16 kg × 15 m/s = 2.4 kg m/s. F = ∆p ÷ ∆t = 2.4 kg m/s ÷ 0.020 s = 120 N, away from the board. The ice is smooth, so this is the only force along the line of motion.

Checked automatically

Average force: 1.2×10^2 N (2 significant figures)
- Common wrong answer 24 : The puck reverses: its velocity changes by 9.0 + 6.0 = 15 m/s.
- Common wrong answer 72 : That force only stops the puck. It is also sent back at 6.0 m/s.
- Common wrong answer 0.12 : Change 20 ms to 0.020 s before you divide.
- Common wrong answer 120000 : Change 160 g to 0.16 kg first.
- Common wrong answer 0.048 : Force = change in momentum ÷ time; do not multiply.

A full answer includes

I changed 160 g to 0.16 kg.
I changed 20 ms to 0.020 s.
I wrote the change in velocity as 9.0 + 6.0 = 15 m/s.
I wrote ∆p = 0.16 × 15 = 2.4 kg m/s.
I wrote F = 2.4 ÷ 0.020.
I wrote 120 N.
I did not use 9.0 − 6.0.

Also accept

120 N

Watch for

“The change in velocity is 9.0 − 6.0 = 3.0 m/s.” — The puck reverses, so its velocity changes by 9.0 + 6.0 = 15 m/s.
“The board only has to stop the puck.” — It also sends it back at 6.0 m/s, so the change in momentum is bigger than a stop.

17. CT2 · Apply

Model answer. (a) ∆p = 0.16 kg × 30 m/s = 4.8 kg m/s. (b) F = 4.8 ÷ 0.016 = 300 N. (c) F = 4.8 ÷ 0.12 = 40 N: 7.5 times smaller, because the stop takes 7.5 times longer. (d) The impulse F∆t equals ∆(mv). The ball has the same mass and goes from 30 m/s to rest both times, so the impulse is the same; only the force and the time differ. (e) 40 N is the average over the whole 0.12 s. The force starts at zero and ends at zero, so in the middle it must be bigger than the average.

A full answer includes

(a) I wrote ∆p = 0.16 kg × 30 m/s.
(a) I wrote 4.8 kg m/s.
(b) I wrote F = 4.8 ÷ 0.016.
(b) I wrote 300 N.
(c) I wrote F = 4.8 ÷ 0.12.
(c) I wrote 40 N.
(d) I wrote that the impulse F∆t equals the change in momentum.
(d) I wrote that the mass and the change in velocity are the same in both catches.
(d) I did not write that moving the hands back reduces the change in momentum.
(e) I wrote that 40 N is the average force over the whole 0.12 s.
(e) I wrote that the force is smaller than the average at the start and the end.
(e) I wrote that so the force must be bigger than the average in the middle.

Also accept

(a) 4.8 kg m/s (b) 300 N (c) 40 N (d) impulse = change in momentum, the same (e) 40 N is the average; the peak is higher

Watch for

“Moving the hands back reduces the ball’s change in momentum.” — The ball still stops from 30 m/s. It is the time that changes.
“The peak force equals the average force.” — The force is zero at the start and the end of the catch, so it must rise above the average in between.