Momentum — answer key

1. CQ1 · Practice

Model answer. Momentum is defined as mass × velocity, p = mv. Mass × acceleration is the resultant force, and force × distance is the work done.

A full answer includes

• I chose mass × velocity.
• I did not choose mass × acceleration.

Also accept

• mass × velocity

Watch for

• “mass × acceleration” — Mass × acceleration is the resultant force, F = ma. Momentum uses the velocity.
• “force × distance” — Force × distance is the work done, measured in joules. Momentum is about mass and velocity.
• “mass ÷ velocity” — Dividing would give a faster object less momentum. A faster object is harder to stop, so momentum is mass multiplied by velocity.

2. CQ2 · Practice

Model answer. Momentum = mass × velocity = 0.80 kg × 2.5 m/s = 2.0 kg m/s.

Checked automatically

• Momentum: 2.0 kg m/s (2 significant figures)
  • Common wrong answer 0.32 : Multiply the mass by the velocity; do not divide.
  • Common wrong answer 3.1 : Multiply the mass by the velocity; do not divide.

A full answer includes

• I wrote p = mv = 0.80 kg × 2.5 m/s.
• I wrote 2.0 kg m/s.
• I did not divide the mass by the velocity.

Also accept

• 2.0 kg m/s

Watch for

• “p = 0.80 ÷ 2.5 = 0.32 kg m/s.” — Momentum is mass multiplied by velocity, not divided.

3. CQ3 · Practice

Model answer. Rearrange p = mv: m = p ÷ v = 54 kg m/s ÷ 1.5 m/s = 36 kg.

Checked automatically

• Mass: 36 kg (2 significant figures)
  • Common wrong answer 81 : Divide the momentum by the velocity; do not multiply.
  • Common wrong answer 0.028 : Divide the momentum by the velocity, not the velocity by the momentum.

A full answer includes

• I rearranged p = mv to m = p ÷ v.
• I wrote m = 54 kg m/s ÷ 1.5 m/s.
• I wrote 36 kg.
• I did not multiply 54 by 1.5.

Also accept

• 36 kg

Watch for

• “m = 54 × 1.5 = 81 kg.” — p = mv, so the mass is the momentum divided by the velocity.
• “m = 1.5 ÷ 54 = 0.028 kg.” — That is the velocity divided by the momentum. Divide the momentum by the velocity.

4. CQ4 · Practice

Model answer. Take their direction of motion as positive. Total momentum before = 60 kg × 4.0 m/s + 40 kg × 1.5 m/s = 240 + 60 = 300 kg m/s. Both move the same way, so both momenta are positive. After, the skaters move together with a mass of 100 kg: 100 kg × v = 300 kg m/s, so v = 3.0 m/s, in the same direction.

Checked automatically

• Velocity after: 3.0 m/s (2 significant figures)
  • Common wrong answer 1.8 : Both move the same way, so add the two momenta.
  • Common wrong answer 2.4 : The second skater moves before, so her 60 kg m/s counts too.
  • Common wrong answer 2.75 : Add the momenta, not the speeds, then divide by the total mass.

A full answer includes

• I wrote the total momentum before as 60 × 4.0 + 40 × 1.5 = 300 kg m/s.
• I wrote the total mass after as 100 kg.
• I wrote 3.0 m/s.
• I did not subtract the second skater’s momentum.

Also accept

• 3.0 m/s

Watch for

• “Total before = 240 − 60 = 180 kg m/s, so v = 1.8 m/s.” — Both skaters move in the same direction, so both momenta are positive. Add them.
• “The second skater is the one being hit, so count her as at rest: v = 2.4 m/s.” — She is moving at 1.5 m/s, so she has 60 kg m/s of momentum before. Add it.
• “Take the mean of the speeds: v = 2.75 m/s.” — Speeds are not conserved. Add the momenta, then divide by the total mass.

5. CQ5 · Practice

Model answer. The mass must be in kilograms: 145 g = 0.145 kg. Rearrange p = mv: v = p ÷ m = 5.8 kg m/s ÷ 0.145 kg = 40 m/s.

Checked automatically

• Speed: 40 m/s (2 significant figures)
  • Common wrong answer 0.040 : Change 145 g to 0.145 kg before you divide.
  • Common wrong answer 0.84 : Divide the momentum by the mass; do not multiply.

A full answer includes

• I changed 145 g to 0.145 kg.
• I wrote v = p ÷ m = 5.8 kg m/s ÷ 0.145 kg.
• I wrote 40 m/s.
• I did not divide by 145.

Also accept

• 40 m/s

Watch for

• “v = 5.8 ÷ 145 = 0.040 m/s.” — Change the mass to kilograms first: 145 g = 0.145 kg.
• “v = 5.8 × 0.145 = 0.84 m/s.” — To find v, divide the momentum by the mass.

6. CQ6 · Practice

Model answer. The 2.0 kg trolley has 2.0 kg × (+1.5 m/s) = +3.0 kg m/s. The 3.0 kg trolley moves left, so it has 3.0 kg × (−1.0 m/s) = −3.0 kg m/s. The momenta are the same size and opposite in direction. Neither has more momentum: the heavier one is slower by just the right amount.

A full answer includes

• I worked out +3.0 kg m/s for the 2.0 kg trolley.
• I worked out −3.0 kg m/s for the 3.0 kg trolley.
• I chose: equal in size and opposite in direction.
• I did not choose that they have the same momentum.

Also accept

• Their momenta are equal in size and opposite in direction.

Watch for

• “The 3.0 kg trolley has more momentum, because it is heavier.” — Momentum depends on the velocity too. The 3.0 kg trolley is slower: 3.0 × 1.0 = 3.0 kg m/s, the same size as the other’s 2.0 × 1.5.
• “The 2.0 kg trolley has more momentum, because it is faster.” — Momentum depends on the mass too. 2.0 × 1.5 = 3.0 kg m/s, the same size as the heavier trolley’s.
• “They have the same momentum, +3.0 kg m/s each.” — The sizes are equal, but one trolley moves left, so its momentum is −3.0 kg m/s.

7. CQ7 · Practice

Model answer. To the left is the negative direction, so v = −0.50 m/s and p = mv = 1.2 kg × (−0.50 m/s) = −0.60 kg m/s.

Checked automatically

• Momentum: -0.60 kg m/s (within ±0.005 kg m/s)
  • Common wrong answer 0.60 : The size is right, but moving to the left makes the momentum negative.
  • Common wrong answer -2.4 : Multiply the mass by the velocity; do not divide.
  • Common wrong answer 2.4 : Multiply the mass by the velocity, and give the sign for moving left.

A full answer includes

• I wrote v = −0.50 m/s.
• I wrote p = 1.2 kg × (−0.50 m/s).
• I wrote −0.60 kg m/s.
• I did not leave out the minus sign.

Also accept

• −0.60 kg m/s

Watch for

• “p = +0.60 kg m/s.” — The size is right, but the trolley moves to the left, so its momentum is negative.
• “p = 1.2 ÷ 0.50 = 2.4 kg m/s.” — Multiply the mass by the velocity; do not divide.

8. CQ8 · Practice

Model answer. With no resultant force from outside, the total momentum of A and B stays the same. A loses 0.36 kg m/s, so B must gain exactly 0.36 kg m/s.

A full answer includes

• I chose: it rises by 0.36 kg m/s.
• I used: the total momentum stays the same.
• I did not choose that some momentum is lost in the collision.

Also accept

• It rises by 0.36 kg m/s.

Watch for

• “It falls by 0.36 kg m/s.” — Then the total would fall by 0.72 kg m/s. The total stays the same, so what A loses, B gains.
• “It stays the same.” — Then the total would fall by 0.36 kg m/s. B pushes on A and A pushes back on B, so B’s momentum changes too.
• “It rises by less than 0.36 kg m/s.” — No momentum is lost in a collision without an outside force. B gains all of it.

9. CQ9 · Practice

Model answer. Total momentum before = 1.5 kg × 4.0 m/s + 0 = 6.0 kg m/s. After, the trolleys move together with mass 2.0 kg, so 2.0 kg × v = 6.0 kg m/s and v = 3.0 m/s.

Checked automatically

• Velocity after: 3.0 m/s (2 significant figures)
  • Common wrong answer 4.0 : The same momentum is shared by a bigger mass, so the velocity falls.
  • Common wrong answer 2.0 : The mass grows from 1.5 kg to 2.0 kg, not to twice as much.

A full answer includes

• I wrote the momentum before as 1.5 kg × 4.0 m/s = 6.0 kg m/s.
• I wrote the mass after as 2.0 kg.
• I wrote 2.0 kg × v = 6.0 kg m/s.
• I wrote 3.0 m/s.
• I did not keep the speed at 4.0 m/s.

Also accept

• 3.0 m/s

Watch for

• “v = 4.0 m/s, the same as before.” — The same momentum is now shared by a bigger mass, so they move more slowly.
• “v = 2.0 m/s, half the speed.” — The mass did not double: it went from 1.5 kg to 2.0 kg.

10. CQ10 · Practice

Model answer. (a) P has momentum 0.60 kg × 2.0 m/s = +1.2 kg m/s; Q moves left, so it has 0.40 kg × (−1.5 m/s) = −0.60 kg m/s. Total = +1.2 − 0.60 = +0.60 kg m/s. (b) The total after is still +0.60 kg m/s. Q now has 0.40 kg × 1.2 m/s = +0.48 kg m/s, so P has 0.60 − 0.48 = +0.12 kg m/s and v = 0.12 kg m/s ÷ 0.60 kg = +0.20 m/s, to the right.

Checked automatically

• (a) Total momentum before: 0.60 kg m/s (within ±0.005 kg m/s)
  • Common wrong answer 1.8 : Q moves to the left: its momentum is −0.60 kg m/s.
• (b) Velocity of P after (to the right is positive): 0.20 m/s (within ±0.005 m/s)
  • Common wrong answer 0.60 : That is the velocity if they stuck together. Q moves off on its own at 1.2 m/s.
  • Common wrong answer 2.2 : Start from the total with Q’s momentum negative: +0.60 kg m/s, not 1.8.
  • Common wrong answer -0.20 : P’s momentum is the total minus Q’s: 0.60 − 0.48 = +0.12 kg m/s, to the right.

A full answer includes

• (a) I wrote P’s momentum as 0.60 × 2.0 = +1.2 kg m/s.
• (a) I wrote Q’s momentum as 0.40 × (−1.5) = −0.60 kg m/s.
• (a) I wrote the total as +0.60 kg m/s.
• (a) I did not add 1.2 and 0.60.
• (b) I wrote 0.60 = 0.40 × 1.2 + 0.60 × v.
• (b) I wrote +0.20 m/s, to the right.
• (b) I did not treat the trolleys as stuck together.

Also accept

• (a) 0.60 kg m/s (b) 0.20 m/s to the right

Watch for

• “The total is 1.2 + 0.60 = 1.8 kg m/s.” — Q moves the other way, so its momentum is negative: subtract it.
• “After the collision, P and Q move together at 0.60 m/s.” — They do not stick here: Q moves off on its own at 1.2 m/s. Take Q’s momentum away from the total.

11. CQ11 · Practice

Model answer. Take right as positive. Before: 0.25 kg × 2.4 m/s = +0.60 kg m/s. After: the small trolley has 0.25 kg × (−0.40 m/s) = −0.10 kg m/s, and the big trolley has 0.50 kg × v. So −0.10 + 0.50 v = 0.60, 0.50 v = 0.70 and v = +1.4 m/s, to the right.

Checked automatically

• Velocity of the 0.50 kg trolley: 1.4 m/s (2 significant figures)
  • Common wrong answer 1.0 : The bounce is to the left, so the small trolley’s momentum after is negative.
  • Common wrong answer 0.80 : The trolleys do not stick together here.

A full answer includes

• I wrote the total before as 0.25 × 2.4 = +0.60 kg m/s.
• I wrote the small trolley’s momentum after as −0.10 kg m/s.
• I wrote −0.10 + 0.50 v = 0.60.
• I wrote +1.4 m/s, to the right.
• I did not write 1.0 m/s.

Also accept

• 1.4 m/s to the right

Watch for

• “v = 1.0 m/s.” — That treats the bounce as forwards. The small trolley moves left, so its momentum after is −0.10 kg m/s.
• “The trolleys move off together at 0.80 m/s.” — They do not stick: the small one bounces back. Use each trolley’s own velocity.

12. CQ12 · Practice

Model answer. Nothing moves before, so the total momentum is zero and stays zero. Y has 1.0 kg × 0.30 m/s = +0.30 kg m/s, so X must have −0.30 kg m/s: v = −0.30 kg m/s ÷ 0.50 kg = −0.60 m/s, 0.60 m/s to the left. The trolleys move off in opposite directions, and the lighter one moves faster.

Checked automatically

• Velocity of X: -0.60 m/s (within ±0.005 m/s)
  • Common wrong answer 0.60 : X moves the opposite way to Y, so its velocity is negative.
  • Common wrong answer -0.30 : Equal and opposite momenta, not speeds: X is half Y’s mass.

A full answer includes

• I wrote the total before as zero.
• I wrote Y’s momentum as 1.0 × 0.30 = +0.30 kg m/s.
• I wrote X’s momentum as −0.30 kg m/s.
• I wrote −0.60 m/s.
• I did not give X the same speed as Y.

Also accept

• −0.60 m/s (0.60 m/s to the left)

Watch for

• “X moves at 0.30 m/s, the same speed as Y.” — The momenta are equal and opposite, not the speeds. X is lighter, so it moves faster.
• “X moves to the right as well.” — The total must stay zero, so X’s momentum is opposite to Y’s: X moves left.

13. CQ13 · Practice

Model answer. The total momentum is zero before and after. The bullet has 0.010 kg × 400 m/s = 4.0 kg m/s forwards, so the rifle has 4.0 kg m/s backwards: v = 4.0 kg m/s ÷ 4.0 kg = 1.0 m/s.

Checked automatically

• Recoil speed: 1.0 m/s (2 significant figures)
  • Common wrong answer 400 : The momenta are equal, not the speeds.
  • Common wrong answer 1000 : Change 10 g to 0.010 kg first.

A full answer includes

• I changed 10 g to 0.010 kg.
• I wrote the bullet’s momentum as 0.010 × 400 = 4.0 kg m/s.
• I wrote v = 4.0 ÷ 4.0.
• I wrote 1.0 m/s.
• I did not give the rifle the bullet’s speed.

Also accept

• 1.0 m/s

Watch for

• “The rifle recoils at 400 m/s too.” — Equal momenta, not equal speeds. The rifle is 400 times heavier, so it moves 400 times more slowly.
• “v = 10 × 400 ÷ 4.0 = 1000 m/s.” — Change 10 g to 0.010 kg before you multiply.

14. CQ14 · Practice

Model answer. Conservation of momentum applies only when no resultant force from outside acts. The ground, part of the Earth, pushes up on the bag from outside it while it stops. The bag’s momentum passes to the Earth, which is so massive that its change in velocity is far too small to notice.

A full answer includes

• I chose: the ground pushes up from outside.
• I chose: the momentum passes to the Earth.
• I did not choose that the momentum turns into heat.

Also accept

• The ground pushes up from outside, and the momentum passes to the Earth.

Watch for

• “Momentum is destroyed when the bag stops, so it is not conserved here.” — It is conserved: the Earth gains what the bag loses. The bag alone has an outside force on it.
• “The bag’s momentum changes into heat and sound as it hits the ground.” — Heat and sound are energy. Momentum is not energy, and it cannot change into energy.
• “The bag had no momentum, because it started falling from rest.” — It started at rest, but it was moving downwards when it hit the ground, so it had momentum.

15. CQ15 · Practice

Model answer. Step 2 is wrong. Both trolleys move to the right, so both momenta are positive: total before = 2.0 × 3.0 + 1.0 × 1.5 = 7.5 kg m/s. Then 3.0 kg × v = 7.5 kg m/s and v = 2.5 m/s. The pupil’s 1.5 m/s cannot be right: after joining, the trolleys must move faster than the slower one and slower than the faster one.

A full answer includes

• I chose step 2.
• I wrote that the 1.0 kg trolley’s momentum is +1.5 kg m/s.
• I worked out the correct answer: 7.5 kg m/s ÷ 3.0 kg = 2.5 m/s.
• I did not choose step 4.

Also accept

• Step 2

Watch for

• “Step 1.” — Any direction can be chosen as positive. Choosing right is fine.
• “Step 3.” — The trolleys stick together, so 3.0 kg moves with one velocity. This step is right.
• “Step 4.” — The algebra is right for the number in step 2. The mistake comes earlier.

16. CT1 · Apply

Model answer. (a) Nothing moves, so the total momentum is zero. (b) Spanner: p = mv = 2.5 kg × 4.0 m/s = 10 kg m/s, away from the station. (c) No resultant force acts from outside, so the total is still zero: the astronaut has 10 kg m/s in the opposite direction, towards the station. v = 10 kg m/s ÷ 125 kg = 0.080 m/s towards the station. (d) The total momentum must stay zero, so the astronaut’s momentum is equal in size and opposite in direction to the spanner’s. Throwing the spanner away from the station sends the astronaut towards it, slowly, because the astronaut is 50 times heavier.

A full answer includes

• (a) I stated that the total is zero.
• (b) I wrote p = 2.5 kg × 4.0 m/s.
• (b) I wrote 10 kg m/s, away from the station.
• (c) I gave the astronaut 10 kg m/s towards the station.
• (c) I wrote v = 10 kg m/s ÷ 125 kg.
• (c) I wrote 0.080 m/s, towards the station.
• (c) I did not write 4.0 m/s.
• (d) I wrote that the total momentum must stay zero.
• (d) I wrote that the astronaut’s momentum is equal in size and opposite in direction to the spanner’s.
• (d) I did not write that the astronaut needs something to push against.

Also accept

• (a) 0 (b) 10 kg m/s (c) 0.080 m/s towards the station (d) total stays zero: equal and opposite momenta

Watch for

• “The astronaut moves at 4.0 m/s too.” — The momenta are equal and opposite, not the speeds.
• “With nothing to push against in space, the astronaut cannot move.” — The astronaut pushes on the spanner, and the spanner pushes back.