Senior 3 practice book · Unit 2 of 10

Linear Momentum, Impulse and Collisions

50 questions that complete the Senior 3 quiz for this unit: 21 core and 29 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

A fast, light object always has more momentum than a slow, heavy one.Momentum p = mv depends on mass and velocity together: a 60 kg runner at 5 m/s (300 kg·m/s) has far more momentum than a 10 g bullet at 800 m/s (8 kg·m/s).
In a crash the heavy vehicle pushes harder on the light one than the light one pushes back.By Newton's third law the two forces are equal and opposite and last for the same time, so the impulses are equal. The light vehicle suffers more because the same change of momentum gives it a larger change of velocity.
Kinetic energy that 'disappears' in a crash has been destroyed.Energy is never destroyed. In an inelastic collision kinetic energy becomes heat, sound and the work of bending metal; the total energy is unchanged.
An airbag protects the driver by making the change of momentum smaller.The driver's change of momentum is the same (from mv to zero). The airbag makes the stopping time longer, so the force F = Δp/Δt is smaller.
A rocket needs air behind it to push against.A rocket pushes exhaust gas backwards; the gas pushes the rocket forwards. The gas gains backward momentum and the rocket gains equal forward momentum, so it works best in empty space.

What this unit covers

Topics marked new are not tested much in the quiz, so this book gives them extra questions.

Linear momentum p = mv as a vector quantity
Impulse and Newton's second law F = Δp/Δt
Force–time graphs: the area is the impulse
Conservation of momentum in collisions (same and opposite directions)
Elastic and inelastic collisions: checking kinetic energy
Explosions, recoil and rocket propulsion
Safety and sport: crumple zones, airbags, seat belts, landing and catching
Momentum–time and velocity–time graphs: the gradient gives the force
Newton's third law in collisions: equal and opposite forces and impulses
Experiments to test conservation of momentum (trolleys, ticker timer, light gates)
Combining momentum and energy: ballistic pendulum, collision followed by a rise or slide
Force from a stream of mass: water jets, conveyor belts, F = (mass per second) × Δv

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 29 harder ones are marked ★ Challenge.

1True or false
A seat belt that stretches a little during a crash is safer than a perfectly rigid one.
Show answer
Answer: True
Stretching lengthens the time in which the passenger stops, so the force on the body is smaller.
Common mistakeThinking a stronger, stiffer belt is always safer ignores that a stiff belt stops the body in a very short time, with a huge force.
2True or false · ★ Challenge
The momentum of a ballistic pendulum block stays constant while it swings upwards after the bullet hits it.
Show answer
Answer: False
During the swing gravity and the string tension act on the block, so its momentum changes (it slows down); only energy is conserved.
Common mistakeApplying momentum conservation everywhere forgets that it needs no external resultant force.
3True or false · ★ Challenge
When a stationary shell explodes, the total kinetic energy of the pieces is zero because their total momentum is zero.
Show answer
Answer: False
The total momentum is zero, but every moving piece has positive kinetic energy; the energy comes from the chemical energy of the explosive.
Common mistakeTreating kinetic energy like momentum forgets that KE is a scalar and cannot cancel.
4True or false
In a collision between a heavy car and a light car, the light car receives a larger impulse than the heavy car.
Show answer
Answer: False
The forces are equal and act for the same time, so the impulses are equal in size; the light car has a larger change of velocity.
Common mistakeConfusing impulse (Δp) with change of velocity makes learners think the light car gets the bigger impulse.
5Multiple choice · ★ Challenge
Whatever its speed, a trolley that runs into an identical trolley at rest and couples to it loses the same fraction of its kinetic energy. What is that fraction?
1. AOne half
2. BNone of it
3. COne quarter
4. DAll of it
Show answer
Answer: A. One half
Mass doubles, so the speed halves: v = u/2. KE after = ½(2m)(u/2)² = ¼mu², which is half of ½mu². Half the KE is lost.
Common mistakeChoosing 'all of it' thinks the trolleys stop; they still move together at half the original speed, so half the KE remains.
6True or false
Two identical moto taxis travel at the same speed in opposite directions along a straight road. Their total momentum is zero.
Show answer
Answer: True
Momentum is a vector: +mv and −mv add to zero, even though both motos are moving.
Common mistakeAdding the sizes (2mv) forgets that momenta in opposite directions have opposite signs.
7Fill in the blank · ★ Challenge
A 0.15 kg ball at rest is given an impulse of 3.0 N s. It moves off at ______ m/s.
Show answer
Answer: 20
Impulse = Δp = mv, so v = 3.0 ÷ 0.15 = 20 m/s.
Common mistakeAnswering 0.45 multiplies impulse by mass; the speed is impulse ÷ mass.
8Multiple choice · ★ Challenge
Tea leaves drop at 5.0 kg/s onto a conveyor belt moving at 2.0 m/s. What extra force must the motor apply to keep the belt at a steady speed?
1. A10 N
2. B2.5 N
3. C0.40 N
4. D20 N
Show answer
Answer: A. 10 N
Each second 5.0 kg must be given 2.0 m/s: Δp/Δt = 5.0 × 2.0 = 10 N.
Common mistakeChoosing 2.5 N divides the mass rate by the speed; the force is (mass per second) × (change of velocity).
9True or false
When a moving trolley hits a stationary trolley on a level, smooth track, the momentum lost by the first trolley equals the momentum gained by the second.
Show answer
Answer: True
The total momentum of the isolated pair stays the same, so whatever one loses the other gains.
Common mistakeThinking the stationary trolley 'has no momentum to share' forgets that it gains momentum during the collision.
10Multiple choice · ★ Challenge
A firework of mass 3.0 kg at rest explodes into three equal pieces. One moves east at 12 m/s and one moves west at 12 m/s. What does the third piece do?
1. AIt moves east at 12 m/s
2. BIt moves upwards at 24 m/s
3. CIt moves west at 6.0 m/s
4. DIt stays at rest
Show answer
Answer: D. It stays at rest
Total momentum is zero before. 1.0 × 12 + 1.0 × (−12) = 0 already, so the third piece must have zero momentum.
Common mistakeChoosing 24 m/s upwards thinks the third piece must carry 'the rest' of the momentum; the first two already cancel.
11Multiple choice
Which of these collisions is closest to being elastic?
1. AA lump of wet clay hitting the hard floor
2. BA car crashing into a concrete wall
3. CA dart sticking into a dartboard
4. DTwo snooker balls striking each other
Show answer
Answer: D. Two snooker balls striking each other
Hard snooker balls bounce apart with almost no loss of kinetic energy; the others stick or crumple, losing most of it.
Common mistakeChoosing the car crash because 'momentum is conserved' mixes up the two ideas; momentum is conserved in all of them, KE only in elastic ones.
12Multiple choice · ★ Challenge
A hose sends 2.0 kg of water each second horizontally at 15 m/s onto a wall. The water stops and runs down the wall. What force does the water exert on the wall?
1. A7.5 N
2. B30 N
3. C225 N
4. D60 N
Show answer
Answer: B. 30 N
Each second the momentum destroyed is 2.0 × 15 = 30 kg·m/s, so F = Δp/Δt = 30 N.
Common mistakeChoosing 60 N treats the water as bouncing back; here it stops, so Δv = 15 m/s, not 30 m/s.
13Short answer · ★ Challenge
A blown-up balloon is let go with its neck open and it shoots across the classroom. Explain this using momentum.
Show answer
Model answer: Before release the total momentum is zero. The stretched rubber pushes air out of the neck backwards, giving the air backward momentum; the balloon gains equal forward momentum so the total stays zero. This is how a rocket works.
Common mistakeSaying the air 'pushes on the room air behind it' is the wrong idea; the balloon moves because the escaping air carries momentum backwards.
14Multiple choice
How can a learner check that a runway has been tilted just enough to compensate for friction?
1. AA trolley left at rest starts rolling down by itself
2. BA trolley given a push stops after a short distance
3. CThe ticker-tape dots get further apart along the tape
4. DA trolley given a small push moves at a steady speed
Show answer
Answer: D. A trolley given a small push moves at a steady speed
When friction is just balanced, there is no resultant force, so a pushed trolley neither speeds up nor slows down: the ticker-tape dots are evenly spaced.
Common mistakeChoosing 'starts rolling by itself' means the slope is too steep; then gravity gives a resultant force and the system is not isolated.
15Fill in the blank · ★ Challenge
In a crash test a 0.50 kg model car changes velocity by 4.0 m/s when it hits a 2.5 kg model lorry. The lorry's velocity changes by ______ m/s (in the opposite direction).
Show answer
Answer: 0.80
Equal and opposite impulses: Δp = 0.50 × 4.0 = 2.0 kg·m/s; Δv = 2.0 ÷ 2.5 = 0.80 m/s.
Common mistakeAnswering 4.0 assumes both vehicles change velocity by the same amount; they change MOMENTUM by the same amount.
16Multiple choice
Which of these has the greatest momentum?
1. AA 10 g bullet flying at 800 m/s
2. BA 60 kg learner running at 5.0 m/s
3. CA 1500 kg car parked by the road
4. DA 0.45 kg football kicked at 20 m/s
Show answer
Answer: B. A 60 kg learner running at 5.0 m/s
p = mv: learner 60 × 5.0 = 300 kg·m/s; bullet 0.010 × 800 = 8 kg·m/s; parked car 0; football 0.45 × 20 = 9 kg·m/s.
Common mistakeChoosing the bullet looks only at speed; momentum needs mass × velocity, and 10 g is tiny.
17True or false · ★ Challenge
If the momentum–time graph of a body is a straight line sloping downwards, a constant resultant force acts against its motion.
Show answer
Answer: True
A constant negative gradient means Δp/Δt is constant and negative: a steady force opposite to the motion (for example braking).
Common mistakeThinking a downward slope means 'no force, just slowing down' forgets that any change of momentum needs a force.
18Multiple choice
A hunter holds a rifle tightly against his shoulder when firing. Why does this reduce the kick he feels?
1. AHolding it tightly stops the bullet gaining momentum
2. BHis body and the rifle recoil together, so more slowly
3. CThe rifle then has less momentum than the bullet
4. DHis shoulder absorbs the bullet's momentum before firing
Show answer
Answer: B. His body and the rifle recoil together, so more slowly
The backward momentum equals the bullet's forward momentum. Shared by the larger mass of rifle + body, it gives a much smaller recoil velocity.
Common mistakeChoosing 'the rifle has less momentum than the bullet' breaks conservation; the backward momentum is always equal to the bullet's.
19Fill in the blank · ★ Challenge
A 1.0 kg ball moving at 6.0 m/s hits a wall and bounces straight back at 4.0 m/s. In the impact, ______ J of kinetic energy is changed into other forms.
Show answer
Answer: 10
KE before = ½ × 1.0 × 6.0² = 18 J; after = ½ × 1.0 × 4.0² = 8 J; 18 − 8 = 10 J.
Common mistakeAnswering 2 J subtracts the speeds first (6 − 4 = 2) and then squares; work out each KE separately.
20Multiple choice · ★ Challenge
In a collision between a 1.0 kg trolley and a 3.0 kg trolley, the momentum of the 1.0 kg trolley changes by −6.0 kg·m/s. What is the change in momentum of the 3.0 kg trolley?
1. A−6.0 kg·m/s
2. B+6.0 kg·m/s
3. C+2.0 kg·m/s
4. D+18 kg·m/s
Show answer
Answer: B. +6.0 kg·m/s
Equal and opposite forces for the same time give equal and opposite impulses: Δp = +6.0 kg·m/s (its velocity changes by 6.0 ÷ 3.0 = 2.0 m/s).
Common mistakeChoosing +2.0 kg·m/s gives the change of VELOCITY of the 3.0 kg trolley, not its change of momentum.
21Multiple choice
Why are the ground areas under children's climbing frames covered with sand or rubber chips?
1. AThe child's momentum is reduced before reaching the ground
2. BSand pushes back with a larger force than concrete
3. CThe child stops over a longer time, so the force is smaller
4. DThe child bounces higher, so the change in momentum is smaller
Show answer
Answer: C. The child stops over a longer time, so the force is smaller
The child's change of momentum is the same, but the soft surface stretches the stopping time, so F = Δp/Δt is smaller.
Common mistakeChoosing 'momentum is reduced before the ground' is wrong: the child arrives with the same momentum; only the stopping time changes.
22Short answer · ★ Challenge
A learner standing still on a skateboard catches a heavy ball thrown horizontally to her. Describe and explain what happens to her.
Show answer
Model answer: She and the ball move off together in the direction the ball was travelling, but more slowly than the ball. The momentum of the ball is shared: m_ball × u = (m_ball + m_learner) × v, so v is much smaller than u.
Common mistakeSaying she stays still because she 'absorbs' the ball ignores conservation: the ball's momentum must go somewhere.
23Fill in the blank
A 1200 kg car has a momentum of 18 000 kg·m/s. Its speed is ______ m/s.
Show answer
Answer: 15
v = p/m = 18 000 ÷ 1200 = 15 m/s.
Common mistakeMultiplying 18 000 by 1200 instead of dividing gives a meaningless number; v = p/m.
24Multiple choice · ★ Challenge
A ticker timer makes 50 dots per second. Before a collision, a length of tape covering 5 dot-intervals measures 10 cm. What is the speed of the trolley?
1. A2.0 m/s
2. B1.0 m/s
3. C0.50 m/s
4. D0.20 m/s
Show answer
Answer: B. 1.0 m/s
Time for 5 intervals = 5 ÷ 50 = 0.10 s; v = 0.10 m ÷ 0.10 s = 1.0 m/s.
Common mistakeChoosing 2.0 m/s takes 5 intervals as 0.05 s; each interval is 1/50 s = 0.02 s.
25True or false
A light gate finds the speed of a trolley by timing how long a card of known length on the trolley takes to pass through the light beam.
Show answer
Answer: True
Speed = card length ÷ time the beam is blocked.
Common mistakeThinking the light gate measures distance travelled confuses it with a ruler or the ticker tape.
26Multiple choice · ★ Challenge
A resultant force of 20 N acts for 3.0 s on a 4.0 kg trolley that starts at rest. What is the final speed of the trolley?
1. A60 m/s
2. B5.0 m/s
3. C1.7 m/s
4. D15 m/s
Show answer
Answer: D. 15 m/s
Impulse FΔt = 20 × 3.0 = 60 N s = Δp; v = 60 ÷ 4.0 = 15 m/s.
Common mistakeChoosing 60 m/s gives the impulse but forgets to divide by the mass to get the velocity.
27True or false
Two impacts whose force–time graphs have different shapes but the same area give the same change of momentum.
Show answer
Answer: True
The area under a force–time graph is the impulse, and impulse equals change of momentum; the shape does not matter.
Common mistakeThinking the taller graph gives more momentum looks at the peak force and ignores the time.
28Short answer · ★ Challenge
Describe how you would use two trolleys, a pin and cork, a ticker timer, a balance and a friction-compensated runway to test whether momentum is conserved when trolleys stick together.
Show answer
Model answer: Weigh both trolleys. Fix a pin on A and cork on B so they stick. Attach tape from the timer to A. Push A into the stationary B. From the tape find A's speed before (dot spacing before the jump) and the common speed after. Check that m_A u = (m_A + m_B)v.
Common mistakeForgetting to measure the masses, or using the same speed before and after, means the momenta cannot actually be compared.
29Short answer · ★ Challenge
In a ballistic pendulum, explain why you must use conservation of momentum (not kinetic energy) for the moment the bullet hits the block, but may use energy conservation for the swing that follows.
Show answer
Model answer: The impact is inelastic: much kinetic energy becomes heat and deformation, so KE is not conserved, but momentum is (no external horizontal force during the short impact). During the swing no energy is lost (ignoring air), so KE → potential energy: ½mv² = mgh.
Common mistakeUsing ½mu² = ½(m + M)v² for the impact assumes an elastic collision and gives a speed far too high.
30Short answer
Explain the difference between force and impulse, and explain how the same impulse can be given by a large force or by a small force.
Show answer
Model answer: Force is a push or pull (N); impulse is force × time for which it acts (N s), equal to the change of momentum. A large force for a short time and a small force for a long time can give the same product FΔt, and so the same change of momentum.
Common mistakeTreating impulse as 'a big sudden force' misses that impulse includes the time; a gentle force acting for long can give the same impulse.
31Multiple choice · ★ Challenge
A moto with rider (total 200 kg) at 10 m/s runs into the back of a 1000 kg car moving at 4.0 m/s in the same direction. They lock together. What is their speed just after?
1. A1.7 m/s
2. B7.0 m/s
3. C5.0 m/s
4. D3.3 m/s
Show answer
Answer: C. 5.0 m/s
Total momentum = 200 × 10 + 1000 × 4.0 = 6000 kg·m/s; v = 6000 ÷ 1200 = 5.0 m/s.
Common mistakeChoosing 1.7 m/s forgets the car's own momentum (it was already moving); add both momenta.
32Short answer
In an inelastic collision, kinetic energy is 'lost'. Where does it go? Give three forms and explain why the total energy is still conserved.
Show answer
Model answer: It becomes heat (internal energy) in the bodies, sound, and the energy used to bend or crush them (deformation). The kinetic energy is changed into these forms, not destroyed, so the total energy is the same before and after.
Common mistakeWriting that energy is 'used up' suggests it disappears; it is transferred to other forms.
33Multiple choice · ★ Challenge
A bat hits a ball. The force on the ball rises steadily from 0 to 600 N in 0.002 s, stays at 600 N for 0.004 s, then falls steadily to 0 in 0.002 s. What impulse does the ball receive?
1. A3.6 N s
2. B4.8 N s
3. C2.4 N s
4. D1.2 N s
Show answer
Answer: A. 3.6 N s
Area = two triangles + rectangle = 2 × ½ × 600 × 0.002 + 600 × 0.004 = 1.2 + 2.4 = 3.6 N s.
Common mistakeChoosing 4.8 N s multiplies the peak force by the whole time (0.008 s), as if the force were 600 N all the time.
34Short answer · ★ Challenge
In an experiment, a 1.0 kg trolley at 0.60 m/s hits a stationary 1.0 kg trolley and they move off together at 0.28 m/s. Is momentum conserved within experimental error? Suggest one reason for the difference.
Show answer
Model answer: Before: 1.0 × 0.60 = 0.60 kg·m/s. After: 2.0 × 0.28 = 0.56 kg·m/s, about 7% less. This is close, so momentum is conserved within error; the small loss is probably due to friction not being fully compensated (or timing errors in reading the tape).
Common mistakeConcluding 'momentum is not conserved' ignores experimental error; a 7% difference is explained by friction.
35Short answer
Explain why a jet of water hitting a wall pushes on the wall, and why the push is larger if the water bounces back instead of running down the wall.
Show answer
Model answer: The wall changes the water's momentum, so it pushes on the water; by Newton's third law the water pushes back on the wall. If the water bounces back, its velocity changes by more (from +v to −v instead of to 0), so Δp each second, and the force, is larger (up to twice).
Common mistakeThinking the force depends only on the water's speed ignores that the force is the rate of CHANGE of momentum.
36Short answer · ★ Challenge
Use Newton's third law to show that the total momentum of two trolleys A and B is the same before and after they collide.
Show answer
Model answer: During contact A pushes B with force F and B pushes A with −F (third law), for the same time Δt. So Δp_B = FΔt and Δp_A = −FΔt. Adding: Δp_A + Δp_B = 0, so the total momentum does not change.
Common mistakeStating 'momentum is conserved' without using equal forces and equal times does not show WHY.
37Short answer
At a school sports day, high jumpers land on a thick foam mattress instead of on the grass. Explain why, using the idea of momentum.
Show answer
Model answer: The jumper must lose the same momentum whatever she lands on. The soft mattress squashes and makes the stopping time much longer, so the average force F = Δp/Δt on her body is much smaller and less likely to injure her.
Common mistakeSaying the mattress 'reduces the momentum' is wrong; it reduces the force by making the time longer.
38Short answer · ★ Challenge
The same ball is dropped from the same height onto a concrete floor and onto a thick gym mat; it stops without bouncing each time. Describe how the two force–time graphs differ, and say what stays the same.
Show answer
Model answer: Concrete: a tall, narrow peak (large force, short time). Mat: a low, wide curve (small force, long time). The areas are equal because the ball loses the same momentum, so the impulse is the same.
Common mistakeSaying the mat 'absorbs momentum' so the area is smaller is wrong; the ball stops in both cases, so Δp and the area are equal.
39Multiple choice
A 40 kg child running at 3.0 m/s jumps onto a 20 kg cart standing still. They move off together. What is their speed?
1. A3.0 m/s
2. B2.0 m/s
3. C1.5 m/s
4. D6.0 m/s
Show answer
Answer: B. 2.0 m/s
Momentum before = 40 × 3.0 = 120 kg·m/s = (40 + 20) × v, so v = 120 ÷ 60 = 2.0 m/s.
Common mistakeChoosing 6.0 m/s divides by the cart's mass only; after jumping on, the child and cart move together, so use 60 kg.
40Multiple choice · ★ Challenge
A 0.020 kg pellet at 150 m/s hits and sticks in a 0.98 kg wooden block hanging on strings. How high does the block swing up? (g = 10 N/kg)
1. A0.45 m
2. B1125 m
3. C0.30 m
4. D0.90 m
Show answer
Answer: A. 0.45 m
Momentum: v = 0.020 × 150 ÷ 1.0 = 3.0 m/s. Energy for the swing: h = v²/2g = 9.0 ÷ 20 = 0.45 m.
Common mistakeChoosing 1125 m uses the pellet's speed in h = v²/2g; the block only moves at 3.0 m/s after the impact.
41Short answer · ★ Challenge
A 2.0 kg trolley moves to the right at 3.0 m/s and an identical trolley moves to the left at 3.0 m/s. Find their total momentum and their total kinetic energy, and explain why one total is zero and the other is not.
Show answer
Model answer: Momentum: 2.0 × 3.0 + 2.0 × (−3.0) = 0. Kinetic energy: 2 × ½ × 2.0 × 3.0² = 18 J. Momentum is a vector, so opposite directions cancel; kinetic energy is a scalar and is never negative, so it cannot cancel.
Common mistakeLearners often give 12 kg·m/s for the total momentum by ignoring direction, or think KE must also be zero.
42Fill in the blank
A horizontal line on a momentum–time graph shows that the resultant force on the body is ______.
Show answer
Answer: zero
A horizontal line has zero gradient, so Δp/Δt = 0 and the resultant force is zero.
Common mistakeSaying the body is at rest is wrong; a horizontal line at a non-zero value means steady motion, not rest.
43Fill in the blank · ★ Challenge
A 2.0 kg trolley at 3.0 m/s hits and sticks to a 1.0 kg trolley at rest, and the pair rolls up a smooth slope. They rise ______ m higher before stopping. (g = 10 N/kg)
Show answer
Answer: 0.20
v = 2.0 × 3.0 ÷ 3.0 = 2.0 m/s; h = v²/2g = 4.0 ÷ 20 = 0.20 m.
Common mistakeAnswering 0.45 uses 3.0 m/s; after sticking, the pair moves at only 2.0 m/s.
44Multiple choice
A 1200 kg car speeds up from 10 m/s to 25 m/s in 6.0 s. What is the average resultant force on it?
1. A5000 N
2. B2000 N
3. C3000 N
4. D18 000 N
Show answer
Answer: C. 3000 N
Δp = 1200 × (25 − 10) = 18 000 kg·m/s; F = Δp/Δt = 18 000 ÷ 6.0 = 3000 N.
Common mistakeChoosing 5000 N uses the final speed only; the force depends on the CHANGE of velocity.
45Multiple choice · ★ Challenge
A 0.010 kg bullet at 300 m/s embeds itself in a 2.99 kg block resting on a smooth table. How much kinetic energy is changed into other forms?
1. A0 J (kinetic energy is kept)
2. B1.5 J
3. Cabout 449 J
4. Dabout 225 J
Show answer
Answer: C. about 449 J
v = 0.010 × 300 ÷ 3.0 = 1.0 m/s. KE before = ½ × 0.010 × 300² = 450 J; after = ½ × 3.0 × 1.0² = 1.5 J; lost ≈ 449 J.
Common mistakeChoosing 1.5 J gives the KE that is left, not the KE that is changed into heat and sound.
46Multiple choice · ★ Challenge
The velocity–time graph of a 2.0 kg trolley is a straight line from 1.0 m/s at t = 0 to 5.0 m/s at t = 4.0 s. What is the resultant force on it?
1. A1.0 N
2. B2.5 N
3. C2.0 N
4. D8.0 N
Show answer
Answer: C. 2.0 N
Δp = 2.0 × (5.0 − 1.0) = 8.0 kg·m/s; F = 8.0 ÷ 4.0 = 2.0 N.
Common mistakeChoosing 1.0 N gives the acceleration (gradient of the v–t graph) and forgets to multiply by the mass.
47Multiple choice
A moving lorry hits a stationary moto. How do the forces they exert on each other compare during the collision?
1. AThe lorry's force on the moto is larger
2. BThe moto's force on the lorry is larger
3. CEqual in size and in the same direction
4. DEqual in size and opposite in direction
Show answer
Answer: D. Equal in size and opposite in direction
By Newton's third law the forces form a pair: equal in size, opposite in direction, acting on different bodies.
Common mistakeChoosing 'the lorry's force is larger' mixes up force with effect; the moto is damaged more because its mass is smaller.
48Multiple choice · ★ Challenge
A 60 kg driver moving at 15 m/s is stopped by hitting the steering wheel in 0.10 s, or by an airbag in 0.60 s. How do the average forces compare?
1. A9000 N and 1500 N: six times less with the airbag
2. B9000 N in both cases, as Δp is the same
3. C1500 N and 9000 N: larger with the airbag
4. D900 N and 150 N: six times smaller with the airbag
Show answer
Answer: A. 9000 N and 1500 N: six times less with the airbag
Δp = 60 × 15 = 900 kg·m/s. F = 900 ÷ 0.10 = 9000 N; F = 900 ÷ 0.60 = 1500 N.
Common mistakeChoosing 'the same in both cases' confuses the change of momentum (equal) with the force (which depends on time).
49Multiple choice
On a graph of the momentum of a body against time, what does the gradient give?
1. AThe resultant force on the body
2. BThe total impulse given to the body
3. CThe acceleration of the body
4. DThe kinetic energy of the body
Show answer
Answer: A. The resultant force on the body
Gradient = Δp/Δt, and by Newton's second law this is the resultant force.
Common mistakeChoosing impulse confuses gradient with area; the area under a FORCE–time graph is the impulse.
50Short answer · ★ Challenge
A moto starting from rest has these momenta: t = 0, 1, 2, 3 s; p = 0, 450, 900, 1350 kg·m/s. Describe the shape of the momentum–time graph and find the resultant force on the moto.
Show answer
Model answer: The points lie on a straight line through the origin, so the momentum increases steadily. Gradient = 1350 ÷ 3 = 450 kg·m/s², so the resultant force is a constant 450 N.
Common mistakeReading the force as 1350 N takes the last momentum value instead of the gradient Δp/Δt.