Senior 4 practice book · Unit 4 of 10

Conservation of Linear Momentum and Collision

50 questions that complete the Senior 4 quiz for this unit: 23 core and 27 challenge. Easy and hard questions are mixed. Try each question, then tap “Show answer”.

Common misconceptions

A body with a lot of kinetic energy must also have a lot of momentum, and the two are conserved together.Momentum (mv) is a vector and is conserved in every collision of an isolated system; kinetic energy (½mv²) is a scalar and is conserved only in elastic collisions.
A soft landing reduces the change in momentum of a falling person.The change in momentum is the same whatever you land on; a soft surface makes the stopping time longer, so the force (Δp/Δt) is smaller.
A rocket moves forward by pushing against the air behind it.The rocket pushes gas backwards and the gas pushes the rocket forwards (Newton’s third law, momentum conservation); it works even better in empty space.
Carrying a heavy load across a level floor is a lot of work against gravity.Work against gravity needs movement along the direction of the force; moving sideways at constant height does no work against gravity, although your muscles tire.

What this unit covers

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

Work done by a force: W = Fd cos θ, negative work and work from a force–distance graph
Kinetic energy, gravitational potential energy and elastic potential energy
Conservation of mechanical energy and energy losses
Power: P = W/t and P = Fv
Efficiency of machines and energy-conversion devices
Linear momentum and Newton's second law as F = Δp/Δt
Impulse and force–time graphs
Impulse and safety: crumple zones, seat belts, airbags, helmets, landing mats
Conservation of momentum in recoil and explosions
Elastic, inelastic and perfectly inelastic collisions in one dimension
Coefficient of restitution and bouncing
Collisions in two dimensions
Thrust: rockets, hoses, wind and helicopters (force = rate of change of momentum)

Go to the questions

Questions (1–50)

Easier and harder questions are mixed, just like in a real exam. The 27 harder ones are marked ★ Challenge.

1True or false
No real machine can have an efficiency greater than 100%.
Show answer
Answer: True
That would mean getting out more useful energy than was put in, which breaks the principle of conservation of energy.
Common mistakeLearners sometimes confuse efficiency with mechanical advantage, which can be much larger than 1 for a lever or pulley.
2True or false · ★ Challenge
When a stationary firework bursts, the total kinetic energy of the pieces is zero, because the total momentum is zero.
Show answer
Answer: False
The momenta of the pieces are vectors that cancel, but each piece has positive kinetic energy, which comes from the chemical energy of the explosive.
Common mistakeThe error is treating kinetic energy like momentum; KE has no direction, so it cannot cancel out.
3True or false · ★ Challenge
A ball with a coefficient of restitution of 0.5 rises to one quarter of its previous height after each bounce.
Show answer
Answer: True
The rebound speed is half the impact speed, and height is proportional to speed squared, so each height is (0.5)² = 0.25 of the previous one.
Common mistakeThe tempting answer "one half" applies e to the height; e applies to the speeds, so the heights fall by e².
4True or false
If two lumps of clay moving towards each other with equal and opposite momenta collide and stick together, they stop dead.
Show answer
Answer: True
Total momentum before is zero, so after sticking the combined lump must have zero momentum: it stops (and all the KE becomes heat and deformation).
Common mistakeSome expect the heavier lump to "win" and carry on; with equal and opposite momenta neither wins, whatever their masses.
5Fill in the blank · ★ Challenge
A ball moving along the x-axis strikes a stationary ball and both move off at angles. After the collision, the y-components of the momenta of the two balls must add up to ______.
Show answer
Answer: zero
There was no momentum along y before the collision, and momentum is conserved in each direction, so the y-components must cancel.
Common mistakeSome learners expect the y-components to equal the x-momentum before; momentum along x and along y are conserved separately.
6True or false
The soft foam lining of a motorcycle helmet protects the rider’s head mainly by increasing the time over which the head is brought to rest.
Show answer
Answer: True
The foam crushes during an impact, so the head stops over a longer time and the force on it (Δp/Δt) is smaller.
Common mistakeMany think the hard shell does all the work; without the crushable foam the head would stop almost instantly and suffer a huge force.
7Fill in the blank · ★ Challenge
An egg dropped onto a pillow does not break, while the same egg dropped onto a concrete floor does, because the pillow increases the stopping ______ and so reduces the force.
Show answer
Answer: time
The egg’s change in momentum is the same in both cases; F = Δp/Δt is much smaller when Δt is long.
Common mistakeAnswering "momentum" is the trap: the egg loses the same momentum on both surfaces.
8True or false
A loaded lorry moving slowly and a bicycle moving fast can have exactly the same momentum.
Show answer
Answer: True
p = mv: a large mass with a small speed can give the same product as a small mass with a large speed (e.g. a 5000 kg lorry at 0.20 m/s and a 100 kg bicycle with rider at 10 m/s both have 1000 kg m/s).
Common mistakeThinking heavier objects always have more momentum forgets that speed matters equally in p = mv.
9Multiple choice · ★ Challenge
A moving trolley collides with an identical trolley at rest, and they couple together. What fraction of the original kinetic energy is lost?
1. ANone of it
2. BOne half
3. COne quarter
4. DAll of it
Show answer
Answer: B. One half
The common velocity is u/2. KE after = ½(2m)(u/2)² = ¼mu², which is half of ½mu², so half is lost.
Common mistakeChoosing "one quarter" compares the speeds squared ((½)² = ¼) but forgets that the moving mass has doubled.
10True or false
While a skydiver falls at constant (terminal) velocity, her gravitational potential energy decreases but her kinetic energy stays the same, so her mechanical energy is not conserved.
Show answer
Answer: True
All the potential energy lost is transferred to the air as heat and movement by air resistance; KE + PE falls steadily.
Common mistakeSome learners say energy is "destroyed" here; total energy is conserved, but mechanical energy is not, because work is done against air resistance.
11Multiple choice · ★ Challenge
A bus speeds up from 20 km/h to 60 km/h. By what factor does its kinetic energy increase?
1. A9 times
2. B3 times
3. C6 times
4. D27 times
Show answer
Answer: A. 9 times
KE = ½mv²; the speed is multiplied by 3, so KE is multiplied by 3² = 9.
Common mistakeChoosing 3 times treats KE as proportional to v; because v is squared, tripling the speed gives nine times the energy.
12Fill in the blank · ★ Challenge
A toy spring stores 4.0 J when it is compressed by 2.0 cm. Compressed by 4.0 cm (still obeying Hooke’s law), it stores ______ J.
Show answer
Answer: 16
E = ½kx², so doubling x multiplies the energy by 2² = 4: 4.0 × 4 = 16 J.
Common mistakeAnswering 8 J treats the energy as proportional to x; the force also grows with x, so the energy grows with x².
13Multiple choice
How does a helicopter stay hovering at one height?
1. AIts spinning blades cancel the pull of gravity on the helicopter
2. BAir pressure below is larger simply because the helicopter is heavy
3. CThe rotor pushes against the ground, which pushes the helicopter up
4. DIt pushes air downwards; the air’s rate of momentum change equals the weight
Show answer
Answer: D. It pushes air downwards; the air’s rate of momentum change equals the weight
The rotor gives a downward momentum to a large mass of air every second; by Newton’s third law the air pushes up on the rotor with a force equal to this rate of change of momentum.
Common mistakeChoosing "cancels gravity" suggests gravity is switched off; gravity still acts, and it is balanced by the upward force from the air.
14True or false
A woman carrying a basket on her head while walking at steady speed along a level road does no work on the basket against gravity.
Show answer
Answer: True
Her upward force on the basket is at 90° to the horizontal displacement, so W = Fd cos 90° = 0 (her muscles still use energy, but no work is done on the basket).
Common mistakeAnswering "false" confuses feeling tired with doing work; physical work needs a displacement along the force.
15Multiple choice · ★ Challenge
A football hits a wall at 8.0 m/s and rebounds along the same line at 6.0 m/s. What is the coefficient of restitution between the ball and the wall?
1. A1.33
2. B0.25
3. C0.56
4. D0.75
Show answer
Answer: D. 0.75
e = speed of separation ÷ speed of approach = 6.0 ÷ 8.0 = 0.75.
Common mistakeChoosing 0.56 uses the ratio of kinetic energies (6² ÷ 8²); e is the ratio of the SPEEDS.
16Multiple choice · ★ Challenge
A fire hose sends out 15 kg of water each second at a speed of 20 m/s. What backward force must the firefighters exert to hold the hose still?
1. A0.75 N
2. B3000 N
3. C300 N
4. D150 N
Show answer
Answer: C. 300 N
Force = rate of change of momentum = (mass per second) × velocity = 15 × 20 = 300 N.
Common mistakeChoosing 150 N brings in a factor of ½ as if working out kinetic energy; thrust is simply mass per second × change in velocity, with no ½.
17Fill in the blank
Electricity bills use the kilowatt-hour; one kilowatt-hour is equal to ______ J.
Show answer
Answer: 3.6 × 10⁶ (3 600 000)
1 kWh = 1000 W × 3600 s = 3 600 000 J.
Common mistakeAnswering 1000 J or 3600 J forgets one of the two factors: 1 kW = 1000 J per second, and 1 hour = 3600 s.
18Multiple choice · ★ Challenge
A truck on the Kigali–Musanze road travels at a steady 20 m/s while its engine delivers 60 kW to the wheels. What is the total resistive force on the truck?
1. A1.2 × 10⁶ N
2. B300 N
3. C3000 N
4. D3.0 N
Show answer
Answer: C. 3000 N
At steady speed the driving force equals the resistance: P = Fv, so F = P/v = 60 000 ÷ 20 = 3000 N.
Common mistakeChoosing 1.2 × 10⁶ N multiplies power by speed; from P = Fv the force is power DIVIDED by speed.
19Multiple choice
How does the crumple zone at the front of a car reduce injuries in a crash?
1. AIt reduces the change in momentum of the passengers
2. BIt makes the car bounce back off the obstacle
3. CIt makes the collision last longer, so the force is smaller
4. DIt makes the collision shorter, so it is over quickly
Show answer
Answer: C. It makes the collision last longer, so the force is smaller
The passengers’ change in momentum is fixed by their speed; crumpling stretches the stopping time, and F = Δp/Δt then gives a smaller force.
Common mistakeChoosing "reduces the change in momentum" is wrong: the car still goes from full speed to rest; only the time (and so the force) changes.
20Fill in the blank · ★ Challenge
A crane of efficiency 40% takes 5.0 kW of electrical power. It lifts a 500 kg load at a steady speed of ______ m/s. (g = 10 N/kg)
Show answer
Answer: 0.40
Useful power = 0.40 × 5000 = 2000 W = mgv, so v = 2000 ÷ (500 × 10) = 0.40 m/s.
Common mistakeUsing all 5.0 kW gives 1.0 m/s; only the useful 40% of the input power raises the load.
21Fill in the blank
The thrust on a rocket equals the mass of gas it throws out each second multiplied by the ______ of the gas relative to the rocket.
Show answer
Answer: speed (velocity)
Thrust = rate of change of momentum of the gas = (Δm/Δt) × v.
Common mistakeAnswering "acceleration" is the usual slip; the gas’s momentum per kilogram depends on its speed, not on an acceleration.
22Multiple choice · ★ Challenge
A trolley is released from rest at the top of a smooth track 20 m high. What is its speed when it passes over the top of a second hump 15 m high? (g = 10 m/s², no friction)
1. A17.3 m/s
2. B20 m/s
3. C10 m/s
4. D5.0 m/s
Show answer
Answer: C. 10 m/s
It has lost 20 − 15 = 5 m of height: ½v² = gΔh, so v = √(2 × 10 × 5) = 10 m/s.
Common mistakeChoosing 17.3 m/s uses the 15 m height of the hump; only the height LOST since the start turns into kinetic energy.
23Fill in the blank
For a perfectly elastic collision the coefficient of restitution is ______, and for a perfectly inelastic collision it is 0.
Show answer
Answer: 1
In a perfectly elastic collision the bodies separate as fast as they approached, so e = 1; if they stick together they do not separate at all, so e = 0.
Common mistakeAnswering 100 or "infinite" is a slip; e is a ratio of two speeds and lies between 0 and 1.
24Multiple choice · ★ Challenge
The momentum of a car increases from 15 000 kg m/s to 30 000 kg m/s in 5.0 s. What is the average resultant force on the car?
1. A6000 N
2. B9000 N
3. C3000 N
4. D10 N
Show answer
Answer: C. 3000 N
F = Δp/Δt = (30 000 − 15 000) ÷ 5.0 = 15 000 ÷ 5.0 = 3000 N.
Common mistakeChoosing 6000 N divides the final momentum by the time; the force depends on the CHANGE of momentum.
25Multiple choice
An electric winch uses 2000 J of electrical energy to raise a load, which gains 1500 J of potential energy. What is the efficiency of the winch?
1. A75%
2. B133%
3. C25%
4. D50%
Show answer
Answer: A. 75%
Efficiency = useful output ÷ input × 100% = 1500 ÷ 2000 × 100% = 75%.
Common mistakeChoosing 133% divides input by output; efficiency can never exceed 100%, so the useful energy must go on top.
26Short answer · ★ Challenge
Volleyball coaches tell players to "follow through" when they spike the ball, keeping the hand in contact with it for longer. Use the idea of impulse to explain why the ball then leaves faster.
Show answer
Model answer: Impulse = force × contact time = change in momentum of the ball. With the same force, a longer contact time gives a larger impulse, so the ball gains more momentum and leaves with a higher velocity.
Common mistakeSaying a follow-through gives "more force" misses the point: the player’s force may be the same; it is the extra TIME that increases the impulse.
27Multiple choice
A 50 kg girl and a 75 kg boy stand still on smooth ice. She pushes him and moves off at 1.5 m/s. What is the boy’s velocity?
1. A1.5 m/s in the opposite direction
2. B2.25 m/s in the opposite direction
3. C1.0 m/s in the same direction
4. D1.0 m/s in the opposite direction
Show answer
Answer: D. 1.0 m/s in the opposite direction
Total momentum stays zero: 50 × 1.5 = 75 × v, so v = 75 ÷ 75 = 1.0 m/s, opposite to the girl.
Common mistakeChoosing 1.5 m/s assumes equal speeds; it is the MOMENTA that are equal and opposite, so the heavier boy moves more slowly.
28Multiple choice · ★ Challenge
On a smooth floor a 2.0 kg puck moving east at 3.0 m/s collides with a 1.0 kg puck moving north at 4.0 m/s, and they stick together. What is their velocity afterwards?
1. A2.4 m/s at 34° north of east
2. B2.4 m/s at 56° north of east
3. C3.3 m/s at 34° north of east
4. D7.2 m/s at 34° north of east
Show answer
Answer: A. 2.4 m/s at 34° north of east
East: 2.0 × 3.0 = 6.0 kg m/s; north: 1.0 × 4.0 = 4.0 kg m/s. p = √(6.0² + 4.0²) = 7.2 kg m/s; v = 7.2 ÷ 3.0 = 2.4 m/s; tan θ = 4.0 ÷ 6.0, θ = 34° north of east.
Common mistakeChoosing 7.2 m/s forgets to divide the total momentum by the combined mass of 3.0 kg.
29Multiple choice · ★ Challenge
A 60 kg passenger travelling at 15 m/s is stopped in a crash. With a seat belt she stops in 0.30 s; without one she hits the dashboard and stops in 0.010 s. What are the average forces on her?
1. A3.0 kN with the belt, 90 kN without
2. B90 kN with the belt, 3.0 kN without
3. C270 N with the belt, 9.0 N without
4. D3.0 kN with the belt, 9.0 kN without
Show answer
Answer: A. 3.0 kN with the belt, 90 kN without
Δp = 60 × 15 = 900 kg m/s. With the belt F = 900 ÷ 0.30 = 3000 N; without it F = 900 ÷ 0.010 = 90 000 N.
Common mistakeChoosing 270 N and 9.0 N multiplies Δp by the time instead of dividing; F = Δp/Δt.
30Multiple choice
Which has the greater momentum: a 0.010 kg bullet moving at 400 m/s, or a 60 kg athlete jogging slowly at 0.10 m/s?
1. AThe athlete: 6.0 kg m/s against 4.0 kg m/s
2. BThe bullet: 4.0 kg m/s against 0.60 kg m/s
3. CThe bullet, because it is so much faster
4. DThey have equal momentum, 4.0 kg m/s each
Show answer
Answer: A. The athlete: 6.0 kg m/s against 4.0 kg m/s
p = mv: bullet 0.010 × 400 = 4.0 kg m/s; athlete 60 × 0.10 = 6.0 kg m/s.
Common mistakeChoosing "the bullet, because it is faster" ignores mass: momentum depends on mass AND velocity.
31Multiple choice · ★ Challenge
A moto taxi skids to a stop over 25 m while the road exerts a friction force of 600 N on it. How much work does friction do on the moto?
1. A+15 kJ
2. B−15 kJ
3. C0 J
4. D−24 J
Show answer
Answer: B. −15 kJ
Friction acts backwards while the moto moves forwards (θ = 180°): W = Fd cos 180° = 600 × 25 × (−1) = −15 000 J.
Common mistakeChoosing +15 kJ ignores the direction: a force opposing the motion does negative work, which removes kinetic energy.
32Short answer
Explain why long jumpers land in a sand pit and high jumpers land on thick foam mats rather than on the hard ground.
Show answer
Model answer: The athlete’s change in momentum on landing is the same on any surface. Sand or a thick mat lets the body sink in and stop gradually, so the stopping time is much longer. Since F = Δp/Δt, the force on the body is much smaller and injuries are avoided.
Common mistakeSaying the mat "absorbs momentum" is a common slip; momentum is still lost completely, only more slowly.
33Short answer · ★ Challenge
A child on a swing is released from rest 1.25 m above the lowest point and passes the lowest point at 4.5 m/s. What percentage of the starting energy has been lost, and where has it gone? (g = 10 m/s²)
Show answer
Model answer: Per kilogram: PE at the start = gh = 10 × 1.25 = 12.5 J; KE at the bottom = ½v² = ½ × 4.5² = 10.1 J. Lost = 12.5 − 10.1 = 2.4 J, which is 2.4 ÷ 12.5 ≈ 19%. It has been transferred to heat and sound by air resistance and friction at the swing’s chains/hinges.
Common mistakeYou do not need the child’s mass: it cancels, so working per kilogram is allowed; forgetting this makes many learners think the question cannot be done.
34Multiple choice
A 0.50 kg ball rolling at 4.0 m/s is given an impulse of 6.0 N s in its direction of motion. What is its new speed?
1. A12 m/s
2. B8.0 m/s
3. C3.0 m/s
4. D16 m/s
Show answer
Answer: D. 16 m/s
Δv = impulse ÷ m = 6.0 ÷ 0.50 = 12 m/s, so v = 4.0 + 12 = 16 m/s.
Common mistakeChoosing 12 m/s gives the change in velocity only; the ball was already moving at 4.0 m/s.
35Short answer · ★ Challenge
Show that Newton’s second law written as F = Δp/Δt becomes F = ma when the mass of the body does not change.
Show answer
Model answer: Δp = mv − mu = m(v − u) when m is constant. So F = Δp/Δt = m(v − u)/Δt. Since (v − u)/Δt is the acceleration a, F = ma.
Common mistakeSome learners write F = mv/t; the law is about the CHANGE of momentum (mv − mu), not the momentum itself.
36Fill in the blank
The area under a force–time graph is equal to the ______ given to the body.
Show answer
Answer: impulse (change in momentum)
Impulse = FΔt, which for a changing force is the area under the F–t graph, and it equals Δp.
Common mistakeTaking the gradient of an F–t graph gives the rate of change of force, not the impulse.
37Multiple choice · ★ Challenge
A force–time graph for a kick shows a constant 40 N for the first 0.30 s, then a straight-line fall to zero at 0.50 s. What impulse is given?
1. A20 N s
2. B16 N s
3. C12 N s
4. D10 N s
Show answer
Answer: B. 16 N s
Impulse = area: rectangle 40 × 0.30 = 12 N s, plus triangle ½ × 40 × 0.20 = 4 N s; total 16 N s.
Common mistakeChoosing 20 N s treats the force as 40 N for the whole 0.50 s; the falling part is a triangle, worth only half of a rectangle.
38Multiple choice
A pump on a tea farm lifts 600 kg of water through a height of 15 m in 2.0 minutes. What useful power does it deliver? (g = 10 N/kg)
1. A45 kW
2. B750 W
3. C75 W
4. D1500 W
Show answer
Answer: B. 750 W
Work = mgh = 600 × 10 × 15 = 90 000 J; time = 2.0 × 60 = 120 s; P = 90 000 ÷ 120 = 750 W.
Common mistakeChoosing 45 kW divides by 2 instead of 120: minutes must be changed to seconds before using P = W/t.
39Short answer · ★ Challenge
A 1.0 kg ball and a 4.0 kg ball have the same kinetic energy. Show that the heavier ball has twice the momentum of the lighter one.
Show answer
Model answer: From KE = ½mv² and p = mv, p² = m²v² = 2m × KE, so p = √(2m KE). With the same KE, p ∝ √m: p(4 kg) ÷ p(1 kg) = √(4/1) = 2. (For example with KE = 8 J: the 1 kg ball has v = 4 m/s, p = 4 kg m/s; the 4 kg ball has v = 2 m/s, p = 8 kg m/s.)
Common mistakeAssuming equal KE means equal momentum ignores that the lighter ball needs a much higher speed; momentum depends on √m at equal energy.
40Short answer · ★ Challenge
A strong wind blowing at 12 m/s hits a wall of area 10 m² at right angles and is brought to rest. The density of air is 1.2 kg/m³. Estimate the force of the wind on the wall.
Show answer
Model answer: Volume of air arriving per second = A × v = 10 × 12 = 120 m³; mass per second = 1.2 × 120 = 144 kg/s. Force = rate of change of momentum = 144 × 12 = 1728 N ≈ 1.7 kN.
Common mistakeUsing F = mv with only one of the two factors of v misses that the speed appears twice: once in the mass arriving per second and once in the momentum each kilogram carries.
41Multiple choice
Which of these is the best example of a nearly elastic collision?
1. AA lump of wet clay hitting the floor
2. BTwo steel balls colliding in a Newton’s cradle
3. CTwo cars crumpling in a crash
4. DA bullet burying itself in a block
Show answer
Answer: B. Two steel balls colliding in a Newton’s cradle
Hard steel balls deform very little and spring back, so almost no kinetic energy is lost; the others all lose a lot of KE to heat and deformation.
Common mistakeChoosing the car crash because the cars "bounce apart" misses the large energy lost in crumpling; elastic means no KE lost.
42Multiple choice
A pendulum bob passes through its lowest point at 2.0 m/s. How high above the lowest point does it rise before stopping? (g = 10 m/s², no air resistance)
1. A0.10 m
2. B0.40 m
3. C2.0 m
4. D0.20 m
Show answer
Answer: D. 0.20 m
mgh = ½mv², so h = v²/(2g) = 2.0² ÷ 20 = 0.20 m.
Common mistakeChoosing 0.40 m forgets the ½ in kinetic energy (h = v²/g).
43Short answer · ★ Challenge
A tennis ball dropped from 1.8 m onto a classroom floor bounces back up to 0.80 m. Find the speed just before and just after the bounce, and the coefficient of restitution. (g = 10 m/s², no air resistance)
Show answer
Model answer: Before: v = √(2gh) = √(2 × 10 × 1.8) = 6.0 m/s. After: v = √(2 × 10 × 0.80) = 4.0 m/s. e = 4.0 ÷ 6.0 = 0.67 (also = √(0.80/1.8)).
Common mistakeWriting e = 0.80 ÷ 1.8 = 0.44 uses the heights directly; e is a ratio of speeds, and speed depends on the square root of height.
44Short answer · ★ Challenge
At a traffic light in Remera, a 1200 kg car moving at 15 m/s runs into the back of an 800 kg car waiting at rest, and the two cars lock together. Find their speed just after the impact and the kinetic energy changed into other forms.
Show answer
Model answer: Momentum: 1200 × 15 = (1200 + 800)v, so v = 18 000 ÷ 2000 = 9.0 m/s. KE before = ½ × 1200 × 15² = 135 000 J; KE after = ½ × 2000 × 9.0² = 81 000 J; lost = 54 000 J (54 kJ), to heat, sound and bending metal.
Common mistakeUsing energy conservation to find the speed (½ × 1200 × 15² = ½ × 2000 × v²) gives 11.6 m/s, which is wrong; in a sticking collision only momentum is conserved.
45Multiple choice
A 70 kg hiker climbs Mount Kigali, rising 400 m above her starting point (g = 10 N/kg). How much gravitational potential energy does she gain?
1. A28 kJ
2. B2.8 MJ
3. C0.28 kJ
4. D280 kJ
Show answer
Answer: D. 280 kJ
ΔPE = mgh = 70 × 10 × 400 = 280 000 J = 280 kJ.
Common mistakeChoosing 28 kJ drops g (70 × 400); the weight in newtons, mg, must be used, not the mass.
46Short answer · ★ Challenge
Trolleys of mass 0.50 kg and 1.5 kg rest against each other with a compressed spring between them. When the spring is released, the lighter trolley moves off at 3.0 m/s. Find the velocity of the heavier trolley and the energy that was stored in the spring.
Show answer
Model answer: Momentum before = 0, so 0.50 × 3.0 = 1.5 × v, giving v = 1.0 m/s in the opposite direction. Energy stored = total KE after = ½ × 0.50 × 3.0² + ½ × 1.5 × 1.0² = 2.25 + 0.75 = 3.0 J.
Common mistakeTaking only the lighter trolley’s KE (2.25 J) forgets that both trolleys move; the spring’s energy goes to both.
47Short answer
In a game of snooker a moving ball strikes a stationary ball off-centre, and the two balls move off at angles. Explain why momentum must be treated in two perpendicular directions to analyse this collision.
Show answer
Model answer: Momentum is a vector, so it is conserved in each direction separately. Along the original line, the components of the two balls’ momenta after the collision add up to the first ball’s momentum. At right angles to it, the total was zero before, so the sideways components of the two balls must be equal and opposite afterwards.
Common mistakeAdding the sizes of the momenta after the collision and comparing with the momentum before ignores direction; magnitudes of vectors at an angle do not simply add.
48Short answer · ★ Challenge
At a hydroelectric station in the Northern Province, 25 000 kg of water flows each second through turbines 120 m below the dam surface. The station is 85% efficient. Find its electrical power output. (g = 10 N/kg)
Show answer
Model answer: Energy lost by the water per second = mgh per second = 25 000 × 10 × 120 = 3.0 × 10⁷ J/s, so input power = 30 MW. Output = 0.85 × 30 MW = 25.5 MW.
Common mistakeForgetting the efficiency gives 30 MW; some energy is always lost as heat and sound in the turbines and generator.
49Short answer
Two learners of the same weight run up the same flight of stairs, one in 8 s and the other in 12 s. Compare the work they do against gravity and the power they develop.
Show answer
Model answer: Both do the same work against gravity, because work = weight × vertical height, and these are the same. The learner taking 8 s develops more power, since P = W/t: 12 ÷ 8 = 1.5 times the power of the slower learner.
Common mistakeSaying the faster learner does more work confuses work with power; time affects only the rate at which the work is done.
50Short answer · ★ Challenge
The force pulling a cart rises steadily from 0 to 80 N over the first 4.0 m, then stays at 80 N for the next 6.0 m. Use the force–distance graph to find the total work done.
Show answer
Model answer: Work = area under the graph. Triangle: ½ × 4.0 × 80 = 160 J. Rectangle: 6.0 × 80 = 480 J. Total = 160 + 480 = 640 J.
Common mistakeUsing 80 × 10 = 800 J assumes the force was 80 N all the way; while the force is rising, the work is only the triangular area.