Answers · Resultant force

Balanced and unbalanced forces — answer key

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

1. FQ1 · Practice

Model answer. The resultant force is 0.20 N east. The two pulls act along the same straight line in opposite directions, so one takes away from the other: 0.35 N − 0.15 N = 0.20 N. The bigger pull is towards the east, so the resultant points east. A resultant is not finished until it has a size and a direction.

A full answer includes

I chose “0.20 N east”.
I subtracted the opposite pulls: 0.35 N − 0.15 N = 0.20 N.
I gave the direction of the bigger pull, east.
I did not add the pulls to get 0.50 N.

Also accept

b: 0.20 N east; 0.35 N − 0.15 N, in the direction of the bigger pull.

Watch for

“a: 0.50 N east: add the two pulls.” — Only forces in the same direction add up. These two pull opposite ways, so subtract: 0.35 N − 0.15 N.
“c: 0.20 N west.” — The size is right, but the resultant points the same way as the bigger pull, and the bigger pull, 0.35 N, is towards the east.
“d: 0.25 N east, halfway between the two pulls.” — The resultant is not an average. It is the single force with the same effect as both: subtract the opposite forces, 0.35 N − 0.15 N.

2. FQ2 · Practice

Model answer. First add the forces that point the same way. Towards the east: 0.25 N + 0.10 N = 0.35 N. Towards the west there is 0.40 N. These two totals are opposite, so subtract: 0.40 N − 0.35 N = 0.05 N. The west side is bigger, so the resultant force is 0.05 N towards the west. With east counted as positive: 0.25 + 0.10 − 0.40 = −0.05 N, and the minus sign means west.

Checked automatically

Size of the resultant force: 0.05 N (within ±0.005 N)
- Common wrong answer 0.75 : You added all three forces. The 0.40 N acts the other way, so it takes away: 0.25 + 0.10 − 0.40.
- Common wrong answer 0.35 : That is the total towards the east. Now take away the 0.40 N that acts towards the west.
- Common wrong answer 0.15 : You left out the 0.10 N east. Add both eastward forces first: 0.25 N + 0.10 N = 0.35 N.
- Common wrong answer 0.55 : You took the 0.10 N away, but it acts east, the same way as the 0.25 N. Add those two first.
- Common wrong answer -0.05 : Your working is right, with east as positive. Give the size here, 0.05 N, and write “west” as the direction.

A full answer includes

Method: I added the forces towards the east, 0.25 N + 0.10 N = 0.35 N.
Method: I took the east total away from the west force, 0.40 N − 0.35 N.
Answer: the resultant is 0.05 N.
Its direction is west, the side with the bigger total.
I did not add all three forces to get 0.75 N.

Also accept

0.05 N towards the west (−0.05 N with east as positive).

Watch for

“The resultant is 0.75 N, because there are three forces to add.” — Only the forces in the same direction add. The 0.40 N points the other way, so it takes away: 0.25 + 0.10 − 0.40.
“The resultant points east, because two of the three forces point east.” — The number of arrows does not decide the direction; their sizes do. East has 0.35 N in all, west has 0.40 N, so the resultant points west.

3. FQ3 · Practice

Options as numbered on the pupil sheet: 1 — stays at rest; 2 — keeps moving east at a steady speed; 3 — speeds up; 4 — slows down; 5 — changes direction, curving towards the north; 6 — stops at once.

Model answer. With no resultant force, the velocity does not change: a puck at rest stays at rest, and a moving puck keeps moving east at a steady speed in a straight line. A resultant force along the motion makes it speed up. A resultant force against the motion makes it slow down; if the force keeps acting, the puck stops and then moves back west. A resultant force at right angles to the motion changes the direction: the path curves towards the north. No force on the air table makes the puck stop at once: a force changes the velocity gradually.

A full answer includes

I matched “at rest, zero resultant” with “stays at rest”.
I matched “moving east, zero resultant” with “keeps moving east at a steady speed”.
I matched “resultant towards the east” with “speeds up”.
I matched “resultant towards the west” with “slows down”.
I matched “resultant towards the north” with “changes direction, curving towards the north”.
I did not use “stops at once” for any of them.

Also accept

at rest, zero – stays at rest; moving, zero – steady speed east; east – speeds up; west – slows down; north – changes direction.

Watch for

“Moving east with zero resultant: it slows down and stops.” — With no resultant force, nothing changes its velocity. On the air table the gaps between the dots stayed equal, 2 cm every 0.1 s, all the way across.
“Moving east with a resultant towards the west: it stops at once.” — A force changes the velocity gradually. In run 4 the puck took 0.6 s to stop, and went 9 cm further, before it came back.
“Moving east with a resultant towards the north: it turns and goes straight north at once.” — The path curves gradually. In run 5 the puck curved north more and more, and it was still moving east as well when it reached the rim.

4. FQ4 · Practice

Model answer. Nothing keeps it moving, and nothing needs to. The two pulls are equal and opposite, 0.10 N each way, so the resultant force is zero. With no resultant force, a moving object continues in a straight line at constant speed. The nudge acted only while the hand touched the puck; after that it was over. Meter W pulls no harder than meter E, so it does not drag the puck along. The air pushes the puck up, balancing its weight; it does not push it west.

A full answer includes

I chose “No force: the resultant force on it is zero”.
I explained that the two pulls are equal and opposite, so the resultant force is zero.
I explained that with no resultant force a moving object continues in a straight line at constant speed.
I did not say that the nudge keeps acting after the hand lets go.

Also accept

b: no force; the resultant force is zero, so the puck keeps a constant velocity.

Watch for

“a: The force of the nudge stays in the puck and keeps it going.” — A force acts only while something pushes or pulls. Once the hand let go, the nudge was over. The puck keeps going because nothing changes its velocity.
“c: Meter W pulls it along, because the puck moves towards it.” — Meter W pulls 0.10 N west and meter E pulls 0.10 N east. They balance, so together they have no effect on the motion.
“d: The air blowing up through the holes pushes it along.” — The air pushes the puck upwards, and that balances its weight. It does not push the puck along the table.

5. FQ5 · Practice

Model answer. (1) balanced and (2) zero: 60 N forwards and 60 N backwards cancel, so the resultant force is 0 N; both forces still act. (3) straight: with no resultant force, a moving object keeps going at a steady speed in a straight line. (4) decreases: a resultant force against the motion slows the canoe down. (5) direction: a force at right angles to the motion turns the canoe. (6) velocity: a velocity has a size and a direction, so it changes when the speed changes and when the direction changes.

A full answer includes

(1) balanced
(2) 0 (zero)
(3) straight
(4) decreases (goes down)
(5) direction
(6) velocity
I did not write “speed” for gap 6: the speed need not change when the canoe turns.

Also accept

balanced; 0; straight; decreases; direction; velocity.

Watch for

“(1) There are no forces on the canoe, because it moves at a steady speed.” — Both forces act: 60 N forwards and 60 N backwards. They are balanced, so they cancel and the resultant force is zero.
“(4) The canoe stops at once.” — A resultant force changes the velocity gradually. The drag of the water slows the canoe down bit by bit.
“(6) speed.” — When the wind turns the canoe, its speed need not change, but its direction does. The quantity that changes in both cases is the one that includes direction.

6. FQ6 · Practice

Model answer. A: 5 N to the left and 5 N to the right cancel, so the resultant is zero. B: the two forces to the left add up to 3 N + 5 N = 8 N, which balances the 8 N to the right. C: 6 N − 4 N = 2 N to the right, so it is unbalanced. D: up and down balance, 20 N each, and left and right balance, 4 N each, so the resultant is zero. E: 12 N − 10 N = 2 N up, so it is unbalanced. The forces are balanced in A, B and D.

A full answer includes

I chose A: 5 N each way.
I chose B: 3 N + 5 N = 8 N to the left balances 8 N to the right.
I chose D: up and down balance, and so do left and right.
I did not choose C: 6 N − 4 N leaves 2 N to the right.
I did not choose E: 12 N − 10 N leaves 2 N up.
I gave the resultant for C, 2 N to the right, or for E, 2 N up.

Also accept

a, b, d: A, B and D are balanced; C has 2 N to the right and E has 2 N up.

Watch for

“B is unbalanced, because it has three forces of different sizes.” — Add the forces that point the same way first: 3 N + 5 N = 8 N to the left, which balances 8 N to the right. The number of arrows does not matter.
“D is unbalanced, because its forces are the biggest.” — Balanced does not mean small. Up and down cancel, and left and right cancel, so the resultant is zero.
“C is balanced, because it has one force each way.” — One force each way balances only when the two are equal. Here 6 N − 4 N leaves 2 N to the right.
“E is balanced, because 12 N and 10 N are nearly the same.” — Nearly equal is not equal. 12 N up and 10 N down leave a resultant of 2 N up, so the box would start to move up.

7. FQ7 · Practice

Model answer. The weight of the beam is 450 kg × 9.8 N/kg = 4410 N, downwards. The cable pulls up with 5000 N. These two forces act along the same line in opposite directions, so subtract: 5000 N − 4410 N = 590 N. The bigger force is the cable’s, so the resultant is 590 N upwards. A beam at rest with an upward resultant force starts to move up, faster and faster. It would rise at a steady speed only if the cable pulled with exactly 4410 N.

Checked automatically

Size of the resultant force on the beam: 590 N (within ±5 N)
- Common wrong answer 9410 : You added the pull of the cable and the weight. They act in opposite directions, up and down, so subtract.
- Common wrong answer 4410 : That is the weight, 450 kg × 9.8 N/kg. Now compare it with the pull of the cable, 5000 N, which acts the other way.
- Common wrong answer 4550 : You took the weight as 450 N. Weight = mass × g = 450 kg × 9.8 N/kg.
- Common wrong answer 500 : You used g = 10 N/kg. With g = 9.8 N/kg the weight is 4410 N.
- Common wrong answer 5000 : That is the pull of the cable alone. The weight acts on the beam too, downwards.

A full answer includes

Method: weight = 450 kg × 9.8 N/kg = 4410 N, downwards.
Method: resultant = 5000 N − 4410 N.
Answer: the resultant is 590 N.
Its direction is upwards, the way of the bigger force.
Starting at rest, the beam speeds up as it rises.
I did not use 450 N as the weight.

Also accept

590 N upwards; from rest, the beam speeds up as it rises.

Watch for

“The cable pulls harder than the weight, so the beam rises at a steady speed.” — A steady speed needs a zero resultant. Here the resultant force is upwards, so the velocity changes: the beam speeds up as it rises.
“The mass is 450 kg, so the weight is 450 N.” — Mass, in kilograms, is not the same as weight, in newtons. Weight = mass × g = 450 kg × 9.8 N/kg = 4410 N.

8. FQ8 · Practice

Model answer. Jamal is wrong that the push stays inside the stone. A force acts only while one object pushes or pulls another: once the hand lets go, the push is over. A moving object does not need a force to keep it moving. If no resultant force acted on it, the stone would continue in a straight line at constant speed. It slows down because a small force does act on it, against its motion: friction from the ice. That is the resultant force, and it points backwards, so it slows the stone down. On smooth ice the friction is small, so the stone slows very gradually. On the air table, with almost no friction, the puck in run 2 did not slow down at all.

A full answer includes

I wrote that a push acts only while the hand touches the stone.
I wrote that once the stone is let go, there is no forward force on it.
I wrote that with no resultant force it would continue in a straight line at constant speed.
I wrote that it slows down because friction from the ice acts against its motion.
I did not say that the stone carries a store of force that gets used up.

Also accept

Wrong: no force is needed to keep it moving; once let go there is no forward push; it slows because friction from the ice acts against its motion.

Watch for

“The stone slows down because it runs out of force.” — A moving object does not carry a store of force. It slows down only when a force acts against its motion, here friction from the ice.
“On perfectly smooth ice the stone would still stop in the end.” — With nothing acting against its motion there would be no resultant force, so the stone would keep a constant speed in a straight line, like the puck in run 2.

9. FQ9 · Practice

Options as numbered on the pupil sheet: 1 — the speed only; 2 — the direction only; 3 — both of them; 4 — neither of them.

Model answer. The runner keeps the same speed but turns, so only the direction changes; the velocity still changes, because velocity includes direction. The ball falls along one straight line and goes faster: only its speed changes. Nothing changes for the train, so the resultant force on it is zero. The car slows down in a straight line: only the speed changes. Both change for the stone: its path curves downwards, so its direction changes, and it goes faster and faster, so its speed changes too.

A full answer includes

Runner on the bend: the direction only.
Falling ball: the speed only.
Train at a constant 30 m/s: neither.
Car braking: the speed only.
Stone off the cliff: both.
I did not choose “neither” for the runner: the direction changes, so the velocity changes.

Also accept

runner: direction only; ball: speed only; train: neither; car: speed only; stone: both.

Watch for

“Runner: neither, because the speed stays at 7.0 m/s.” — Velocity includes direction. Going round a bend, the direction changes, so the velocity changes and there must be a resultant force.
“Train: it needs a forward resultant force to keep going at 30 m/s.” — At a constant speed in a straight line the resultant force is zero: the push of the engine just balances the forces against the motion.
“Stone: the direction only, because the path curves.” — The stone also falls faster and faster, so its speed changes too. Both parts of its velocity change.

10. FQ10 · Practice

Model answer. Row B. Hovering still, the drone is at rest, so the resultant force is zero: the upward force equals the weight. Rising at a steady speed in a straight line, its velocity does not change, so again the resultant force is zero and the upward force equals the weight. No extra force is needed to keep it moving up. Rising faster and faster, its velocity changes, so there must be a resultant force upwards: the upward force is bigger than the weight.

A full answer includes

I chose row B.
I explained that rising at a steady speed is a constant velocity, so the resultant force is zero.
I explained that a zero resultant needs the upward force to equal the weight.
I did not choose row A: a steady rise does not need a bigger upward force.

Also accept

B: equal, equal, bigger; a steady rise has a zero resultant force.

Watch for

“Row A: rising at a steady speed needs an upward force bigger than the weight.” — A steady rise is a constant velocity, so the resultant force is zero: the upward force equals the weight. Only speeding up needs a bigger upward force.
“Row C: even hovering, the upward force must be bigger than the weight.” — Hovering still is being at rest, so the resultant force is zero: the upward force equals the weight. The drone needs no extra force to stay where it is.
“Row D: the upward force equals the weight in all three cases.” — With equal forces the velocity cannot change. To rise faster and faster, the upward force must be bigger than the weight.

11. FQ11 · Practice

Model answer. The pulls of meters W and E act along one line in opposite directions, so the resultant is the difference between them. It points west, so meter W is the bigger pull: W − 0.10 N = 0.15 N. So W = 0.10 N + 0.15 N = 0.25 N. Check: 0.25 N west and 0.10 N east leave 0.15 N west.

Checked automatically

Pull of meter W: 0.25 N (within ±0.005 N)
- Common wrong answer 0.05 : That leaves a resultant of 0.05 N towards the east, not 0.15 N west. Meter W must beat meter E by 0.15 N: 0.10 + 0.15.
- Common wrong answer 0.15 : That is the resultant. Meter W must also cancel the 0.10 N east: 0.15 N + 0.10 N.
- Common wrong answer 0.1 : Equal pulls give a zero resultant. Meter W must be 0.15 N bigger than meter E.

A full answer includes

I saw that meter W must pull harder, because the resultant points west.
Method: W − 0.10 N = 0.15 N.
Answer: W = 0.25 N.
I did not give 0.15 N, the size of the resultant.

Also accept

0.25 N (0.25 N − 0.10 N = 0.15 N towards the west).

Watch for

“W = 0.15 N, the size of the resultant.” — With 0.15 N west and 0.10 N east, the resultant would be only 0.05 N west. Meter W must cancel the east pull and then leave 0.15 N more.

12. FQ12 · Practice

Model answer. (a) In run 1 the puck moves the same distance every 0.1 s, along a straight line, so it moves at a constant speed in a straight line. (b) In run 2 it goes further every 0.1 s, so it speeds up. The resultant force acts along the motion, towards the east. (c) In run 3 it goes less far every 0.1 s, so it slows down while still moving east. The resultant force acts against the motion, towards the west. (d) In run 1 the puck moves 3 cm = 0.03 m every 0.1 s, so its speed is 0.03 m ÷ 0.1 s = 0.30 m/s.

A full answer includes

(a) Run 1: equal gaps, so a constant speed in a straight line.
(b) Run 2: the gaps grow, so the puck speeds up.
(b) The resultant force is along the motion, towards the east.
(c) Run 3: the gaps shrink, so the puck slows down.
(c) The resultant force is against the motion, towards the west.
(d) Method: 3 cm = 0.03 m in 0.1 s, so speed = 0.03 m ÷ 0.1 s.
(d) Answer: 0.30 m/s.
I did not give the direction of run 3’s force as east, the way the puck moves.

Also accept

(a) constant speed in a straight line; (b) speeds up, resultant east; (c) slows down, resultant west; (d) 0.30 m/s.

Watch for

“(a) The gaps are equal, so the puck is not moving.” — Equal gaps mean the puck moves the same distance in every 0.1 s: it moves at a constant speed.
“(c) The resultant force is towards the east, because the puck is still moving east.” — The puck moves east but slows down, so the resultant force acts against its motion, towards the west.
“(d) The speed is 30 m/s.” — 3 cm is 0.03 m, not 3 m: 0.03 m ÷ 0.1 s = 0.30 m/s.

13. FT1 · Apply

Model answer. (a) The rocket is at rest, so the resultant force on it is zero. Its weight, 2.0 × 10⁶ N, acts down, and the launch pad pushes up on it with an equal force: the forces are balanced. (b) After lift-off the thrust acts up and the weight acts down, along the same line, so subtract: 3.0 × 10⁶ N − 2.0 × 10⁶ N = 1.0 × 10⁶ N, upwards. (c) The resultant force is along the motion, so the rocket speeds up as it rises. (d) The weight and the air resistance both act downwards, so they add. The rocket rises at a steady speed in a straight line, so its velocity does not change and the resultant force is zero. So the thrust, upwards, is equal to the weight and the air resistance together. No extra force is needed to keep it rising; a bigger thrust would make it speed up again.

A full answer includes

(a) I stated that the resultant force is zero.
(a) I explained that the weight, 2.0 × 10⁶ N down, is balanced by the push of the pad, 2.0 × 10⁶ N up.
(b) Method: 3.0 × 10⁶ N − 2.0 × 10⁶ N.
(b) Answer: 1.0 × 10⁶ N, upwards.
(b) I did not add the forces to get 5.0 × 10⁶ N.
(c) It speeds up as it rises, because the resultant force is along its motion.
(d) The thrust is equal to the weight and the air resistance together.
(d) I explained that a steady speed in a straight line needs a zero resultant force.
(d) I did not say that the thrust must be bigger to keep the rocket rising.

Also accept

(a) zero, weight balanced by the pad; (b) 1.0 × 10⁶ N up; (c) speeds up; (d) the thrust equals the weight and the air resistance together.

Watch for

“(a) There are no forces on the rocket before lift-off.” — Its weight acts on it all the time. The pad pushes up with an equal force, so the forces are balanced and the resultant is zero.
“(b) The resultant is 5.0 × 10⁶ N: add the thrust and the weight.” — The thrust acts up and the weight acts down: opposite directions, so subtract.
“(d) The thrust must still be bigger than the weight and the air resistance, or the rocket would stop rising.” — A steady speed in a straight line is a constant velocity, so the resultant force is zero. A thrust bigger than the downward forces would make the rocket speed up.