Answers · Resultant force

F = ma and circular motion — answer key

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

1. FQ22 · Practice

Model answer. F = ma, so a = F ÷ m = 330 N ÷ 150 kg = 2.2 m/s². The acceleration is in the same direction as the resultant force, forwards, so the go-kart speeds up by 2.2 m/s every second.

Checked automatically

Acceleration of the go-kart: 2.2 m/s² (within ±0.02 m/s²)
- Common wrong answer 49500 : You multiplied the force by the mass. F = ma, so a = F ÷ m = 330 N ÷ 150 kg.
- Common wrong answer 0.45 : You divided the mass by the force. a = F ÷ m: the force goes on top, 330 ÷ 150.
- Common wrong answer 330 : That is the force. Divide it by the mass to get the acceleration.

A full answer includes

Method: a = F ÷ m = 330 N ÷ 150 kg.
Answer: a = 2.2 m/s².
Its direction is forwards, the direction of the resultant force.
I did not multiply the force by the mass.

Also accept

2.2 m/s² forwards.

Watch for

“a = F × m = 49 500 m/s².” — No go-kart could speed up that fast. F = ma, so a = F ÷ m: divide the force by the mass.
“a = m ÷ F = 0.45 m/s².” — The force goes on top: a = F ÷ m = 330 ÷ 150. A bigger force must give a bigger acceleration.

2. FQ23 · Practice

Model answer. Her acceleration is towards the west. In F = ma the mass is a scalar, so the acceleration always points the same way as the resultant force. She is still moving east, but her acceleration points west, against her motion, so she slows down. This is a deceleration. The acceleration does not wait until she stops: it points west from the moment her foot drags.

A full answer includes

I chose “Towards the west, like the force”.
I explained that the acceleration is always in the same direction as the resultant force.
I explained that an acceleration against her motion means she slows down.
I did not choose “towards the east” because she is moving east.

Also accept

b: towards the west, the direction of the resultant force; she slows down.

Watch for

“a: Towards the east, because she is moving east.” — The acceleration points the way the resultant force points, not the way she moves. An acceleration against the motion slows her down.
“c: None, because she is slowing down, not speeding up.” — Slowing down is a change of velocity, so it is an acceleration too, here towards the west.
“d: Towards the east until she stops, then towards the west.” — The resultant force points west from the moment her foot drags, so her acceleration does too. That is why she slows down.

3. FQ24 · Practice

Model answer. F = ma needs the mass in kilograms: 800 g = 0.800 kg. F = ma = 0.800 kg × 4.5 m/s² = 3.6 N. The resultant force points the same way as the acceleration, forwards along the track.

Checked automatically

Resultant force on the car: 3.6 N (within ±0.05 N)
- Common wrong answer 3600 : You used the mass in grams. Change it to kilograms first: 800 g = 0.800 kg.
- Common wrong answer 0.18 : You divided the mass by the acceleration. F = ma: multiply, 0.800 × 4.5.
- Common wrong answer 5.6 : You divided the acceleration by the mass. F = ma: multiply, 0.800 × 4.5.
- Common wrong answer 180 : You divided the mass in grams by the acceleration. F = ma: multiply, with the mass in kilograms.

A full answer includes

Method: 800 g = 0.800 kg.
Method: F = ma = 0.800 kg × 4.5 m/s².
Answer: F = 3.6 N.
Its direction is the direction of the acceleration, forwards along the track.
I did not use 800 as the mass in F = ma.

Also accept

3.6 N forwards.

Watch for

“F = 800 × 4.5 = 3600 N.” — That uses the mass in grams. F = ma needs kilograms: 800 g = 0.800 kg, so F = 0.800 × 4.5 = 3.6 N.
“The resultant force is 0.800 N, the mass of the car.” — Mass is measured in kilograms, not newtons. The resultant force is the mass times the acceleration: 0.800 × 4.5.

4. FQ25 · Practice

Options as numbered on the pupil sheet: 1 — the acceleration doubles; 2 — the acceleration halves; 3 — the acceleration stays the same; 4 — the acceleration becomes four times as big.

Model answer. a = F ÷ m. Doubling the force on the same mass doubles the acceleration, as in run B. Doubling the mass with the same force halves it. Doubling both leaves a = F ÷ m unchanged, as in run C. Halving the force halves the acceleration, and halving the mass doubles it.

A full answer includes

I matched “force doubled, same mass” with “the acceleration doubles”.
I matched “mass doubled, same force” with “the acceleration halves”.
I matched “force and mass both doubled” with “the acceleration stays the same”.
I matched “force halved, same mass” with “the acceleration halves”.
I matched “mass halved, same force” with “the acceleration doubles”.
I did not use “four times as big” for any of them.

Also accept

2F – doubles; 2m – halves; 2F and 2m – the same; F/2 – halves; m/2 – doubles.

Watch for

“A bigger mass gives a bigger acceleration.” — A heavier trolley is harder to speed up: with the same force it accelerates less. In run C, two pucks with 0.20 N accelerated at only 0.50 m/s².
“Doubling both the force and the mass makes the acceleration four times bigger.” — The two changes cancel: a = 2F ÷ 2m = F ÷ m. Run C, with both doubled, followed run A’s line exactly.

5. FQ26 · Practice

Model answer. Path B. While the string pulls the puck towards the pin, it keeps changing the puck’s direction. Once the string is cut, there is no force towards the centre any more, and no resultant force at all. So the puck carries on in a straight line at constant speed, in the direction it had at X. At the top of the circle it was moving to the left, so it moves off straight to the left, along the tangent. It does not fly straight out from the pin (A), keep going round (C) or curve outwards (D).

A full answer includes

I chose path B.
I explained that once the string is cut, there is no resultant force on the puck.
I explained that with no resultant force it goes on in a straight line at constant speed, in the direction it had at X.
That direction is along the tangent to the circle.
I did not choose path A: no force pushes the puck outwards.

Also accept

b: path B, along the tangent; with no resultant force it goes straight on, in its direction at X.

Watch for

“Path A: it flies straight outwards, away from the pin.” — No force pushes the puck outwards. Once the string is cut, nothing acts on it, so it keeps the velocity it had at X, which was to the left.
“Path C: it keeps going round for a while, then slows down.” — Going round needs a force towards the centre all the time. Without the string the direction stops changing at once, and the puck goes straight on.
“Path D: it curves outwards.” — A curving path needs a force. With the string cut there is no resultant force, so the path is a straight line.

6. FQ27 · Practice

Model answer. (1) centre: the wire pulls the aeroplane towards the post, at the centre of the circle. (2) right angles: the pull is along the wire, and the aeroplane moves at right angles to the wire. (3) direction: a force at right angles to the motion turns the aeroplane without changing its speed. (4) acceleration: the direction keeps changing, so the velocity keeps changing; a changing velocity is an acceleration, in the same direction as the resultant force, towards the centre. (5) tangent: with the wire gone there is no resultant force, so the aeroplane goes straight on, along the tangent.

A full answer includes

(1) centre
(2) right angles
(3) direction
(4) acceleration
(5) tangent
I did not write “radius” for gap 5.

Also accept

centre; right angles; direction; acceleration; tangent.

Watch for

“(4) It has no acceleration, because its speed is steady.” — Its direction keeps changing, so its velocity keeps changing. A changing velocity is an acceleration, even at a steady speed.
“(1) edge, because the aeroplane is pulled outwards.” — The wire is tied to the post, so it pulls the aeroplane inwards. Nothing pulls it outwards.
“(5) radius: it flies straight out from the post.” — With no force on it, the aeroplane keeps the velocity it had, which was at right angles to the wire.

7. FQ28 · Practice

Model answer. F = ma, so m = F ÷ a = 7.0 N ÷ 1.2 m/s² = 5.83… kg, which is 5.8 kg to 2 significant figures. Check: 5.8 kg × 1.2 m/s² = 7.0 N.

Checked automatically

Mass of the ball: 5.8 kg (2 significant figures)
- Common wrong answer 8.4 : You multiplied the force by the acceleration. m = F ÷ a = 7.0 ÷ 1.2.
- Common wrong answer 0.1714 : You divided the acceleration by the force. m = F ÷ a: the force goes on top.

A full answer includes

Method: m = F ÷ a = 7.0 N ÷ 1.2 m/s².
Answer: m = 5.8 kg, to 2 significant figures.
I did not multiply the force by the acceleration.

Also accept

5.8 kg.

Watch for

“m = F × a = 8.4 kg.” — Check by putting your mass back into F = ma: 8.4 × 1.2 is more than 7.0 N. Divide the force by the acceleration.
“m = a ÷ F = 0.17 kg.” — The force goes on top: m = F ÷ a = 7.0 ÷ 1.2. Check: 0.17 kg × 1.2 m/s² is far less than 7.0 N.

8. FQ29 · Practice

Model answer. First find the resultant force. The two pulls act along one line in opposite directions, so subtract: 0.35 N − 0.05 N = 0.30 N, towards the east. Then a = F ÷ m = 0.30 N ÷ 0.40 kg = 0.75 m/s². The acceleration is in the same direction as the resultant force, towards the east.

Checked automatically

Acceleration of the pucks: 0.75 m/s² (within ±0.01 m/s²)
- Common wrong answer 1 : You added the two pulls. They are opposite, so the resultant is 0.35 N − 0.05 N = 0.30 N.
- Common wrong answer 1.5 : You used the mass of one puck. Two pucks have a mass of 0.40 kg: 0.30 ÷ 0.40.
- Common wrong answer 0.875 : You used only meter E’s pull. Meter W pulls the other way: the resultant is 0.30 N.
- Common wrong answer 0.12 : You multiplied the force by the mass. a = F ÷ m = 0.30 ÷ 0.40.
- Common wrong answer 1.33 : You divided the mass by the force. a = F ÷ m: the force goes on top.

A full answer includes

Method: resultant force = 0.35 N − 0.05 N = 0.30 N.
The resultant force is towards the east.
Method: a = F ÷ m = 0.30 N ÷ 0.40 kg.
Answer: a = 0.75 m/s².
The acceleration is towards the east, the same direction as the resultant force.
I did not use 0.35 N, the bigger pull, as F.

Also accept

0.75 m/s² towards the east.

Watch for

“Use the bigger pull, 0.35 N, in F = ma.” — F = ma uses the resultant force. Meter W pulls the other way, so the resultant is 0.35 N − 0.05 N = 0.30 N.

9. FQ30 · Practice

Options as numbered on the pupil sheet: 1 — Tests 1 and 2; 2 — Tests 1 and 3; 3 — Tests 1 and 4; 4 — Tests 2 and 3.

Model answer. Tests 1 and 2 have the same mass and radius; test 2 is faster, 4.5 m/s, and needs the bigger pull, 27 N. So a bigger force goes with a bigger speed. Tests 1 and 3 have the same mass and speed; test 1 has the smaller circle, 1.5 m, and the bigger pull, 12 N, against 9.0 N. So a bigger force goes with a smaller radius. Tests 1 and 4 have the same speed and radius; test 4 has twice the mass and needs the bigger pull, 24 N. So a bigger mass needs a bigger force. Tests 2 and 3 differ in both speed and radius, so they cannot show one rule on its own.

A full answer includes

I matched “bigger speed” with tests 1 and 2: 3.0 → 4.5 m/s, and the pull rises from 12 N to 27 N.
I matched “smaller radius” with tests 1 and 3: the smaller circle, 1.5 m, has the bigger pull, 12 N.
I matched “bigger mass” with tests 1 and 4: 2.0 → 4.0 kg, and the pull rises from 12 N to 24 N.
I compared only tests in which one quantity changes and the others stay the same.
I did not choose tests 2 and 3: the speed and the radius both change.

Also accept

speed – tests 1 and 2; radius – tests 1 and 3; mass – tests 1 and 4.

Watch for

“Tests 2 and 3 show the radius rule, because their radii differ.” — Their speeds differ too, so you cannot tell which change made the pull different. Compare tests in which only the radius changes: 1 and 3.
“A bigger circle needs a bigger pull.” — Test 3 has the bigger circle, 2.0 m, and the smaller pull, 9.0 N. At the same speed, a tighter circle needs a bigger force.
“A heavier cart needs a smaller pull, because it is harder to move.” — Being harder to move means a bigger force is needed to change its direction: in test 4, twice the mass needed 24 N, twice the 12 N of test 1.

10. FQ31 · Practice

Model answer. Correct: the force on the puck points towards the centre, it is at right angles to the velocity, and the speed stays the same while the velocity keeps changing. Wrong: there is no outward force balancing the string, so the resultant force is not zero; it is the string’s pull, towards the centre. A constant speed does not mean zero resultant force when the direction is changing. And the pull does not make the puck go faster: it acts at right angles to the motion, so it changes only the direction.

A full answer includes

I chose a: the force points towards the centre of the circle.
I chose b: the force is at right angles to the velocity.
I chose c: the speed stays the same, but the velocity keeps changing.
I did not choose d, e or f.
If I explained d: nothing pushes the puck outwards; the only horizontal force is the pull of the string, inwards.
If I explained e: the direction keeps changing, so the velocity changes and there must be a resultant force.
If I explained f: the pull is at right angles to the motion, so it changes the direction, not the speed.

Also accept

a, b, c.

Watch for

“d: An outward force balances the pull of the string.” — If the forces balanced, the resultant would be zero and the puck would go in a straight line. It goes round, so there must be a resultant force, inwards.
“e: The resultant force is zero, because the speed is constant.” — In a circle, the direction keeps changing, so the velocity changes. That needs a resultant force, towards the centre.
“f: The pull of the string makes the puck go faster and faster.” — The pull is at right angles to the motion, so it does not speed the puck up or slow it down. It changes only the direction.

11. FQ32 · Practice

Model answer. Bigger on the tighter bend. Ravi and his bicycle have the same mass on both bends, and he rides at the same speed. On the tighter bend his direction of motion changes more quickly, so his velocity changes more quickly: he is accelerating towards the centre at a bigger rate. A faster change of velocity needs a bigger resultant force, towards the centre of the bend. So, with mass and speed constant, a smaller radius goes with a bigger force: a bigger force towards the centre makes the radius smaller.

A full answer includes

I chose: bigger on the tighter bend.
I explained that his mass and speed are the same on both bends.
I explained that on the tighter bend his direction, and so his velocity, changes more quickly.
I explained that a smaller radius, at the same speed and mass, needs a bigger force towards the centre.
I did not choose “the same”: the force depends on the radius as well as on the speed and the mass.

Also accept

a: bigger; at the same speed and mass, a tighter (smaller) circle needs a bigger force towards the centre.

Watch for

“b: Smaller, because the tighter bend is shorter.” — The length of the bend does not matter. On a tighter bend his direction changes more quickly at the same speed, so the force towards the centre must change his velocity faster.
“c: The same, because his speed and mass are the same.” — At the same speed and mass, the force still depends on how tight the circle is. Compare how quickly the direction changes on each bend.

12. FT3 · Apply

Model answer. (a) The wire pulls the ball towards the centre of the circle, towards the athlete’s hands. This force is at right angles to the ball’s motion, so it changes the ball’s direction. (b) With the same ball and circle, the pull grew from about 1400 N to about 2400 N as the speed rose from 15 m/s to 20 m/s. With the same mass and radius, a bigger speed needs a bigger force towards the centre, so at 25 m/s the pull must be bigger than 2400 N. (c) Once the wire is let go, there is no force towards the centre. Seen from above, the ball moves off in a straight line along the tangent to the circle, in the direction it was moving at that moment. (d) With the same speed and radius, a smaller mass needs a smaller force, so the 4.0 kg hammer needs a smaller pull. (e) Weight = mass × g = 7.26 kg × 9.8 N/kg = 71 N, downwards. With only its weight acting, a = F ÷ m = 71 N ÷ 7.26 kg = 9.8 m/s², downwards, the same direction as the weight.

A full answer includes

(a) Towards the centre of the circle, along the wire.
(a) I did not say that the wire pulls the ball outwards.
(b) Bigger than 2400 N.
(b) With the same mass and radius, a bigger speed needs a bigger force towards the centre.
(c) A straight line along the tangent, the way the ball was moving at the moment of release.
(c) I did not say that it flies straight outwards from the centre.
(d) A smaller pull.
(d) With the same speed and radius, a smaller mass needs a smaller force.
(e) Method: weight = 7.26 kg × 9.8 N/kg = 71 N.
(e) Method: a = F ÷ m = 71 N ÷ 7.26 kg.
(e) Answer: a = 9.8 m/s², downwards.

Also accept

(a) towards the centre; (b) bigger than 2400 N; (c) along the tangent; (d) smaller; (e) 71 N, 9.8 m/s² downwards.

Watch for

“(a) The wire pulls the ball outwards, away from the athlete.” — The wire is pulled tight between the athlete and the ball. On the ball it pulls inwards, towards the athlete, towards the centre of the circle.
“(c) The ball flies straight outwards, away from the centre.” — With no force towards the centre, the ball keeps the velocity it had at release, which was along the tangent, not along the wire.
“(e) The acceleration is 71 m/s².” — Use F = ma with the mass: a = 71 N ÷ 7.26 kg = 9.8 m/s². Any object with only its weight acting accelerates at g.