Answers · Acceleration, falling and drag

Falling and drag — answer key

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

1. FQ11 · Practice

Model answer. a, b and c. The submarine pushes through sea water, the cyclist through air and the ball through oil; each liquid or gas pushes back against the motion, so drag acts. Drag in air, as on the cyclist, is called air resistance. The space probe (d) and the stone in the empty tube (e) move through nothing, so there is nothing to push aside and no drag.

A full answer includes

I chose a, b and c.
I explained that drag acts when an object moves through a liquid or a gas, which pushes back against the motion; in empty space or in a tube with the air pumped out there is nothing to push aside, so there is no drag.

Also accept

a, b, c

Watch for

“Drag only acts in air.” — Drag acts in any liquid or gas. In water or oil it is even bigger than in air at the same speed.
“b: there is no drag on the cyclist, because air is too thin.” — Air is a gas: drag in air is air resistance, the push you feel riding into the wind.

2. FQ12 · Practice

Model answer. (1) small; (2) increases; (3) equal; (4) terminal velocity. Air resistance grows with speed. While it is smaller than her weight she keeps speeding up, by less and less; when it equals her weight the forces balance and her speed stays the same.

A full answer includes

(1) small
(2) increases (gets bigger)
(3) equal
(4) terminal velocity

Also accept

small; increases; equal; terminal velocity.

Watch for

“(3) bigger” — If the air resistance were bigger than her weight she would slow down. At a constant speed the two forces are equal.
“(4) top acceleration” — At this point her speed stays the same, so her acceleration is zero. The constant speed is the terminal velocity.

3. FQ13 · Practice

Options as numbered on the pupil sheet: 1 — Air resistance is very small: the ball speeds up with an acceleration close to 9.8 m/s²; 2 — Air resistance is smaller than the weight but growing: the ball speeds up by less each second; 3 — Air resistance is equal to the weight: the ball falls at its terminal velocity.

Model answer. P: the ball is still slow, so the air resistance is very small and it speeds up with an acceleration close to g, 9.8 m/s² (the line is steep). Q: faster, so the air resistance is bigger, but still smaller than the weight: the ball still speeds up, but by less each second (the line is less steep). R: the air resistance has grown as big as the weight; the forces balance and the speed stays at 20 m/s, the terminal velocity (the line is flat).

A full answer includes

I matched P: air resistance very small, acceleration close to 9.8 m/s²; Q: air resistance smaller than the weight but growing, still speeding up by less each second; R: air resistance equal to the weight, terminal velocity.

Also accept

P very small air resistance; Q smaller but growing; R equal to the weight, terminal velocity.

Watch for

“R: the ball has stopped, because the line is flat.” — A flat line at 20 m/s means a steady 20 m/s. The ball is still falling fast.
“P: the air resistance is biggest at the start, because the ball has just been let go.” — Air resistance grows with speed. At P the ball is slow, so the air resistance is small.

4. FQ14 · Practice

Model answer. b. Each 0.10 m takes almost the same time, about 1.43 s, so the ball moves at a constant speed. It is falling at its terminal velocity: in the oil the drag grows so quickly that the forces on the ball soon balance, long before the first mark. If it were speeding up, each 0.10 m would take less time than the one before.

A full answer includes

I chose b.
I explained that the times for equal distances are almost the same, so the ball is moving at a steady speed: its terminal velocity, reached because the drag of the oil grew until the forces on the ball balance.

Also accept

b; equal times for equal distances mean a steady speed, the terminal velocity.

Watch for

“The ball is speeding up, because it is falling.” — A falling object speeds up only while the forces on it do not balance yet. Equal times for equal distances mean a steady speed.
“There is no drag on the ball, because its speed is constant.” — Drag is large here: it is what stops the ball speeding up, so that the forces on the ball balance.

5. FQ15 · Practice

Model answer. Mean time = (3.06 + 2.98 + 3.02) s ÷ 3 = 9.06 s ÷ 3 = 3.02 s. Speed = distance ÷ time = 0.15 m ÷ 3.02 s = 0.050 m/s. You could also use the whole distance: 0.45 m ÷ 9.06 s = 0.050 m/s. The nearly equal times show a steady speed: the bead's terminal velocity in the syrup.

Checked automatically

Mean time for one gap: 3.02 s (within ±1%)
- Common wrong answer 9.06 : That is the total time for all three gaps. Divide it by 3.
- Common wrong answer 3.06 : That is only the first time. Add all three times and divide by 3.
Speed of the bead: 0.050 m/s (within ±5%)
- Common wrong answer 20 : You divided the time by the distance. Speed = distance ÷ time.
- Common wrong answer 0.017 : You divided one gap by the total time. Use one gap and the mean time, or all three gaps and the total time.

A full answer includes

Mean time = (3.06 + 2.98 + 3.02) s ÷ 3 = 3.02 s.
Speed = 0.15 m ÷ 3.02 s = 0.050 m/s. The times are almost equal, so the bead is at its terminal velocity.

Also accept

3.02 s; 0.050 m/s (0.05 m/s)

Watch for

“Speed = 3.02 ÷ 0.15 = 20 m/s.” — That is seconds per metre. Speed is metres per second: divide the distance by the time.
“Speed = 0.15 ÷ 9.06 = 0.017 m/s.” — 9.06 s is the time for all three gaps, 0.45 m. Divide one gap by one mean time, or the whole distance by the whole time.

6. FQ16 · Practice

Model answer. A: she speeds up, quickly at first and then by less each second, because the air resistance grows with her speed but is still smaller than her weight. B: her speed is constant at 50 m/s, her terminal velocity: the air resistance has grown as big as her weight, so the forces balance. C: the parachute opens, the air resistance becomes much bigger than her weight, and she slows down (decelerates); as she slows, the air resistance gets smaller. D: the air resistance has fallen until it equals her weight again, so she falls at a new, lower terminal velocity of 6 m/s until she lands.

A full answer includes

A: speeding up, by less and less each second (a changing acceleration): the air resistance is smaller than her weight but grows as her speed grows.
B: a constant 50 m/s, her terminal velocity: the air resistance equals her weight, so the forces balance.
C: slowing down, a deceleration, after the parachute opens: the air resistance is now bigger than her weight; it gets smaller as she slows.
D: a new, lower terminal velocity of 6 m/s: the air resistance equals her weight again.

Also accept

A speeding up, air resistance < weight; B terminal velocity, equal; C slowing, air resistance > weight; D lower terminal velocity, equal.

Watch for

“C: she moves upwards when the parachute opens.” — Her speed falls but never goes below zero: she keeps falling, only more slowly.
“B: no forces act on her, because her speed is constant.” — Both forces still act and both are large. They are equal and opposite, so they balance.

7. FQ17 · Practice

Model answer. c. At terminal velocity the raindrop falls at a constant speed. Its speed is not changing, so the forces on it balance: the air resistance pushing up is equal to its weight pulling down. Both forces still act; if the weight were bigger it would speed up, and if the air resistance were bigger it would slow down.

A full answer includes

I chose c.
I explained that at terminal velocity the speed does not change, so the forces balance: the air resistance upwards equals the weight downwards, and both still act.

Also accept

Watch for

“No forces act on the raindrop, because its speed is constant.” — Its weight still pulls it down. A constant speed means the forces balance, not that they are gone.
“Its weight must be bigger than the air resistance, because it is still falling.” — A bigger weight would make it speed up. Moving down at a steady speed needs equal forces.

8. FQ18 · Practice

Model answer. a, b and e. The canopy pushes aside far more air at the same speed, so the air resistance becomes much bigger than her weight (a) and she slows down: a deceleration (b). As her speed falls, the air resistance gets smaller (e), until it is as big as her weight again: the forces balance at a new, lower terminal velocity, about 5 m/s. She never moves upwards (c), and her weight does not change (d).

A full answer includes

I chose a, b and e.
I explained that the open parachute makes the air resistance much bigger than her weight, so she slows down; as she slows the air resistance gets smaller until it equals her weight again, and she falls at a new, lower terminal velocity.

Also accept

a, b, e

Watch for

“c: she shoots upwards when the parachute opens.” — She keeps falling the whole time; she only slows down. In films the camera falls past her, so she seems to go up.
“d: the parachute makes her weight smaller.” — Her weight stays the same. What changes is the air resistance at each speed.

9. FQ19 · Practice

Model answer. a, b and c. Drag grows with speed (a). A bigger or more spread-out shape, like an open parachute, has to push aside more air (b). A liquid such as water is much thicker than air, so at the same speed the drag is much bigger (c). In a vacuum there is nothing to push aside, so there is no drag at all (d). Making the object heavier without changing its size or shape does not change the drag at a given speed (e): it changes the weight, which is a different force.

A full answer includes

I chose a, b and c.
I explained that drag depends on the speed, the size and shape, and what the object moves through; a heavier object of the same size and shape pushes aside the same air, so it meets the same drag at the same speed. I did not choose d or e.

Also accept

a, b, c

Watch for

“e: a heavier object meets more drag.” — Drag depends on the speed, the size and shape, and the liquid or gas, not on the weight. A heavier ball of the same size meets the same drag at the same speed.
“d: in a vacuum the drag is biggest, because nothing holds the object up.” — In a vacuum there is nothing to push aside, so there is no drag at all.

10. FT2 · Apply

Model answer. (a) The driving force stays the same, but the air resistance grows as the car goes faster. At 60 m/s the air resistance has become as big as the driving force: the forces balance and the speed stops rising, just as a skydiver reaches terminal velocity when the air resistance matches her weight. (b) The roof box makes the car push aside more air at each speed, so the air resistance is bigger at every speed and matches the driving force at a lower speed: the top speed is lower. (c) Both curves start at 0 with nearly the same steep gradient, then get flatter; without the box the curve levels off at 60 m/s, with the box it levels off lower.

A full answer includes

(a) As the speed grows, the air resistance on the car grows, until it is as big as the driving force; the forces balance, so the speed stops changing.
(b) The top speed is lower: the box makes the air resistance bigger at each speed, so it equals the driving force at a lower speed.
(c) Labelled axes; a curve from 0 that is steep at first, then flatter, levelling off at 60 m/s; with the box, a curve that levels off lower.

Also accept

(a) air resistance grows until it equals the driving force; (b) lower top speed, more air resistance; (c) curves levelling off

Watch for

“(a) The car stops speeding up because the engine's force runs out.” — The driving force stays the same. What grows is the air resistance, until it matches the driving force.
“(b) The top speed stays the same, because the engine is the same.” — The engine is the same, but the air resistance at each speed is bigger, so the forces balance at a lower speed.