Answers · Acceleration, falling and drag

Acceleration, falling and drag — final check key

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

1. FK1 · Final check

Model answer. (a) The line rises from 0 to 12 m/s in 8 s: a = 12 m/s ÷ 8 s = 1.5 m/s². (b) From 8 s to 28 s the line is flat at 12 m/s: the speed does not change, so the acceleration is zero. (c) The line falls from 12 m/s to 0 in 6 s: a = (0 − 12) m/s ÷ 6 s = −2.0 m/s², a deceleration of 2.0 m/s². (d) Friction between the solid surfaces of the pads and the discs: it acts against the motion, so it slows the wheels, and it heats the pads and the discs. (e) The bolt starts from rest and gains 9.8 m/s every second: after 3.0 s its speed is 9.8 × 3.0 = 29.4 m/s, about 29 m/s. A heavier bolt would have the same speed, because in free fall every object has the same acceleration, g ≈ 9.8 m/s².

A full answer includes

(a) Gradient = (12 − 0) m/s ÷ 8 s = 1.5 m/s².
(b) Zero: the line is flat, so the speed stays at 12 m/s.
(c) (0 − 12) m/s ÷ (34 − 28) s = −2.0 m/s²: a deceleration of 2.0 m/s².
(d) Friction: it acts against the motion, slowing the discs and the wheels, and it heats the pads and the discs.
(e) Speed = 9.8 m/s² × 3.0 s = 29.4 m/s (29 m/s). No: in free fall every object has the same acceleration, so a heavier bolt has the same speed.

Also accept

(a) 1.5 m/s²; (b) 0; (c) −2.0 m/s², 2.0 m/s²; (d) friction, opposes motion and heats; (e) 29.4 m/s, no

Watch for

“(b) The acceleration is 12 m/s² because the line is at 12.” — 12 m/s is a speed. A flat line means the speed is not changing, so the acceleration is zero.
“(e) A heavier bolt falls faster.” — In free fall every object has the same acceleration, about 9.8 m/s², so after 3.0 s every bolt has the same speed.

2. FK2 · Final check

Model answer. (a) At the start both lines are equally steep, about 9.8 m/s². The balls are still slow, so the air resistance on them is tiny and both are almost in free fall, with the acceleration g. (b) Y speeds up quickly at first, then by less and less, and from about 2 s its speed stays at 9 m/s. As Y goes faster the air resistance grows; Y is light, so the air resistance soon equals its weight. Then the forces balance and Y falls at its terminal velocity, 9 m/s. (c) X has the bigger terminal velocity. The balls are the same size, so at any speed the air resistance on them is about the same. X is much heavier, so the air resistance has to grow much bigger, at a much higher speed, before it equals X's weight. (d) At 5 s the air resistance on X is still smaller than its weight, so it is still speeding up, though by less each second. (e) Without air there is no air resistance: both balls are in free fall and accelerate at 9.8 m/s², so both graphs are the same straight line through the origin.

A full answer includes

(a) Both lines start with the same steepness, about 9.8 m/s²: their speeds are small, so the air resistance is small and both are almost in free fall.
(b) Y speeds up, then by less and less each second, and from about 2 s falls at a constant 9 m/s: its air resistance grows with speed until it equals its weight; then the forces balance and Y is at its terminal velocity.
(c) X. The balls are the same size, so at the same speed they meet the same air resistance; X is heavier, so it must go much faster before the air resistance is as big as its weight.
(d) The air resistance on X is still smaller than its weight (though growing).
(e) Both the same straight line from 0, gradient 9.8 m/s²: with no air there is no air resistance, so both are in free fall with the same acceleration.

Also accept

(a) same, ~9.8 m/s², tiny air resistance; (b) levels off at 9 m/s, air resistance = weight; (c) X, heavier; (d) air resistance < weight; (e) one straight line, 9.8 m/s²

Watch for

“(a) X has the bigger acceleration from the start, because it is heavier.” — At the start both lines are equally steep: with little air resistance, every object accelerates at about 9.8 m/s².
“(c) Y has the bigger terminal velocity, because it reaches it sooner.” — Y reaches its terminal velocity sooner because it is lower: 9 m/s. X is still speeding up at 30 m/s.