Question bank › A-level Physics › Mechanics and materials

Motion in a straight line and projectile motion (RP3)

Answer on paper first, then reveal the mark scheme one mark at a time.

1Core4 marks

A car travelling at 24 m s$^{-1}$ brakes with uniform deceleration and comes to rest in 3.2 s.

(a) Calculate the deceleration of the car. [2]

(b) Calculate the braking distance. [2]

(a) $a = \dfrac{v - u}{t} = \dfrac{0 - 24}{3.2}$(1)
(a) 7.5 m s$^{-2}$(1)
(b) $s = \dfrac{(u+v)}{2}t = \dfrac{24 + 0}{2} \times 3.2$ or any correct suvat route(1)
(b) 38 m allow 38.4 m(1)
Answer: (a) 7.5 m s$^{-2}$ (b) 38 m
Examiner tip: Write the suvat list first (s, u, v, a, t) — it shows which equation to pick and earns method marks even if you slip later.

2Standard6 marks

A ball is kicked from level ground with a velocity of 18 m s$^{-1}$ at 40° above the horizontal. Air resistance can be ignored.

(a) Calculate the time the ball is in the air. [2]

(b) Calculate the horizontal distance travelled before the ball lands. [2]

(c) Air resistance is now taken into account. State and explain its effect on the range and on the maximum height. [2]

(a) vertical component $u_v = 18\sin40° = 11.6$ m s$^{-1}$(1)
(a) $t = \dfrac{2u_v}{g} = \dfrac{2 \times 11.57}{9.81} = 2.4$ s allow 2.36 s(1)
(b) horizontal component $u_h = 18\cos40° = 13.8$ m s$^{-1}$(1)
(b) range $= 13.79 \times 2.36 = 33$ m allow 32.5 m; ecf from (a)(1)
(c) range is reduced because the horizontal velocity decreases (horizontal deceleration due to drag)(1)
(c) maximum height is reduced because drag acts downwards with weight while rising, so greater vertical deceleration(1)
Answer: (a) 2.4 s (b) 33 m (c) both reduced
Examiner tip: Horizontal and vertical motion are independent: only the vertical component goes into the time of flight.

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