Physics (AS & A Level) · May / June 2016 · Paper 21 · Question 2
9702
As printed in the paper
The question as text
A ball is thrown from a point P with an initial velocity u of 12 m s–1 at 50° to the horizontal, as illustrated in Fig. 2.1. P horizontal path of ball 50° X =12 m s–1 Q Fig. 2.1 The ball reaches maximum height at Q. Air resistance is negligible.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
2(a)(i)[1]
horizontal component (= 12 cos 50°) = 7.7 m s–1 [1]
2(a)(ii)[1]
vertical component (= 12 sin 50° or 7.7 tan 50°) = 9.2 m s–1 [1]
2(b)[2]
v2 = u2 + 2as and v = 0 or mgh = ½mv2 or s = v2 sin2θ / 2g 9.22 = 2 × 9.81 × h hence h = 4.3 (4.31) m [2]
alternative methods using time to maximum height of 0.94 s: s = ut + ½at2 and t = 0.94 (s) (C1) s = 9.2 × 0.94 – ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = vt – ½at2 and t = 0.94 (s) (C1) s = ½ × 9.81 × 0.942 hence s = 4.3 m (A1) or s = ½(u + v)t and t = 0.94 (s) (C1) s = ½ × 9.2 × 0.94 hence s = 4.3 m (A1)
2(c)[4]
t (= 9.2 / 9.81) = 0.94 (0.938) s horizontal distance = 0.938 × 7.7 (= 7.23 m) displacement = [4.32 + 7.232]1/2 = 8.4 m [4]
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.