Physics (AS & A Level) · Feb / March 2016 · Paper 22 · Question 2
9702
As printed in the paper
Show the answer and mark scheme
2(a)[1]
change in velocity / time (taken) or rate of change of velocity
2(b)(i)[1]
vX = (24 / 1.5) = 16 (m s–1)
2(b)(ii)[2]
tan 28° = vY / vX or vX = v cos 28° and vY = v sin 28° vY = 16 tan 28° or vY = 16 × (sin 28° / cos 28°) so vY = 8.5 (m s–1)
2(b)(iii)[2]
v = u + at t = (0 – 8.5) / (–9.81) = 0.87 (s)
2(b)(iv)[2]
straight line from positive vY at t = 0 to negative vY at t =1.5 s line starts at (0, 8.5) and crosses t-axis at (0.87, 0) and does not go beyond t = 1.5 s. A1
2(c)(i)[2]
(v 2 = u 2 + 2as) 0 = 8.52 + 2(–9.81)s or (s = ut + ½at 2) s = 8.5×0.87 + ½ × (–9.81) × 0.872 or (s = vt – ½at 2) s = 0 – ½×(–9.81)×0.872 or (s = ½(u + v)t or area under graph) s = 0.5 × 8.5 × 0.87 s = 3.7 (m)
2(c)(ii)[2]
∆EP = mg∆h (allow E = mgh) m = 22 / (9.81 × 3.7) = 0.61 (kg)
2(d)[1]
acceleration (of freefall) is unchanged / not dependent on mass, and so no effect (on maximum height) or explanation in terms of energy: (initial) KE ∝ mass, (∆)KE = (∆)PE, (max) PE ∝ mass, and so no effect (on maximum height)
Make a free account to save your answers and see your weak topics and ask the AI why a wrong answer is wrong.