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E N D - O F -TO P I C Q U E S T I O N S

Solutions for Topic 2 – Mechanics
1. a) P
V

A

b) horizontal speed = 15 × cos 45 = 10.6 m s–1



v2 = u2 + 2as; v2 = 10.62 + 2 × 9.8 × 25 = 112 + 490 = 602 v = ± 24.5 m s–1 (positive value is correct one to use)
2 so speed is √ ? 10.6 ?? + 24.52??

vertical speed = 15 × sin 45 = 10.6 m s–1 upwards

___________

= 27 m s–1 2 2. a) (i) h = _ ?? v ? ???= 3.2 m 2g (ii) t = _ ?? u ??= 0.80 s g?? b) time to go from top of cliff to the sea = 3.0 – 1.6 = 1.4 s s = 8.0 × 1.4 + 5.0 × (1.4)2 = 21 m; 3. travels vertically 1.25 m in 0.5 s; 2s?? g = ??_ ? t2 to give g = 10 (±1) m s–2 4. a) (i) Zero (ii)
(normal) reaction (normal) reaction

force downwards on pedals weight/gravity force/mg



(iii)  The drag force is equal to the forward force; the net force is zero and therefore the acceleration is zero. resistive force _ b) (i) acceleration = __ ???? = ?? 40??? ?= 0.57 m s–2 mass??? 70 (ii) v2 = u2 + 2as; 0 = 64 – (2 × 0.57 × s); s = 56 m (iii) air resistance or bearing friction or effectiveness of brakes depends on speed; air resistance reduced as speed drops, estimate will be too low, stopping distance will be further

5. The net force on the car is 0.3 × 1000 = 300 N. There is an additional drag force of 500 N. T = 300 + 500 = 800 N. 6. T1 sin 60 = T2 sin 30 T1 cos 60 + T2 cos 30 = 3800 T1 = 1900 N; T2 = 3300 N

? Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute

1

E N D - O F -TO P I C Q U E S T I O N S

7. a) power is 0.66 kW (read off from graph) b) power = frictional force × speed force = _ ?? 660 ??? ?= 330 N 2 8. a) use the area under the graph as this is v × t b) (i)
ball at t = 2.0s drag force

weight

Earth’s surface





(ii)

25 20 v/m s-1 15 10 5 0 0 2 4 6 8 10



the acceleration of the ball is equal to the gradient of the graph gradient = _ ?? 25 – 6? ??? 4.8 – 0 = 4.0 m s–2 (iii) The net force on the ball is 2 N, the weight is 4.9 N, so the difference between these is the magnitude of the drag force = 2.9 N. (iv) At 5.0 s the gradient is smaller and therefore the acceleration is less than at 2.0 s. The weight is constant and therefore the drag force is greater. c) gain in kinetic energy = _ ?? 1????× 0.5 × 252 = 156 J 2 loss in gravitational potential energy = 0.5×9.8×190 = 931 J change (loss) in energy = 931 – 156 = 775 J
normal reaction/N driving force/ thrust/OWTT friction/drag/ air resistance

t/s



9. a) (i)

weight/gravity force/W (do not accept gravity)





output power _ b) input power = ??__ ?? ?? ?? = ?? 70? ??? 0.35 efficiency c) height gained in 1 s = 6.2 sin(6) = 0.648 (m) = 200 kW

(ii) zero





= 5.4 × 104 W p 1.6 × ?? 104? d) F = ? _ ?? v?? ?? ? = _ ?? ? 6.2 = 2.6 kN

rate of change of PE = 8.5 × 103 × 9.81 × 0.648

()

? Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute

2

E N D - O F -TO P I C Q U E S T I O N S

10. a) (i) momentum before = 800 × 5 = 4 000 N s conservation of momentum gives v = 2.0 m s–1 momentum after = 2 000v

(ii) KE before = 400 × 25 = 10 000 J? KE after = 1 000 × 4 = 4 000 J loss in KE = 6 000 J; b) transformed / changed into heat (internal energy) and sound 11. a) momentum of object = 2 × 103 × 6.0 momentum after collision = 2.4 × 103 × v

b) KE of object and bar + change in PE = 0.5 × 2.4 × 103 × 25 + 2.4 × 103 × 10 × 0.75 use ΔE = Fd, 4.8 × 104 = F × 0.75 F = 64 kN 81 ___ 12. a) time = ???? ??? ?? = 4.8 × 107 s 2.2 × 10–25 × 77 × 1018 b) rate of change of momentum of the xenon atoms = 77 × 1018 × 2.2 × 10–25 × 3.0 × 104 = 0.51 N = mass × acceleration

v = 5.0 m s–1

use conservation of momentum, 2 × 103 × 6.0 = 2.4 × 103 × v



where mass = (540 + 81) kg acceleration of spaceship = _ ?? 0.51??? ? 621 = 8.2 × 10–4 m s–2 F???? c) a = _ ?? m since m is decreasing with time, then a will be increasing with time

d) change in speed = area under graph = (8.0 × 4.8) × 102 + _ ?? 1???(4.8 × 1.4) × 102 2 final speed = (8.0 × 4.8) × 102 + _ ?? 1???( ? 4.8 × 1.4) × 102 + 1.2 × 103 5.4 × 103 m s–1 2 ( ? 350 × 2.62)? __ 13. a) centripetal force = ?? ?? ??? = 410 N 5.8 tension = 410 + (350 × 9.8) = 3800 N b) idea of use of area under graph distance = _ ?? 1????× 0.15 × 2.6 2 = 0.195 m c) idea of momentum as mv total change (= 2.6 × 350) = 910 N s



? Oxford University Press 2014: this may be reproduced for class use solely for the purchaser’s institute

3


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