Page 55 - SP015 Past Years PSPM Chapter 1 -5 Ver.2020
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PSPM SP015
FINAL ANSWER FOR CHAPTER 1: PHYSICAL QUANTITIES AND MEASUREMENT
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1. Length : m, Temperature : K, Velocity : m s , Force : N
3. (b) If r is multiplied by 1 , there will be no change its homogeneity because 1 is a dimensionless
2 2
1
constant @ is without unit.
2
ˆ
4. A B ( 7. 32 i 10 ˆ ) j N
5. (a) A B . 7 27 N m (b) The angle between A and B = 0
6. R = 3.61 m
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7. 5 km h = 1.39 m s
FINAL ANSWER FOR CHAPTER 2: KINEMATICS OF LINEAR MOTION
1. Distance is the actual (total) distance travelled along the path between two points and it is a scalar
quantity.
Displacement is the shortest distance from initial point to final point in a straight line and it is a vector
quantity.
2. No.
3. Speed is the total distance travelled divided by time. Velocity is the rate of change of displacement.
4. (a) No. If the velocity is constant, the speed must also be constant.
(b) Speed. Speedometer only measures the magnitude not direction.
5. (a) Average velocity is displacement per unit time.
(b) Instantaneous velocity is the velocity at a particular instance or a particular position.
6. Instantaneous acceleration is the acceleration at a particular instance of time.
7. Constant velocity or at rest because acceleration is the rate of change of velocity.
8. Yes. A man in a car that is moving at changing velocity.
9. Yes. It is possible if the object is moving at a constant non-zero velocity.
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10. (b) v = 18 m s (c) a = 6 m s
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11. (a) t = 0.47 s (b) v = 2.55 m s
12. (a) t = 3 s and t = 5 s
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13. (a) 0 s to 2 s: Car moves forward at uniform velocity 10 m s . 2 s to 4 s: Car stops.
(c) Total distance travelled = 20 m
14. (a) Zone A: Distance increase, Speed increase, Acceleration constant
Zone B: Distance increase, Speed constant, Acceleration zero
Zone C: Distance no change, Speed zero, Acceleration zero
(b) Zone A
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15. (a) v 1 4 = –1.67 m s (b) v 1 4 = 3 m s (c) v instantaneous = –7 m s
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(d) a nstantaneous = 0 m s
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16. (a) From C to D. (b) a = 20 m s (c) v = 23.3 m s
17. Sketch v-t graph with positive gradient and intercept at u. Use the area under the graph to derive equation,
s ut 1 2 at
2
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18. (a) t = 400 s = 0.111 h (b) a p = 0.125 m s (c) v = 50 m s = 180 km h
(e) F = 3600 N
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19. v = 3.6 m s
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20. s = 6.17 10 m
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21. (b) v = 24 m s
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22. (b) I: a = 0.32 m s II: a = 0 m s III: a = –0.4 m s (c) Total distance = 1780 m
23. (a) v 10 . 17 s m 1 , a 0 s m 2
(b) Continuity graph: 0 < t < 8 curve, 8 < t < 24 straight line, 24 < t < 45 curve
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24. a = 2.2 m s , a = 1.1 m s , a = 1.7 m s
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25. (a) a = –1.11 m s , t = 32.8 s (b) d > s, d = 369.6 m, s = 168 m
26. (a) s = 26 m
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(b) The particle move the slowest from segment A to B (v = 2 m s ).
(c) The particle return to its starting point twice.
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