Page 57 - SP015 Past Years PSPM Chapter 1 -5 Ver.2020

PSPM SP015

FINAL ANSWER FOR CHAPTER 3: MOMENTUM AND IMPULSE

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1. Momentum is the product of mass and velocity. SI unit of momentum: kg m s
2. (a) Momentum is a vector quantity.
(b) Force is proportional to change of momentum. p head-on > p behind  F head-on is larger
3. (a) Impulse is defined as the change of momentum.
(b) Area below the F-t graph represents impulse.
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4. J = 480 N s, u = 33.33 m s
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5. p = 0.22 kg m s
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6. (b) J = 1.2 kg m s
7. (a) t = 4.8 ms (b) J = 0.4 N s (c) F = 83.33 N
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8. (a) J = –160 kg m s (b) F = 200 N
9. t = 0.2 s
10. Principle of conservation of linear momentum states that the total momentum of an isolated (closed)
system is conserved.
Principle of conservation of linear momentum states that when the net external force is zero, the total
linear momentum of the system is constant.
11. Principle of conservation of kinetic energy. Kinetic energy is conserved when collision is elastic and not
conserved when collision is ineleastic.
12. The total momentum is conserved and the total kinetic energy is conserved.
13. Total momentum is conserved and total kinetic energy is NOT conserved.
14. One (1) similarity: Total momentum is conserved in both collision.
One (1) difference: Total kinetic energy is conserved in elastic collision and not conserved in inelastic
collision.
15. (a) Force = rate of change of momentum (b) Kinetic energy
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16. v A = –1.33 m s , v B = 2.67 m s
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17. v = –4.4 m s
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18. v A = –0.67 m s , v B = 1.33 m s

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20. (a) v b = –4 m s , v r = 8 m s (b) It is a perfect elastic collision.
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21. (a) v B = 0.47 m s at the same direction of u A .
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22. (a) J =  8 kg m s (b) F =  800 N (c) Elastic collision. Show: K i = K f = 40 J
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23. (b) K total = mu
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24. (a) v A = 0.82 m s , v B = 2.82 m s (b) F = ± 4.7 N
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25. (a) v B = 4.17 m s (b) v B = –1.83 m s
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26. (a) v Q = 1.67 m s (b) Show: K i  K f
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27. v = 1 m s
28. (a) m P = 90 kg (b) K after collision = 300 J
m boy v boy
29. v canoe  
m canoe
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30. (a) v AB = 4.6 m s (b) K loss = 15.7 J
31. (a) K f = 12.5 J (b) v = 1.25 m s -1
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32. (b) p Ri = 0.08 kg m s , p Bi = 0 kg m s , p Rf = 0.05 kg m s , p Bf = 0.074 kg m s
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(c) v B = 0.37 m s ,  B = 82.59 below the x-axis
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33. v 1 = 44.4 m s and v 2 = 77.9 m s
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34. (a) v = 4.83 m s at 43.6 from horizontal.
(b) K i = 286 J and K f = 140 J. Since K i  K f  Inelastic collision
35. h = 0.2 m
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36. (a) (i) p = –0.8 kg m s @ 0.8 kg m s (ii) F = –80 N @ 80 N
(a) (iii) It is an elastic collision.
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(b) (i) p = –0.6 kg m s @ 0.6 kg m s (ii) K = –3 J
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37. (a) v pb = 0.022u p (b) u p = 127 m s (c)  m = 53
(d) Inelastic collision bacause K p  K pb
38. (b) y = 4.62 m
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39. (a) v = 0.28 m s (b) J = 1.4 N s (c) s = 0.06 m
40. (a) J = 5.4 N s (b) W = 27 J
(c) If the ball hits the plane of the racquet at an angle, the impulse delivered to the ball will decrease.
Only the perpendicular component of the momentum (velocity) contributes to the impulse.
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