Page 78 - SP015 Past Years PSPM Chapter 6 -14 Ver 2020

PHYSICS PSPM SEM 1 1999 - 2017

FINAL ANSWER FOR CHAPTER 6: CIRCULAR MOTION

1. Centripetal acceleration is the acceleration of an object moving in circular path whose direction is
toward the centre of the circular path and whose magnitude is equal to the square of the speed divided by
the radius.
2. Yes, it does because there is a change in velocity.
3. Acceleration is the rate of change of velocity. In circular motion, velocity is always perpendicular to
radius of the circular path. So, the change in velocity is always pointing towards the centre of the circular
path.
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5. a = 3.38  10 m s
6. Centripetal force is the force needed to keep an object in circular motion.
Centripetal force is the net force required to maintain a body in a circular motion.
7. (a) Since the direction of the velocity is constantly changing, the body accelerates. According to
Newton’s second law, a force F = ma, acts on the body causing it to accelerate.
mv 2
(b) F  acting towards the centre of the circle.
r
8. (a) A satellite orbiting earth.  Gravitational force.
(b) A ball attached to a string that swirls horizontally.  Tension
9. (a) Frictional force. (b) Gravitational force.
2
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2
10. (a) At A, mg  R  mr  ; At B, sin R   mr  (b)  max = 2.21 rad s
11. F c = 31.25 N, T max = 55.78 N, T min = 6.73 N
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12. (b)  max = 63.17, v max = 5.3 m s
1 g tan 
13. f 
2 r
mv 2


14. (b) F  mg cos  N  (c) Normal force (N = 0)
net
r
15. (a) T = 31.84 N (b) T = 28.16 N
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16. (a) v max = 19.4 m s
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17. (a) v = 6.26 m s (b)  = 3.13 rad s (c) T = 0.29 N
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(d) v mid = 5.27 m s (e) Lower velocity
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18. (a) F net = 3.15  10 N
(b) The force that enables the car to successfully negotiate the curve is friction between tyre and road.
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19. (b) v = 1.19 m s (c) T = 1.32 s
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20. (b) v = 1.63 m s (c)  = 4.53 rad s
(d) If the angle remains unchanged but a longer string is used, the angular velocity will decrease
because  is inversely proportional to length of the string  .
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21. (a) a c = 17.2 m s (b) N = 665.1 N
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22. (a)  = 1.46 rad s (b) a = 1.54 m s (c) F = 92.4 N
(d) Travelling in a straight line, tangent to the circle the instant they let go.
23. (b) By leaning towards the centre, the tyres tend to skid away from the centre. Frictional force prevents
the skid. The force f is the centripetal force for the cyclist to be in circular motion.
v 2
(c)  
rg
24. (a)  = 53.7 (b) T = 4.1 N
2
(c) If the speed is increased,  or r will increase because v  r tan 
25. (a) T a = 112.1 N (b) T b = 60.2 N

FINAL ANSWER FOR CHAPTER 7: GRAVITATION

1. Newton’s law of gravitation states that the gravitational force of attraction between two masses (m 1 and
m 2 ) is directly proportional to the product of the masses and inversely proportional to the square or the
distance (r) between them.
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2. F = 3.53  10 N
3. (b) x = 6.74 m
4. Gravitational field strength is the gravitational force per unit mass.
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