Page 3 - 20. Sci Phy Answer Key 2023 (AK016)

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Section C Free response questions (c) Average velocity = 58.2/0.75 = 77.6 km/h
1. (a) A scalar quantity is a quantity that has a Revision 2
magnitude only while a vector quantity is a
quantity that has both the magnitude and 1. (a) It is a rate of change of velocity.
direction. (b) An object experiences a change in the velocity./
(b) Total distance = 240 + 160 = 400 m An object experiences a change in direction.
Take the direction towards east as positive. (c) Deceleration
Displacement = 240 + (–160)
u
= 80 m due east 2. v -= 80 - =45 35 km/h
.
35 km/h = 9 722 m/s
2. (a) Set up a simple pendulum as shown. Make v - u 9 722 2
.
the pendulum swing (or oscillate). Record the a = t = 4 = 243 m/s.
time t for 20 complete oscillations by using a
stopwatch. Repeat the experiment a few times 3. v =+ at = + 0 (. )( = 10 m/s
u
2 05)
to obtain the average value of t. Find the time
for one complete oscillation, T: Distance travelled in 5 seconds = 1 ´ ´ ´10 5 = 25 m
T= t/20 = Time for one complete oscillation 2
Remaining distance of 10 m/s = - 125 25 = 100 m
100
Time taaken for 100 m = = 10 s
10
Total time taken =+10 = 15 s
5
Revision 3
1. (a) B (b) C
bob (c) A (d) D
2. (a) C (b) D
(b) (i) No (ii) Yes (iii) No (c) E (d) B
(e) A
Unit 2: Kinematics 8 - 0
Revision 1 3. (a) v = 2 - 0 = 4 m/s
1. (a) T [The object may be travelling in a circular (b) He is at rest at a distance 8 m from the
path and returns to the original position.] starting point.
(b) T [When the object moves in the opposite
direction to its original direction.] (c) v = 12 -8 = 4 =1 m/s
(c) F [It is defined as a rate of change of 10 -6 4
displacement.] John’s speed is 1 m/s.
(d) F [It measures the instantaneous speed.]
(e) T Revision 4
(f) F [It is called motion.]
(g) T 1. (a) D (b) E
(c) G (d) C
2. (a) Speed = 85 000/(35 × 60) = 40.5 m/s
(b) Time = 30 000/11.9 = 2521 s = 42.0 min (e) A (f) B
(c) Total distance = 1200 × 8 = 9600 km (g) F
3. (a) (i) Distance = 24 m/s × (15 × 60) s = 2. (a) I: The object moves with a non-uniform
21600 m = 21.6 km increasing acceleration.
(ii) Distance = 30 m/s × (30 × 60) s = II: The object moves with a non-uniform
54000 m = 54.0 km decreasing acceleration.
(b) Total distance = 21.6 + 54.0 = 75.6 km III: The object moves with a uniform speed
Total time taken = 45 minutes = 0.75 h with zero acceleration.
Average speed = 75.6/0.75 = 101 km/h IV: The object moves with a non-uniform
decreasing deceleration.
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