Page 28 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics

11. (a) v (b) s 2. B : Velocity = gradient of graph
u At Q, gradient = 0.
3. C : Increase in velocity = area between graph and
time-axis
1
= (12)(4) m s = 24 m s –1
–1
0 t 0 t 2
4. B : Shaded area under graph > total area under graph
ds Speed 2
v = = gradient of (s-t) graph
dt Area = distance
or s =v dt = area under (v-t) graph. u travelled in time t/2
12. v u/2
0 t/2 t Time

0 t 1 2
5. C : y = ut – gt
2
Gradient of y-t graph decrease until zero at the
4 highest point.
2
2
6. (a) v = u + 2as
1. B 2. D 3. C 4. C 5. C = 24.0 + 2(–9.81)(–60.0)
2
1 v = 41.9 m s –1
6. K
4 1
(b) s = (u + v)t
7. (a) v /m s –1 2
1
–60.0 m = (24.0 + (–41.9)t
6.9 2
t = 6.70 s
4.0 v H
Alternative:
1
0 t /s s = ut + at 2
1.41 2
1
–60.0 m = (24.0)t + (–9.81)t 2
2
Solving t = 6.70 s
–6.9 v V
7. (a) x = (v cos θ)t

1
(b) Maximum height = 2.43 m s = ut + at 2
2
8. (a) 45° 1 2

4
(b) 2.55 × 10 m = (v sin θ)t – gt
2
(c) 6.37 × 10 m (b) Vertical motion,
3
1
9. 714 m –10.0 = (12.0 sin 60 )t – (9.81)t 2
o
1 2
2
10. (a) 0.639 s (Use s = ut + gt for vertical motion) 2
2 9.81t – 20.8t – 20 = 0
–1
(b) 5.01 m s (Use 3.2 m = vt) 20.8 ± 20.8 – 4(9.81)(–20)
2
–1
(c) v x = 5.01 m s , v y = 0 + (9.81)(0.639) t = 2(9.81)
= 6.27 m s –1 = 2.84 s
–1
2
V = v x + v y = 8.03 m s Horizontal motion,
2
o
R = (12.0 cos 60 )(2.84) m
v y
At an angle to the horizontal, tan ( ) = 51.4° = 17.0 m
–1
v x
8. (a)
5
v sin θ v
1. D 2. A 3. C 4. B 5. B P(x, y)
y
STPM Practice 2
θ v cos θ x
1
1. C : s = ut + at 2 x
2
1
–12.0 m = (6.0)t – (10.0)t 2
2
t = 2.26 s
57
02 STPM PHY T1.indd 57 4/9/18 8:19 AM