Page 29 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM Chapter 2 Kinematics

x 10. (a) Refer to page 34
x = (v cos θ)t, t = v cos θ

–1
–1
–1
1 (b) (i) 54 km h = 54 × 10 3 m s = 15 m s
60 × 60
y = (v sin θ)t – gt 2 Distance travelled under deceleration

2
1
x
(
x
= (v sin θ) v cos θ ) – g ( v cos θ ) 2 = (25 – 15 × 0.60) m = 16 m
2
2
2
v = u + 2as
gx 2
= x tan θ – 2 2 2

2 2v cos θ a = 0 – 15 m s = –7.0 m s -2
-2
2(16)
(b) (i) Vertical motion

(ii)
1 s (m)
s = ut + at 2
2 25
1
15.0 = 0 + (9.81)t 2 16
2
2(15.0)
t = s = 1.75 s
9.81 0 t
(ii) Distance of boat from the cliff u sin θ
2
2
= 800m – (12.0)(1.75) m (c) (i) H = 2.0 m = 2g
= 779 m u sin2θ
2
R = 22.0 m =
g
(iii) v(1.75) = 779 m Solving, θ = 10.3 and u = 35 m s –1
o
v = 445 m s –1
22.0
(ii) T = = 0.64 s
(iv) V x = 445 m s –1 ucos θ
1
–1
V y = 0 + (9.81)(1.75) m s –1 11. (a) v max = (8)(4) m s = 16 m s –1
= 17.2 m s –1 2
1
(b) Speed = 16 – (2)(2) m s = 14 m s –1
–1
2
V = 445 + 17.2 m s –1 2
2
–1.
–1
= 445.3 m s . 12. (a) 140 m s at 19° to the horizontal
V y (b) 9.1 s
-1
At an angle tan ) to the horizontal = 2.2°
( V x 13. (a) Acceleration is the rate of change of velocity.
Eg. A body accelerates from rest.
9. (a) (i) v x = 8.0 m s –1
(b) (i) 6.26 m s
–1
v = 8.0 ms –1
x
60º (ii) 0.0157 s
14. (a) Refer to page 33
v v y
(b) (i) OA – constant acceleration
AB – constant deceleration

v x BC – constant acceleration
Cos 60° = v (ii) 1.67 m s –2
(iii) 1.20 m
8.0
v = m s = 16.0 m s –1 (iv) 0.65 m
–1
cos 60°
(ii) v y = v x tan 60° = 13.9 m s –1 15. (a) 157.5 m
Vertical motion: (b) 1.0 s
2
v y = 0 + 2(9.81)H
2
13.9 2
H = m = 9.85 m
2(9.81)
(b) If there is air resistance, path of the ball is
shown in the diagram.
New path
Cliff
θ

The angle θ  60°
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02 STPM PHY T1.indd 58 4/9/18 8:19 AM