Page 38 - Pra U STPM 2021 Penggal 1 - Physics

Physics Term 1 STPM STPM Model Paper (960/2)

(b) p (iii)
Extension (mm)
(ii) 3.0 Q
(i)
P
(iii)
0 V 2.0
V 0 2V 0

Section C 1.0
18. (a) Friction between the road and the tyres.
Constant. 0.3 O Force (N)
Air resistance. Increases as speed increases. 0 8.0 16.0 24.0
20 – 0
–2
(b) (i) a = m s = 4.0 m s –2 Permanent extension = 0.3 mm
5.0 (iv) Energy dissipated
(ii) F = ma = 3.84 × 10 N
3
(iii) Forward force > friction + air = area between loading and unloading
resistance graphs
–3
= 5 × 10 J
As the speed increases, air resistance
increases. 20. (a) (i) Q = ∆U + W
To maintain a constant acceleration Q : heat supplied,
the forward force must increase. ∆ U : change in internal energy
(iv) W : work done by the gas
Air resistance (ii) Isothermal expansion, temperature
constant.
Forward force ∆ U = 0
Q = 0 + W
Heat supplied Q used to do work
Friction
against external pressure when the
(c) Power = Fv gas expands.
25 × 10 3 (b) (i) I: V = constant, p ∝ T
F = N = 1.25 × 10 N
3
20 p = 3p , hence T = 3T 0
1
0
1
Stress (ii) II: p = constant, V ∝ T
19. (a) (i) Young’s modulus =
Strain When T = 3T , V = V
F/A (ke)L kL 1 0 0 V
(ii) Y = = = Hence when T = T , V = 0
e/L Ae A 2 0 2 3
(b) (i) OP: Elastic deformation. (iii) I : ∆V = 0, W = 0
1
When the wire is stretched, atoms are V
II : W = (3p )( 0 – V ) = – 2p V
displaced from the mean positions. 2 0 3 0 0 0
When force is removed, atoms return V 2
to the original positions. III: W = nRT ln ( V ) and (p V = nRT )
3
0
0
0
PQ: Plastic deformation. 1
V
When the wire is stretched beyond = p V ln ( 0 ) = p V (ln3)
0 0 V /3 0 0
the elastic limit, atomic planes slide 0
over each other. When the force W = W + W + W 3
1
2
is removed, atoms do not return = (ln3 – 2) p V = – 0.90p V
0 0 0 0
completely to their original positions.
1 YA p
(ii) k = =
gradient L
3p II
20 20 0
Y = ( )( ) N m –2
–3 2
2.5 × 10 –3 π(0.60 × 10 )
III
= 2.1 × 10 N m –2 I
10
p 0
V
0 V /3 V 0
0
328
MODEL PAPER PHY T1.indd 328 4/9/18 8:32 AM