solutions-to-mock-physics-promo-exam

```Solutions to Mock Physics Promo Exam
Section A
C
D
D
B
D
C
C
D
A
C
B
B
B
C
C
B
B
A
B
B
C
C
B
A
Section B
1(a)
The principle of conservation of momentum states that the total momentum of a
system remains constant provided there is no resultant force acting on the
system.
[2]
(b)(i)
(ii)
1.
In an elastic collision, the total kinetic energy of the system is conserved.
[1]
2.
In a head-on collision, the subsequent motion of the particles takes place
along the same straight line.
[1]
1.
velocity of separation = v
2.
Let vA = final speed of neutron
vB = final speed of deuteron
[1]
By conservation of momentum,
v  v A  2v B  (1)
velocity of approach  velocity of separation
v  0  v B  v A  (2)
[4]
(2)  2  (1)
v  3v A
1
vA   v
3
2.
(a)
45
[1]
(b)
Using s = ut + 1/2at2
20 = (u cos)t
t = 20 / (u cos)
Using v = u + gt
0 = u sin  - gt1
t1 = (u sin )/ g
[1]
2t1 = t
2(u sin 45) / g = 20 / (u cos 45)
u = 14.0 m s-1
[1]
(c)
(d)
[1]
[1]
[1]
[1]
(e)
3.
Horizontal:
t2 = 10 / (14.0 cos 45)
t2 = 1.01 s
Vertical:
Using s = ut + 1/2at2
s = (14 sin 45)(1.01) + (1/2)(-9.81) (1.01)2
s = 4.99 m &gt; 2.40 m
Pass is not intercepted.
[1]
[1]
(a)
The specific latent heat of fusion of a substance is defined as the quantity of
heat required per unit mass to change the substance from the solid phase to
the liquid phase without a change in temperature.
[2]
(b)
The contact surface between the ice and the heater is larger when crushed
ice is used.
[1]
This maximizes the heat transferred from heater to ice. Less heat is lost to
the surrounding.
[1]
(c)
Since the temperature of ice is lower than the room temperature, heat is
absorbed from the surroundings and the ice melts.
[2]
(c)
(i)
(ii)
(iii)
4.
[1]
Heat supplied by heater
=Pxt
= 15 x (20 x 60)
= 1.8104 J
Mass of ice melted by absorbing heat from the heater
= m2 – m1
= 8.4 – 3.0
= 5.4 g
E = m Lf
or
P x t = (m2 – m1)Lf
4
1.810 = 0.0054 Lf
Lf = 3.33 x 106 J kg-1
(a)
correct v and a in diagram
(b)
FSP =
(c)
Centripetal force = gravitational force
GMm
r2
mr2
[1]
[1]
[1]
[1]
[1]
[2]
[1]
=
GMm
r2
[1]

(d)
GM
r3
[1]
Since  = 2/T =&gt; r3  T2

rJ3
T J2
=
rE3
[1]
T E2
3
 rJ =
5.
=
(1.50 x1011 ) 3 (11 .9) 2
(1) 2
 rJ = 7.82  1011 m
[1]
(a)
a = - 2 x
[1]
(b)
(i)
[1]
A = 4.0 cm
k=
0.20 x9.81
= 39.24 N m-1
0.05
B =w = (
(ii)
(iii)
2
) 
T
[1]
k
39.24

M
0.20
vmax = w x0
T
[1]
[1]
= 14.0 x 0.04 = 0.56 m s-1
[1]
2
2

 0.449 s
w 14.0
[1]
No. of oscillations made in 25 s =
25
 55.7
0.449
[1]
Kinetic
Energy / J
(iv)
0.0314
-0.04
Shape :
Labelling &amp; axis :
0.04
displacement / m
1]
[1]
Section C
1(a) (i)horizontal component = 20 cos 45o = 14 ms-1
vertical component = 20 sin 45o = 14 ms-1
[2]
[1]
(ii) change in velocity can be found by vector triangle or by resolution of vectors.
45o
v
v-u
900
-u
Correct triangle [1]
Magnitude of change in velocity v-u = (202 +202) = 28 ms-1
Direction is perpendicular from the wall
[1]
[1]
1(b) (i) The conditions for static equilibrium :
The resultant force acting on the body is zero [1] and
the resultant torque about any axis on the body is zero [1].
(i)
The three forces in equilibrium should meet at a point since the resultant is zero.
So R acts along the direction of the spine. [1]
800 N
Use cos 30o = (&frac12; R) / 800
30
o
R = 1390 N
120o
800 N
R
Method + substitution = 2 marks
(ii)
1. Work out moments about the pivot. The resultant torque about pivot should
be zero.
[Concept= 1 mark]
0.6 L cos 60o x W + 0.85L cos 60o x 50 = 0.6L sin 30o x F
Each term in the equation should be given credit [ Substitution = 2 mark, one for
clockwise, one for anticlockwise moments]
F = 870 N
[1]
2. The force R will not act along the spine[1]. The resultant force along the ydirection has increased although the resultant force along the horizontal
direction remains constant. So angle of R from horizontal will be increased [1]
[correct Answer with appropriate explanation = 2 marks)
2 (a)
(i)
(ii)
(iii)
(b)
(i)
(ii)
(c)
(d)
PV = nRT
(1.50 x 105)(1.30 x 10-3) = n x 8.31 x 300
n = 0.0782
[1]
[1]
PV = nRT
(1.00 x 105)(1.80 x 10-4) = n x 8.31 x 300
n = 0.00722
[1]
ni + np = nf
0.0782 + 0.00722 = (Pf x 1.30 x 10-3) / (8.31 x 300)
Pf = 1.64 x 105 Pa
[1]
[1]
Work done on the air = area under graph
= &frac12; (1.50 + 1.25)105 x (0.13 x 10-4) + &frac12; (1.25 + 1.11)105 x (0.10 x 10-4)
+ &frac12; (1.11 + 1.00)105 x (0.10 x 10-4)
= 4.02 J
(accept 3.81 to 4.22 J)
∆U = q + w
= 6.1 + 4.02
= 10.1 J (accept 9.9 to 10.3 J)
Increase in temperature.
[1]
[1]
[1]
[1]
[1]
[1]
Sources must be coherent
[1]
Sources must have roughly the same amplitude
[1]
Separation between the sources is several wavelengths
[1]
Sources must be either unpolarised or have the same plane of polarisation
[1]
D
(i)
using x 
[1]
a
6.0  10 7 2.0 
a
= 2.4  10-4 m
[1]
5.0  10 3

(ii)



Change 1
From x 

λD
a
As a increases, fringe separation increases.
[1]
[1]
Change 2
The bright fringes are less bright; intensity reaching the screen
decreases.
[1]
The narrowed slits allow a decreased amount of light to pass through
for interference.
[1]
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