# 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.

(b)(i)
(ii)
1.
In an elastic collision, the total kinetic energy of the system is conserved.

2.
In a head-on collision, the subsequent motion of the particles takes place
along the same straight line.

1.
velocity of separation = v
2.
Let vA = final speed of neutron
vB = final speed of deuteron

By conservation of momentum,
v  v A  2v B  (1)
velocity of approach  velocity of separation
v  0  v B  v A  (2)

(2)  2  (1)
v  3v A
1
vA   v
3
2.
(a)
45

(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

2t1 = t
2(u sin 45) / g = 20 / (u cos 45)
u = 14.0 m s-1

(c)
(d)




(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.


(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.

(b)
The contact surface between the ice and the heater is larger when crushed
ice is used.

This maximizes the heat transferred from heater to ice. Less heat is lost to
the surrounding.

(c)
Since the temperature of ice is lower than the room temperature, heat is
absorbed from the surroundings and the ice melts.

(c)
(i)
(ii)
(iii)
4.

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







=
GMm
r2


(d)
GM
r3

Since  = 2/T =&gt; r3  T2

rJ3
T J2
=
rE3

T E2
3
 rJ =
5.
=
(1.50 x1011 ) 3 (11 .9) 2
(1) 2
 rJ = 7.82  1011 m

(a)
a = - 2 x

(b)
(i)

A = 4.0 cm
k=
0.20 x9.81
= 39.24 N m-1
0.05
B =w = (
(ii)
(iii)
2
) 
T

k
39.24

M
0.20
vmax = w x0
T


= 14.0 x 0.04 = 0.56 m s-1

2
2

 0.449 s
w 14.0

No. of oscillations made in 25 s =
25
 55.7
0.449

Kinetic
Energy / J
(iv)
0.0314
-0.04
Shape :
Labelling &amp; axis :
0.04
displacement / m
1]

Section C
1(a) (i)horizontal component = 20 cos 45o = 14 ms-1
vertical component = 20 sin 45o = 14 ms-1


(ii) change in velocity can be found by vector triangle or by resolution of vectors.
45o
v
v-u
900
-u
Correct triangle 
Magnitude of change in velocity v-u = (202 +202) = 28 ms-1
Direction is perpendicular from the wall


1(b) (i) The conditions for static equilibrium :
The resultant force acting on the body is zero  and
the resultant torque about any axis on the body is zero .
(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. 
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

2. The force R will not act along the spine. 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 
[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


PV = nRT
(1.00 x 105)(1.80 x 10-4) = n x 8.31 x 300
n = 0.00722

ni + np = nf
0.0782 + 0.00722 = (Pf x 1.30 x 10-3) / (8.31 x 300)
Pf = 1.64 x 105 Pa


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.






Sources must be coherent

Sources must have roughly the same amplitude

Separation between the sources is several wavelengths

Sources must be either unpolarised or have the same plane of polarisation

D
(i)
using x 

a
6.0  10 7 2.0 
a
= 2.4  10-4 m

5.0  10 3

(ii)



Change 1
From x 

λD
a
As a increases, fringe separation increases.


Change 2
The bright fringes are less bright; intensity reaching the screen
decreases.

The narrowed slits allow a decreased amount of light to pass through
for interference.

```