Youth New World Mock Exam 2016
Physics Paper 1 Section B Marking Scheme
1.
(a) 0.3  4100   25  0   m  3.34 105
m  0.0921 kg
1M – for mcT + 1M – for ml
1A
(b) No, the temperature of the tea will not further decrease.
1A
Because the temperature of iced tea and melting ice are the same.
1A
OR There is no heat transfer between the iced tea and melting ice.
(1A)
(c)
(i)
(ii)
(iii)
2.
Room temperature OR the temperature of the surrounding
1A
The time when all the ice melts
1A
The temperature increases after the time t1.
1A
It is because heat is absorbed from the surrounding until the
temperature increases to T0. 1A
The temperature stops rising at T0.
1A
(a) PV  nRT
n
1M
1105  0.006
R  273  25
n  0.242 mol
(b)
1A
Number of moles of helium molecules in the tank

6 105  0.03
R  273  25 
(c)
 7.26 mol
Number of helium balloons
1A
5
 7.26   0.242
6
1M
 25
1A
Temperature increases when the tank is placed near fire.
1A
Gas molecules will move faster OR the kinetic energy of the gas
molecules increases.
1A
The gas molecules will collide with the inner wall of the tank more
frequently and vigorously.
1A
The average force acting on the wall and so the pressure increases.
3.
a
work done = F s
= (15) (0.07)
= 1.05 J
b
work done = K.E.
(1.05) = (0.5) (0.1) v2
V = 4.58 ms-1
c (i)
v = u + at
(-4.9) sin 300 = (4.9) sin 300 – (9.81) t
t = 0.499 s
(ii)
1A
1A
1M : equation
1M : u = (4.9) sin 300
1A
Range = velocity × time of flight
= (4.9) sin 300 × (0.499)
= 2.12 m
(iii)
1M
1M
1A
Suggested method:
-
Start the projection at a higher position.
Increase the angle of projection to 450
(any angle between 300 and 600 is also
accepted)
2A (1 marks for each
reasonable answer)
4.
a
at
t=3s
1A
b (i)
normal reaction (N)
1A : weight and
normal reaction
1A : friction
friction (f)
(ii)
correct direction
correct label
weight (W)
A = (0.5) / (4 – 3) = 0.5 m s-2
(iii)
N – mg = ma
N – (60)(9.81) = (60) (0.5 sin 300)
N = 604 N
(iv)
Displacement (d) = area of v-t graph
= (3 + 4)(0.5) / 2
= 1.75 m
1A
1M
1M : a = 0.5 sin 300
1A
1M : 1.75 m
Mechanical energy = P.E. + K.E.
= mgh + (1/2) m v2
= (60)(9.81)(1.75 sin300) +
(1/2)(60)(0.5)2
= (515.0) + (7.5)
= 523 J
1M : correct K.E.
1M : correct P.E.
path difference △ = 4.25 – 3.57 = 0.68 m
1M
1A
5.
a (i)
△ = 0.68 m = 2 
 = 0.34 m
(ii)
1M
1A
speed v = f 
= (950) (0.34) = 323 m s-1
1A
(iii)
The position of the second loud sound moves towards
1A
X.
b (i)
The sound becomes much softer, even closed to zero,
as P is destructive interference after the interchange.
1A
1A
(ii)
The sound becomes louder but not as loud as part (a),
as there is no interference at P.
1A
1A
(c)
Reflection by the wall can be avoided if the experiment
is conducted at outdoor. A clear pattern of interference
will be observed.
1A
1A
a
After raining, a lot of water droplets were formed in the
sky.
1A
b
Ray b is red light.
1A
c (i)
n1 sinθ1 = n2 sinθ2
(1.00) sin 600 = (1.33) sin r
1M
6.
r = 40.60
(ii)
7.
1A
d = 600 – r = 600 – 40.60 = 19.40
1A
For water-air interference, the critical angle c is:
c = sin-1(1/1.33) = 48.80, and
which is greater than r (40.60).
So, the reflection at T is not total internal reflection and
the statement is incorrect.
1A
1A
8.
9.
string
pulley
Signal Generator
vibrator
Correct arrangement - 1A
With label – 1A
weight
Set up the equipment as above. Switch on the signal generator and
adjust the frequency until a standing wave was formed on the string.
Record the frequency and the corresponding weight used. Varying
1A
1A
1A
the weight by adding the number of slot-weight, adjust the frequency
1A
of signal generator again until the same pattern of standing appeared.
10.
(a)
137
55
OR
137
55
(b)
(i)
0 
Cs  137
56 Ba  1 e
1M – for
0
Cs  137
56 Ba  1 
half-life =
(1M – for
ln 2
k
0 
1
e
0
1
+ 1A
 + 1A)
1M
= 952125248 s
= 30.2 years
1A
6
(ii)
(iii)
 1  30.2
A  2.7 10   
2
1M
 2.35 1016 Bq
1A
16
As 40 years is just about 1.3 times half-life of Caesium-137.
So it is not long enough for all of the radioactive materials to
decay.
1A
So the claim is agreed.
1A