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Answers to JC2 Preliminary Examination Paper 2 (H2 Physics)
Suggested Solutions:
No. / Solution1(a) / The principle of conservation of momentum states that the total momentum of a system remains constant, provided that no external resultant force acts on it.
1(b)(i) / The gradients of the graph represent the forces acting on the trucks, and the forces on the trucks must be equal and opposite in direction, in accordance to Newton’s third law.
1(b)(ii) / Total initial momentum of system
kgms−1
Total final momentum of system
kgms−1
Since the total momentum of the system remains constant, momentum is conserved.
OR
Gain in momentum by truck B
kgms−1
Loss in momentum by truck A
kgms−1
Since the gain in momentum by truck B is equal to the loss in momentum by truck A, momentum is conserved.
1(b)(iii) / By Newton’s second law of motion,
N
1(b)(iv) / Total initial K.E. of system
J
Total final K.E. of system
J
Since K.E. is not conserved, it is an inelastic collision.
1(c) / In an event of a collision, a driver in a vehicle experiences change in momentum . To reduce the impact of force F acting on the driver, the duration of impact must be long, as seen from the equation representing Newton’s second law of motion . The seat belts and air bags help lengthen the time of impact, and hence ensure greater safety.
2(a) / Simple harmonic motion is defined as an oscillatory motion taking place in which the acceleration of an object is directly proportional to the displacement of the object from its equilibrium position, and the acceleration is always directed towards that position.
2(b) / From Fig. 2.1, we can deduced that .
Therefore, velocity can be written as
2(c)(i) / Resonance occurs when natural frequency of rubber duck = driving frequency of water wave.
ms−1
2(c)(ii) / More water waves per unit time would imply a higher driving frequency. Since the natural frequency of the rubber duck is different from that of the driving frequency of the water wave, the amplitude of oscillation of the rubber duck would initially be reduced.
3(a) / The two-source light interference pattern consists of bright and dark fringes. The bright fringe (maxima) is due to constructive interference and the dark fringe (minima) is due to destructive interference.
The principle of superposition is needed to explain interference and it assumes the wave theory of light. It states that the resultant displacement of waves is due to the vector sum of the individual displacements of waves at that point. Thus, to obtain the interference patterns, the wave theory of light has to be applied.
3(b)(i)
1. / There is minimum intensity at position O.
3(b)(i)
2. / The pattern has non-zero minima.
3(b)(ii) / Fringe separation,
m
Since ,
m
4(a) / The electric potential at a point in an electric field is defined as the work done in bringing unit positive charge from infinity to that point.
4(b)(i) / E.P.E. of system
J
4(b)(ii) / By the conservation of energy,
loss in E.P.E. of system = gain in K.E. of system
ms−1
4(b)(iii)1. / The sign of the third charge is positive.
In order for the either of the negative charges to be stationary, the resultant force must be zero. Since charge X exerts a repulsive force on charge Y, the third charge must exert an attractive force on charge Y, hence it must be a positive charge.
4(b)(iii)2. /
C
5(a) /
Trivalent atoms results in empty acceptor levels to the energy band diagram, just above the valence band.
Because the new levels are so near the valence band, at room temperature, valence electrons are raised into the new empty levels, creating more holes in the valence band than electrons in conduction band.
5(b)(i) /
direction of flow of electron
5(b)(ii) /
5(c)(i) /
5(c)(ii)
5(c)(iii) / A cycle from 0 s to 1.0 s is the same as from 0.75 s to 1.75 s.
For 0.75 s < t < 1.25 s, V.
For 1.25 s < t < 1.75 s, V.
Hence, over one complete cycle,
V.
5(c)(iv) / V
6(a)(i) / Anti-clockwise.
6(a)(ii) / The rotating disc cuts the magnetic field and there is change in magnetic flux. By Faraday’s law, an e.m.f. that is proportional to the rate of change of magnetic flux is induced across the rim and axle of the plate. Increasing the speed of rotation increases the rate of change of magnetic flux and hence induces an increasing e.m.f. across the disc.
6(b) / Direct current.
6(c)(i) / This means that an extremely low or negligible resistance path is connected across the terminals.
6(c)(ii) / Charge = area under the I-t graph
= C
[Acceptable range: C to C]
6(c)(iii) / Maximum power
W
6(d)(i) / Units of k
= Units of
= Units of (where A is area and F is magnetic flux)
= Units of (since units of r2 = units of A)
= s−1
6(d)(ii) / Frequency or angular frequency.
6(e)(i) / Electrical energy from the circuit is converted to kinetic energy of the rotating disc.
6(e)(ii) / Kinetic energy of the rotating disc is converted into electrical energy in the circuit.
7 / Aim : To investigate how the first rebound height of a ball off a concrete ground depends on the air pressure inside the ball.
Independent variable : air pressure inside the ball, p, measured by pressure gauge
Dependent variable : first rebound height of the ball, h, measured by
vertically clamped metre rule
Controlled variables : - diameter of the ball and material from which the ball is made
OR use the same ball
- angle of projection of ball
- initial height of ball above ground
- initial velocity of ball
Diagram :
Procedure :
a) Set up the apparatus as shown above.
b) Pump some air into the ball using the bicycle pump and record p using the pressure gauge.
c) Release the ball from rest from the table top. Allow the ball to bounce once and measure the first rebound height h using the vertically clamped metre rule.
d) Repeat the steps to obtain 6 sets of readings for p and h.
e) Use equation and calculate lg h and lg p.
f) Plot a graph of lg h versus lg p, and determine the value of n from the gradient of the graph.
Further details :
Precautions to improve accuracy :
- Use plumbline to ensure that the metre rule is clamped vertically.
- Before taking each h reading, allow the ball to rebound off the ground and obtain an approximate h reading. This allows one to position one’s eye at a suitable level when the actual h measurement is taken, to reduce parallax error.
- For each p reading, repeat the measurement of h at least 2 times and take the average h to reduce random errors.
(For extra reading only)
Safety precautions :
- Ensure the retort stand and metre rule do not obstruct one’s movement.