Thursday, September 10, 2015

Physics 9702 Doubts | Help Page 197

  • Physics 9702 Doubts | Help Page 197

Question 953: [Dynamics > Momentum]
(a) State the principle of conservation of momentum.

(b) A ball X and a ball Y are travelling along the same straight line in the same direction, as shown in Fig.1. 

Ball X has mass 400 g and horizontal velocity 0.65 m s–1.
Ball Y has mass 600 g and horizontal velocity 0.45 m s–1.
Ball X catches up and collides with ball Y. After the collision, X has horizontal velocity 0.41 m s–1 and Y has horizontal velocity v, as shown in Fig.2.

(i) total initial momentum of the two balls,
(ii) velocity v,
(iii) total initial kinetic energy of the two balls.

(c) Explain how you would check whether the collision is elastic.

(d) Use Newton’s third law to explain why, during collision, the change in momentum of X is equal and opposite to the change in momentum of Y.

Reference: Past Exam Paper – November 2014 Paper 23 Q4

Solution 953:
(a) The principle of conservation of momentum states that, for a system (of interacting bodies), the total momentum remains constant provided there is no resultant force acting (on the system)

{Total momentum = momentum of X + momentum of Y.
Momentum p = mv}
Total initial momentum = m1v1 + m2v2 = (0.4×0.65) + (0.6×0.45) = 0.26 + 0.27 = 0.53 Ns

{Total momentum before collision = Total momentum after collision}
0.53 = (0.4×0.41) + (0.6×v)
Velocity v = 0.366 / 0.6 = 0.61 m s–1

KE = ½ mv2
Total initial KE = ½×0.4×(0.65)2 + ½×0.6×(0.45)2 = 0.0845 + 0.06075 = 0.15 (0.145) J

EITHER Check if the relative speed of approach equals the relative speed of separation
OR Total final kinetic energy equals the total initial kinetic energy

{Newton’s 3rd law:} The forces on the two bodies (or on X and Y) are equal and opposite.
{Force is defined as the rate of change of momentum. F = Δp / Δt where Δp is the change in momentum and Δt is the time of collision, which is the same for both spheres.
Δp = F Δt. Newton’s 3rd law states that F is equal and opposite for the spheres. We know that Δt is the same for both spheres. So, Δp is also equal and opposite for the spheres.}
The time (of collision) is the same for both forces and force is defined as change in momentum / time.

Question 954: [Dynamics > Momentum]
A resultant force of 10 N acts on a body for a time of 2.0 s.
Which graph could show variation with time t of the momentum p of the body?

Reference: Past Exam Paper – June 2014 Paper 13 Q11

Solution 954:
Answer: B.
A graph of momentum against time is given. So, we need a relationship that relates these 2 quantities.

Newton’s 2nd law of motion: Force, F = Δp / Δt
Change in momentum, Δp = FΔt

Gradient of graph = Δp / Δt. Thus, the gradient of the graph represents the force.
The gradient of the graph should have a magnitude equal to 10N.
For choice B, gradient = (20 – 0) / (0 – 2) = (-) 10N

Question 955: [Current of Electricity > Resistance]
Diagram shows part of a circuit.

What is the resistance between the points P and Q due to the resistance network?
A 1.3 Ω                       B 4.0 Ω                       C 10 Ω                        D 37 Ω

Reference: Past Exam Paper – November 2010 Paper 12 Q35

Solution 955:
Answer: B.
Consider the parallel combination of the right section of the network.
Combined resistance = [¼ + ¼ + ¼ + ¼]-1 = 1 Ω

Now, consider the two 6.0 Ω resistors in parallel at the lower part of the left section. The combined resistance of these 2 is in series with the 3.0 Ω resistor.
Combined resistance = 3 + [1/6 + 1/6]-1 = 3 + 3 = 6 Ω

The above calculated resistance is in parallel with the 6.0 Ω resistor at the top.
Total combined resistance on the left section = [1/6 + 1/6]-1 = 3 Ω

Overall resistance between P and Q = 3 + 1 = 4 Ω

Question 956: [Current of Electricity > Thermistor]
A thermistor and another component are connected to a constant voltage supply. A voltmeter is connected across one of the components. Temperature of the thermistor is then reduced but no other changes are made.
In which circuit will the voltmeter reading increase?

Reference: Past Exam Paper – June 2008 Paper 1 Q36

Solution 956:
Answer: D.
The resistance of a thermistor decreases as the temperature increases.
The resistance of an LDR decreases as the light intensity increases.

For circuit A, the voltmeter reading remains constant (and equal to the e.m.f. of the supply) since all the 3 components are connected in parallel to each other.

The temperature of the thermistor is reduced and no other changes are made. So, its resistance increases. From Ohm’s law, V = IR, an increase in the resistance of the thermistor would cause the p.d. across it to increase.

Thus, for the voltmeter reading to increase, the voltmeter should be connected in parallel to the thermistor (so that it reads the p.d. across the thermistor).

Question 957: [Pressure > Liquid]
A submarine carries a pressure meter so that the crew can work out how far they are below the surface of the sea. At surface, the meter indicates a pressure of 100 kPa. The density of seawater is 1030 kg m–3.
What is the depth below the surface when the meter reads 450 kPa?
A 34.6 m                     B 44.5 m                     C 340 m                      D 437 m

Reference: Past Exam Paper – November 2007 Paper 1 Q18

Solution 957:
Answer: A.
Pressure P in a liquid = hρg

Since the pressure at the surface is not zero, we should account for the change in pressure.
ΔP = Δhρg

Change in depth Δh = ΔP / ρg = (450000 – 100000) / (1030 × 9.81) = 34.6m

(The depth is the surface is 0.)


  1. Please consider answering ALL of the following questions:
    4/O/N/02 Q.5(b),Q.6(c)(i)
    6/O/N/02 Q.11(a)(b)
    6/O/N/03 Q.9
    04/M/J/04 Q.8(a),(b)(i),(ii)1.
    06/M/J/04 Q.9(b)(iii),Q.11(b)
    06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)
    04/M/J/05 Q.7(a)
    06/O/N/05 Q.8(b),Q.10(a)
    04/M/J/06 Q.6(a),(c),Q.7(b)
    06/M/J/06 Q.14(b)
    04/O/N/06 Q.3(c)
    06/O/N/06 Q.3(b)
    05/M/J/07 Q.2(d)
    04/O/N/07 Q.7(b)(i),(c),Q.10(c)
    04/M/J/08 Q.3(c)(ii),Q.5(b),Q.9(b),Q.11(a)(ii)
    04/O/N/08 Q.7(c)
    04/M/J/09 Q.11(b)(iii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    42/O/N/09 Q.7(b)(ii)
    41/M/J/10 Q.6(a),Q.7(a)
    51/M/J/10 Q.2(d)
    41/M/J/11 Q.8(a)

    1. For 42/O/N/09 Q.7(b)(ii), go to


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