Sunday, August 16, 2015

Physics 9702 Doubts | Help Page 187

  • Physics 9702 Doubts | Help Page 187

Question 912: [Oscillations > Simple harmonic motion]
A student sets out to investigate the oscillation of a mass suspended from the free end of a spring, as illustrated in Fig.1.

The mass is pulled downwards and then released. The variation with time t of the displacement y of the mass is shown in Fig.2.

(a) Use information from Fig.2
(i) to explain why the graph suggests that the oscillations are undamped,
(ii) to calculate angular frequency of the oscillations,
(iii) to determine maximum speed of the oscillating mass.

(i) Determine resonant frequency f0 of the mass-spring system.
(ii) The student finds that if short impulsive forces of frequency ½ f0 are impressed on the mass-spring system, a large amplitude of oscillation is obtained. Explain this observation.

Reference: Past Exam Paper – November 2002 Paper 4 Q3

Solution 912:
(i) It has a constant amplitude.

Period T = 0.75s
Angular frequency ω = 2π / T = 8.4 rad s–1

EITHER use of gradient         OR v = ωy0
{y0 is the amplitude}
Maximum speed v = {8.4 × 0.02 =} 0.168 m s–1

(i) Resonant frequency f0 {= 1/T = 1/0.75} = 1.3Hz

(ii) At ½ f0, a ‘pulse’ is provided to the mass on alternate/some oscillations. So, the ‘pulses’ build up the amplitude.

Question 913: [Current of Electricity]
(a) Distinguish between potential difference (p.d.) and electromotive force (e.m.f.) in terms of energy transformations.

(b) Two cells A and B are connected in series with a resistor R of resistance 5.5 Ω, as shown in Fig.1.

Cell A has e.m.f. 4.4 V and internal resistance 2.3 Ω. Cell B has e.m.f. 2.1 V and internal resistance 1.8 Ω.
(i) State Kirchhoff’s second law.
(ii) Calculate current in the circuit.
(iii) On Fig.1, draw an arrow to show the direction of the current in the circuit. Label this arrow I.
(iv) Calculate
1. p.d. across resistor R,
2. terminal p.d. across cell A,
3. terminal p.d. across cell B.

Reference: Past Exam Paper – November 2011 Paper 23 Q4

Solution 913:
Potential difference (p.d.) is the energy transformed from electrical to other forms, per unit charge.
Electromotive force (e.m.f.) is the energy transformed from other forms to electrical per unit charge.

(i) Kirchhoff’s second law states that the sum of e.m.f.s (in a closed circuit) is equal to the sum of potential differences.

4.4 – 2.1 = I × (1.8 + 5.5 + 2.3)
Current I = 0.24 A

(iii) The arrow (labelled) I should be shown anticlockwise.

1. V = I × R = 0.24 × 5.5 = 1.3(2) V

2. VA = 4.4 – (I × 2.3) = 3.8(5) V
{Voltage loss in cell A = I × 2.3}

EITHER VB = 2.1 + (I × 1.8)                         OR VB = 3.8 – 1.3
{Here, we need to add (I × 1.8) to the e.m.f. of cell B because the current is flowing towards the positive terminal of cell B.}
VB = 2.5(3) V

Question 914: [Dynamics > Moments]
Which pair of forces acts as a couple on the circular object?

Reference: Past Exam Paper – November 2008 Paper 1 Q14

Solution 914:
Answer: A.
A couple consists of 2 forces, equal in magnitude. But acting in opposite directions, so that they produce a resultant turning effect. Thus, these 2 forces produces a resultant moment, while the resultant force is zero.

Choice B: both forces are in the same direction + their magnitudes are not equal
Choice C: different magnitudes
Choice D: directions are not opposite to each other

Question 915: [Kinetic theory of Gases]
(i) State the basic assumption of the kinetic theory of gases that leads to the conclusion that potential energy between the atoms of an ideal gas is zero.
(ii) State what is meant by the internal energy of a substance.
(iii) Explain why an increase in internal energy of an ideal gas is directly related to a rise in temperature of the gas.

(b) A fixed mass of an ideal gas undergoes a cycle PQRP of changes as shown in Fig.1.

(i) State the change in internal energy of the gas during one complete cycle PQRP.
(ii) Calculate work done on the gas during the change from P to Q.
(iii) Some energy changes during the cycle PQRP are shown in Fig.2.

Complete Fig.2 to show all of the energy changes.

Reference: Past Exam Paper – November 2010 Paper 41 & 42 Q2

Solution 915:
(i) There are no forces (of attraction or repulsion) between the atoms / molecules / particles.

(ii) The internal energy of a substance is the sum of kinetic and potential energy of the atoms / molecules due to their random motion.

(iii) The (random) kinetic energy increases with temperature. There is no potential energy. (since the gas is ideal) So, an increase in temperature increases the internal energy.

(i) Change in internal energy = Zero

(ii) Work done = pΔV = (4.0x105)(6x10-4) = 240J


{ΔQ = ΔU – ΔW
+ΔQ is the amount of heat/energy supplied to the gas.
+ΔU is the increase in internal energy.
ΔW is the work done ON the gas. (If + ΔW was used, it would have been ‘work done BY gas’ – but this is not what we are calculating here. The table label is ‘work done ON gas’ – so, we used a negative sign in the equation to account for positive values in the table’s numerical values for this column.)
ΔW = p ΔV

For P to Q: ΔQ = - 600 J. As calculate above, ΔW = + 240 J. ΔU = - 600 + 240 = - 360 J

For Q to R: ΔQ = +720 J. ΔV = 0, so ΔW = 0. ΔQ = ΔU – 0 = +720 J

For R to P: ΔQ = +480 J. As answered in (b)(i), the change in internal energy during one complete cycle is zero. Let the change in internal energy for this change be ΔU. Consider the changes in internal energy for P to Q and Q to R. -360 + 720 + ΔU = 0. So, ΔU = -360 J. ΔW = ΔU – ΔQ = -360 – 480 = -840 J}


  1. For 42/O/N/10 Q.2(b)(iii) How to get the values of the energy changes in the table? Which method should I use?

  2. Please consider answering ALL of the following questions before October:
    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.4(c),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.10(c)(ii),Q.11(b)(iii)
    41/O/N/09 Q.6(a),(b)(i),Q.10
    42/O/N/09 Q.5(a),Q.7(b)(ii)
    41/M/J/10 Q.6(a),Q.7(a)
    42/M/J/10 Q.6(a)(ii)
    51/M/J/10 Q.2(d)
    42/O/N/10 Q.3(c)

    1. For 04/M/J/04 Q.4(c), see solution 916 at

  3. 2004 O/N paper01 Q15
    the kinetic energy of a particle is increased by a factor of 4,
    by what factor does the speed increase?

    1. See solution 67 at

  4. Hi,
    Why is the change in internal energy during one complete cycle in pqrp is zero?

  5. From the first law of thermodynamics, the internal energy depends on the state of the system, that is, its temperature, pressure and volume.

    in a complete cycle, the final state of the system (P, V and T) and the same as its initial state. So, there is no change.
    change in internal energy = 0

  6. Can anyone tell me how is the change in internal energy of the gas during one complete cycle PQRP is zero?


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