Friday, March 20, 2015

Physics 9702 Doubts | Help Page 90

  • Physics 9702 Doubts | Help Page 90

Question 462: [Waves > Stationary waves]
Speed of a transverse wave on a stretched string can be changed by adjusting tension of the string. A stationary wave pattern is set up on stretched string using an oscillator set at frequency of 650 Hz.

How must wave be changed to maintain the same stationary wave pattern if the applied frequency is increased to 750 Hz?
A Decrease the speed of the wave on the string.
B Decrease the wavelength of the wave on the string.
C Increase the speed of the wave on the string.
D Increase the wavelength of the wave on the string.

Reference: Past Exam Paper – June 2013 Paper 11 Q28

Solution 462:
Answer: C.
One complete wave is obtained (as shown in the diagram). This corresponds to 1 wavelength.
The length of the string is unchanged. To maintain the same wave pattern, that is, for only one complete wave to be obtained again, the wavelength of the wave must remain constant.

Speed of wave, v = fλ
The wavelength λ needs to be kept constant and the frequency is increased to 750Hz.
Wavelength λ = v / f
For λ to have the same value, the speed of the wave, v, needs to be increased.

Question 463: [Forces > Hooke’s law]
Spring of unextended length 40 mm is suspended from a fixed point. Load of 16 N is applied to the free end of the spring. This causes spring to extend so that its final length is five times its original length. The spring obeys Hooke’s Law.
What is the energy stored in the spring due to this extension?
A 1.3 J                         B 1.6 J                         C 2.6 J                         D 3.2 J

Reference: Past Exam Paper – June 2013 Paper 13 Q15

Solution 463:
Answer: A.
Original (unextended) length, L = 40mm

Load applied, F = 16N

Extension, e = 5(40) – 40 = 160mm
Energy stored in spring = ½ Fx = 0.5 (16) (160x10-3) = 1.28 = 1.3J

Question 464: [Kinematics]
Two balls roll from rest down the same slope, but ball X reaches the bottom of the slope in half the time it takes Y. If ball X was accelerating at a ms-2, then Y's acceleration, in ms-2, was
A ¼ a                          B ½ a                           C 2 a                            D 4a

Solution 464:
Answer: A.
Consider ball X.
Initial speed, u = 0. Acceleration = a. Time taken = t. Let the distance travelled = s
Equation for uniformly accelerated motion: s = ut + ½ at2
For ball X: s = 0 + 0.5at2 = 0.5at2

Now, consider ball Y.
The distance travelled is the same (and equal to s). Initial speed, u = 0. Time taken = 2t. Let the acceleration = x.
Equation for uniformly accelerated motion: s = ut + ½ at2
For ball Y: s = 0 + 0.5x (2t)2 = 2xt2

Since the distance s is the same for both cases, we can equate both equations.
2xt2 = 0.5at2
Acceleration of ball Y, x = 0.5a / 2 = ¼ a

Question 465: [Matter > Young modulus]
Lift is supported by two steel cables each of length 20 m.
Each of the cables consists of 100 parallel steel wires, each wire of cross-sectional area 3.2 × 10–6 m2. Young modulus of steel is 2.1 × 1011 N m–2.
Which distance does lift move downward when a man of mass 70 kg steps into it?
A 0.010 mm                B 0.020 mm                C 0.10 mm                  D 0.20 mm

Reference: Past Exam Paper – November 2013 Paper 13 Q22

Solution 465:
Answer: C.
Hooke’s law: F = ke

The lift is supported by 2 steel cables, each of which consists of 100 parallel steel wires. So, there is a total of 200 parallel steel wires in parallel.

For parallel spring,
Effective spring constant, keff = k1 + k2 + k3 + …

First, we need to find the spring constant, k for 1 steel wire.

For 1 wire,
Young modulus, E = stress / strain = (F/A) / (e/L) = FL / Ae
Hooke’s law: F = ke
Young modulus, E = (ke)L / Ae = kL / A
Spring constant, k = EA / L = (2.1 × 1011) (3.2 × 10–6) / 20 = 33600 Nm-1

Effective spring constant, keff = 200k
Hooke’s law: F = keff e
Mass of person = 70kg. Weight = mg = 700N           (take g = 10 ms-2)
Extension, e = F / keff = 700 / (200 × 33600) = 0.00010m = 0.10 mm

Question 466: [Work, Energy, Power]
Diagram shows lift system in which the elevator (mass m1) is partly counterbalanced by a heavy weight (mass m2).

At what rate does motor provide energy to the system when elevator is rising at a steady speed v? (g = acceleration of free fall)
A ½ m1v2                     B ½ (m1 – m2)v2                      C m1gv            D (m1 – m2)gv

Reference: Past Exam Paper – November 2010 Paper 11 Q18

Solution 466:
Answer: D.
The rate at which the motor provides energy is the power of the motor.

The system works as such: the mass m2 works downwards under gravity, causing the elevator of mass m1 to move up. However, the motor also provides some energy in moving the elevator up.

The (gravitational) potential energy of the masses are changing continuously and at a steady rate, since they are moving at a steady speed.

GPE of elevator = m1gh1        where h1 is its height above the ground
Rate of change of GPE of elevator = m1gh1 / t = m1gv

GPE of heavy weight = m2gh2            where h2 is its height above the ground
Rate of change of GPE of heavy weight = m2gh2 / t = m2gv

Let the power of the motor be P.
m1gv = m2gv + P
Power of motor = (m1 – m2)gv

Choices A and B have units of energy, not power. If B was correct, then there would have no need for the heavy weight.


  1. Salam,
    I got another solution of solution 465,
    I used (F/A)/(x/l) and got the write answer. Please respond to whether I can use that or else , will I get the wrong answer another time?

    1. could you provide the complete calculations?

  2. E=(F/A) / (X/L)

    Area = 2 * 100 * 3.2x10^-6 = 6.4 * 10^4

    stress=686.7/6.4 * 10^4

    length = 20

    So I used this to solve

    1. It's OK. However when they are not parallel, we cannot calculate the total area as you did.

  3. ok.. And all I wanted to say is thanks for all your support..This site realllllyyy helped me.. Jazakallah

  4. in question 466, what do you mean by 'if B was correct,then there would have been no need for the heavy weight'?

    1. B is definitely wrong as the system does not only involve KE, but also GPE.

  5. And I seriously wish i had found this site sooner....its so helpful


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