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Question 435: [Waves]

Fig shows variation with distance x along a wave of its displacement d at a particular time. 

Wave is a progressive wave having speed of 330ms^-1.

(a)

(i) Use Fig to determine wavelength of the wave:

(ii) Hence calculate frequency of the wave:

(b) Second wave has same frequency and speed as wave shown in Fig but has double the intensity. Phase difference between the 2 waves is 180^o.

On axes of Fig, sketch graph to show variation with distance x of displacement d of this second wave:

Reference: Past Exam Paper – June 2004 Paper 2 Q2



Solution 435:

(a)

(i) Wavelength, λ = 0.6m

(ii) Frequency, f (= v / λ) = 330 / 0.60 = 550Hz

(b)

The amplitude is shown as greater than a but less than 2a and constant. It should have a correct phase (graph is inverted - phase difference of 180^o)

{Intensity I is proportional to A². If intensity I is doubled, the new amplitude would become √2 A ≈ 1.4A}

Question 436: [Waves]
Graph represents a sinusoidal wave in the sea, travelling at speed of 8.0 m s^–1, at one instant of time.
Maximum speed of oscillating particles in the wave is 2πaf, where a is the amplitude and f is the frequency.

Object P of mass 2.0 × 10^–3 kg floats on the surface.
What is maximum kinetic energy of P due to wave? Assume that its motion is vertical.
A 0.026 mJ                  B 4.0 mJ                      C 39 mJ                       D 64 mJ

Reference: Past Exam Paper – November 2007 Paper 1 Q23



Solution 436:

Answer: B.

Speed of a wave: v = fλ

From the graph, the wavelength λ = 50m.

Frequency f = v / λ = 8 / 50 = 0.16Hz

From the graph, the amplitude a = 2m.

Maximum speed v_max = 2πaf = 2π (2)(0.16) = 0.64π ms^-1

Maximum kinetic energy = ½ mv_max² = 0.5 (2.0 × 10^–3) (0.64π)² = 0.0040 = 4.0mJ











Question 437: [Dynamics > Newton’s laws of motion]

(a)

(i) Define force.

(ii) State Newton’s third law of motion.

(b) 2 spheres approach one another along line joining their centres, as illustrated. 

When they collide, average force acting on sphere A is F_A and average force acting sphere B is F_B.

Forces acts for time t_A on sphere A and time t_B on sphere B.

(i) State relationship between

1. F_A and F_B

2. t_A and t_B

(ii) Use answers in (i) to show that change in momentum of sphere A is equal in magnitude and opposite in direction to change in momentum of sphere B.

(c) For spheres in (b), variation with time of momentum of sphere A before, during and after the collision with sphere B is shown. 

Momentum of sphere B before collision also shown.

Complete Fig to show variation with time of momentum of sphere B during and after collision with sphere A.

Reference: Past Exam Paper – June 2010 Paper 22 Q3



Solution 437:

(a)

(i) Force is defined as the rate of change of momentum.

(ii) Newton’s third law of motion states that the force on a body A is equal in magnitude to the force on a body B (from body A). The forces are in opposite directions and are of the same kind.

(b)

(i)

1. F_A = - F_B

2. t_A = t_B

(ii) Change in momentum, Δp = F_At_A = - F_Bt_B

(c)

For the graph, the momentum change occurs at the same time for both spheres. The final momentum of sphere B is to the right and of magnitude 5Ns.

Question 438: [Energy]
Solid rubber ball has diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface. It falls vertically and then bounces back up so that maximum height reached by the top of the ball is 45 cm, as shown.

If kinetic energy of the ball is 0.75 J just before it strikes the surface, what is its kinetic energy just after it leaves the surface?
A 0.36 J                       B 0.39 J                       C 0.40 J                       D 0.42 J

Reference: Past Exam Paper – June 2013 Paper 11 Q17



Solution 438:
Go to
A solid rubber ball has a diameter of 8.0 cm. It is released from rest with the top of the ball 80 cm above a horizontal surface.








Question 439: [Current of Electricity]
Circuit is set up with LDR and a fixed resistor as shown.

Voltmeter reads 4 V.
Light intensity is increased.
What is a possible voltmeter reading?
A 3 V                         B 4 V                         C 6 V                          D 8 V

Reference: Past Exam Paper – June 2007 Paper 1 Q34



Solution 439:

Answer: A.

The resistance of an LDR decreases as the light intensity incident on it increases.

From Ohm’s law: V = IR
The p.d. across a component is dependent on its resistance. Since the resistance of the LDR decreases, the p.d. across it will also decrease.