9702/Oct Nov/13/2017/Q22

The graph shows how the displacement of a particle in a wave varies with time.

Which statement is correct?
A The wave has a period of 2 s and could be either transverse or longitudinal.
B The wave has a period of 2 s and must be transverse.
C The wave has a period of 4 s and could be either transverse or longitudinal.
D The wave has a period of 4 s and must be transverse.

Solution:

Answer: C

The period of the wave is 4s, so option A and B are incorrect.

The wave could be either transverse or longitudinal because the graph is not represent the actual shape of the longitudinal wave. Because the graph is only represent the displacement of the wave during specific time.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q22

9702/Oct Nov/13/2017/Q25

A source of sound of frequency F at point Z is moving at a steady speed. The pattern of the
emitted wavefronts is shown.

Which row describes the frequencies of the sound heard by stationary observers at X and Y?

Solution:
Answer: C
Let v be the speed of sound in air. A source of sound has a frequency F and wavelength λ. The source moves towards an observer at a speed v_s.

The period of oscillation of the source of sound is T (= 1/F). In the time of one oscillation the source moves towards the observer X a distance v_sT. Hence the wavelength is shortened by this distance. The wavelength of the sound received by the observer is λ − v_sT.

Hence the frequency observed f_o = v / (λ - v_s T)

                                         = v / [(v/F)-(v_s / F) ]

                                         = (Fv/(v-v_s))

The source would move away from a stationary observer at position Y on the right-hand side. The observed wavelengths would lengthen.

For a source of sound moving away from an observer the observed frequency can be shown to be           f_o = Fv/(v + v_s)

*The frequency is increased when the source moves towards the observer and the frequency is decreased when the source moves away from the observer.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q25

9702/Oct Nov/13/2017/Q29

A beam of laser light is directed towards a narrow slit.

After emerging from the other side of the slit, the light then falls on a screen.
What is the pattern of light seen on the screen?

Solution:

Answer: B

The laser light will undergo diffraction after going through the narrow slit.

Please refer to this "Laser Diffraction and Interference" video for further understanding.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q29

9702/Oct Nov/13/2017/Q28

An electromagnetic wave travels in a straight line through a vacuum. The wave has a frequency
of 6.0 THz.

What is the number of wavelengths in a distance of 1.0 m along the wave?

A             5.0 × 10^–5

B             2.0 × 10¹

C             2.0 × 10⁴

D             5.0 × 10⁷

Solution:
Answer: C

λ = v/f

= (3.0 x 10⁸ ms^-1)/ (6.0 x 10¹² s^-1)

= 5.0 x 10^-5 m

No. of λ = 1.0m/ (5.0 x 10^-5 m)

= 20 000

= 2.0 x 10⁴

Reference: PYQ - Oct/Nov 2017 Paper 13 Q28

9702/Oct Nov/13/2017/Q37

Three identical cells each have electromotive force (e.m.f.) E and negligible internal resistance.
The cells are connected to three identical resistors, each of resistance R, as shown.

What is the potential difference between P and Q?

Solution:

Answer: C

Two of the cells are providing an e.m.f. in one direction and the other in the opposite direction. So,

Overall e.m.f. = E + E – E = E

The resistors are connected in series.

Total resistance = R + R + R = 3R

From Ohm’s law, I = V / R

Current in circuit = overall e.m.f. / total resistance = E / 3R

Consider the potentials in the section PQ.

Let P be at a potential of E.

The e.m.f. of the cell is E. That is, the potential difference between the terminals of the cell is E. If point P is at a potential of 0, then the point between the positive terminal of the battery and the resistor is E.

Current in the circuit = E / 3R

p.d. V across a resistor = IR = (E/3R) × R = E / 3

Thus, potential at point Q = E – E/3 = 2E / 3

In another wording, potential difference between 2 points means the ‘difference in potential’ between the 2 points.

p.d. between P and Q = potential at Q – potential at P

p.d. between P and Q = (2E / 3) – 0 = 2E / 3

Reference: PYQ - Oct/Nov 2017 Paper 13 Q37

9702/Oct Nov/13/2017/Q36

Two cells, each with electromotive force (e.m.f.) E, but different internal resistances r1 and r2, are
connected in series to a resistor R. The reading on the voltmeter is 0 V.

What is the resistance of R?

Solution:
Answer: B


Total e.m.f = E+E = 2E

Total resistance = r1 + r2 + R

Current, i = (2E)/ (r1+r2+R) ------------------(i)

P.D. across 1st cell

V = E - ir1

0 = E - ir1

This implies,

i = E/r1 ------------------(ii)

From (i) and (ii), we have

=> r1 + r2 + R = 2r1

R = r1 - r2

Reference: PYQ - Oct/Nov 2017 Paper 13 Q36

9702/Oct Nov/13/2017/Q10

Two railway trucks of masses m and 3m move towards each other in opposite directions with speeds 2v and v respectively. These trucks collide and stick together.

What is the speed of the trucks after the collision?

Solution:
Answer: A

In order to solve this question we need to use the principle of conservation of momentum which states:

The momentum in a closed system remains constant before and after a collision or explosion.

I.E.

momentum before=momentum after

And ingeneral momentum is calculated using: P=mv where P is the momentum, m is the mass of the object, and v is the velocity of the object.

Hence the total momentum before the collision is:

P1+P2
=(m x 2v) + (3m x (-v))

NOTE: notice the negative v on the second truck as it is moving in the opposite direction to the first truck
=-mv

After the collision the the trucks stick together so the total mass becomes 4m and the combined trucks move at an unknown speed of v_​after

We will solve the equation for conservation of momentum to determine v_after​:

mv=4mv_after

cancelling out the m's:

v=4v_after

and rearranging to make v_​after the subject of the equation:

v_after​=0.25v

Which is our final answer.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q10

9702/Oct Nov/13/2017/Q9

A snooker ball of mass 200 g hits the cushion of a snooker table at right-angles with a speed of 14.0ms^–1.

The ball rebounds with half of its initial speed. The ball is in contact with the cushion for 0.60s.

What is the average force exerted on the ball by the cushion?

A.            2.3 N

B.            7.0 N

C.            2300 N

D.            7000 N

Solution:

Answer: B.

Reference: PYQ - Oct/Nov 2017 Paper 13 Q9