A particle of mass m, travelling with speed u, collides with a stationary particle of mass M.

Question 18

A particle of mass m, travelling with speed u, collides with a stationary particle of mass M. The velocities of the two particles before and after the collision are shown.

Which vector diagram correctly shows the momenta before and after the collision?

Reference: Past Exam Paper – March 2016 Paper 12 Q10

Solution:

Answer: B.

This question is on collisions in 2D.

 

(Momentum = mass × velocity)

From the conservation of momentum,

Sum of momentum before collision = Sum of momentum after collision

Sum of momentum before collision = mu + 0 = mu

Momentum is a vector quantity and so, we need to account for the direction.

Before the collision, the momentum is mu horizontally to the right.

From the conservation of momentum, adding the momenta of the 2 particles after collision should give a resultant = mu horizontally to the right

After collision,

Momentum of particle m = mu / 2 at an angle β below the horizontal

Momentum of particle M = Mu / 3 at an angle α above the horizontal

mu = Sum of momentum after collision

Consider the options available.

Option A: The momenta are at the wrong angles. [incorrect]

Option B: correct

Option C: wrong angles + the negative of the momentum vectors have been taken (the direction of the vectors are opposite than they should be) [incorrect]

Option D: the negative of the momentum vectors have been taken [incorrect]

A cathode-ray oscilloscope (c.r.o.) is used to determine the frequency of the sound emitted by a loudspeaker. The trace produced on the screen of the c.r.o. is shown in Fig. 4.1.

Question 18

(a) State what is meant by the frequency of a progressive wave. [2]

(b) A cathode-ray oscilloscope (c.r.o.) is used to determine the frequency of the sound emitted by a loudspeaker. The trace produced on the screen of the c.r.o. is shown in Fig. 4.1.

Fig. 4.1

The time-base setting of the c.r.o. is 250 μs cm^-1.

Show that the frequency of the sound wave is 1600 Hz. [2]

(c) The loudspeaker in (b) emits the sound in all directions. A person attaches the loudspeaker to a string and then swings the loudspeaker at a constant speed in a horizontal circle above his head.

An observer, standing a large distance away from the loudspeaker, hears sound of maximum frequency 1640 Hz. The speed of sound in air is 330 m s^-1.

(i) Determine the speed of the loudspeaker. [2]

(ii) Describe and explain, qualitatively, the variation in the frequency of the sound heard by the observer. [2]

[Total: 8]

Reference: Past Exam Paper – November 2016 Paper 21 & 23 Q4

Solution:

(a) Frequency of a progressive wave is the number of oscillations of a point on the wave per unit time.

(b)

{Starting from the first peak (crest) on the left,

2 waves occupies 5 cm

1 wave occupies 5/2 = 2.5 cm

Time-base setting is 250 μs cm^-1. That is, each cm represents 250 μs.}

Period T = 2.5 × 250 (μs) (= 625 μs)

{Frequency = 1 / T

625 μs = 6.25 × 10^-4 s}

Frequency = 1 / (6.25 × 10^-4)             or 1 / (2.5 × 250 × 10^-6) = 1600 Hz

(c)

(i)

{Since the source of wave is moving relative to the observer, Doppler effect occurs.

When the source moves towards the observer (approaching), the observed wavelength decreases and so, the observed frequency increases. From the Doppler effect formula, the negative sign is used.

Observed frequency, f₀= 1640 Hz

Frequency of source, fs = 1600 Hz (as calculated above)

Speed of sound, v = 330 m s^-1

Speed of source of wave (loudspeaker) = v_s }

For maximum frequency:

fo = fsv / (v – vs)

1640 = (1600 × 330) / (330 – vs)

vs = 8(.0) m s^-1 (8.049 m s^-1)

(ii)

{Note that as the loudspeaker is moved in circle, at some point it is moving closer to the observer and at some point, it is moving further away from the observer.}

The loudspeaker moving towards the observer causes a rise in/higher frequency and the loudspeaker moving away from the observer causes a fall in/lower frequency