• Physics 9702 Doubts | Help Page 102

Question 523: [Work, Energy, Power]
A space vehicle of mass m re-enters the Earth’s atmosphere at an angle θ to the horizontal. Because of air resistance, the vehicle travels at a constant speed v.
The heat-shield of the vehicle dissipates heat at a rate P, so that the mean temperature of the vehicle remains constant.
Taking g as the relevant value of the acceleration of free fall, which expression is equal to P?
A mgv
B mgv sin θ
C ½ mv²
D ½ mv² sin² θ

Reference: Past Exam Paper – N95 / I / 6



Solution 523:
Answer: B.
Since the vehicle is travelling at constant speed, the net acceleration (and thus net force) is zero. This means that the gravitational pull is balanced by the air resistance.

A constant speed also means that the kinetic energy of the vehicle stays constant throughout the travel. Heat is dissipated at a rate P. From the conservation of energy, this heat (energy) dissipation comes only (since KE is constant here) from the loss in gravitational potential energy of the vehicle as it approaches the Earth’s surface, losing its height.

The vehicle enters the Earth’s atmosphere at an angle θ to the horizontal with a speed v. The motion of the vehicle is of course towards the Earth’s surface (downwards at an angle θ to the horizontal), not upwards.
Horizontal component of speed = v cosθ
Vertical component of speed = v sinθ

As said before, the rate of heat (energy) dissipated, P is due to the rate of loss of G.P.E of the vehicle.
Rate of loss of G.P.E = Δ(mgh) / Δt = mg (Δh / Δt)               [since m and g are constant]

But, the rate of change of the vertical height of the vehicle, (Δh / Δt), is the vertical component of speed, v sinθ.
Rate of heat dissipated, P = Rate of loss of G.P.E = mgv sinθ










Question 524: [Waves > Stationary waves]
Variation with distance x of the intensity I along a stationary sound wave in air is shown by the following graph.

Speed of sound in air is 340 m s^–1.
What is frequency of the sound wave?
A 1700 Hz                  B 2270 Hz                   C 3400 Hz                   D 6800 Hz

Reference: Past Exam Paper – November 2014 Paper 13 Q27



Solution 524:

Answer: A.

The graph shows the variation with distance x of the intensity I (NOT ‘displacement’) of a stationary sound wave in air. The intensity depends on the square of the amplitude. Thus, at an antinode, where the displacement is maximum, the intensity would have the highest value.

But for a stationary wave, the distance between 2 successive nodes or antinodes is not equal to the wavelength, but half the wavelength.

Distance between two nodes / antinodes = 10.0 cm

Wavelength λ = 2 (10) = 20.0 cm = 0.2 m

Speed of wave, v = fλ

Frequency f = v / λ = 340 / 0.2 = 1700 Hz

Question 525: [Waves > Graph]
When sound travels through air, air particles vibrate. Graph of displacement against time for a single air particle is shown.

Which graph best shows how kinetic energy of the air particle varies with time?

Reference: Past Exam Paper – November 2014 Paper 11 & 12 Q23



Solution 525:
Go to
When sound travels through air, the air particles vibrate. A graph of displacement against time for a single air particle is shown.

Question 526: [Waves > Interference > Diffraction grating]

Monochromatic light is directed at diffraction grating as shown.

Which diagram shows all possible directions of the light, after passing through the grating, that give maximum intensity?

Reference: Past Exam Paper – November 2012 Paper 11 Q29



Solution 526:

Answer: C.

Maximum intensity is due to constructive interference.

First of all, there is always a region of maximum intensity in the direction in which the light is incident. [A is incorrect] Additionally, maximum intensities are not limited to the region just mentioned and regions close to it, but maximum intensity can also be observed at different angles. [D is incorrect] There are different orders.

For diffraction grating: d sinθ = nλ

sinθ = nλ / d

angle θ = sin^-1 (nλ / d)

when n = 1, θ = sin^-1 (1λ / d)

when n = 2, θ = sin^-1 (2λ / d)

when n = 3, θ = sin^-1 (3λ / d)

…
Finally, the angle of diffraction from the 0^th order maxima is smaller and the difference in angle of diffraction of the 2^nd order maxima from the 1^st order maxima.
That is, (θ₁ – θ₀ = θ₁) < (θ₂ – θ₁)
The same is true as the order n increases. This is due to the sine function. [B is incorrect]

Anyway, for small angles, the difference in angles could be the same. Choice B would still be wrong.









Question 527: [Kinematics]

Ball is thrown from A to B as shown in Fig.

The ball is thrown with initial velocity V at 60° to horizontal.

Variation with time t of the vertical component V_v of the velocity of ball from t = 0 to t = 0.60 s is shown in Fig.

Assume air resistance is negligible.

(a)

(i) Complete Fig for the time until the ball reaches B.

(ii) Calculate maximum height reached by the ball.

(iii) Calculate horizontal component V_h of the velocity of the ball at time t = 0.

(iv) On Fig, sketch variation with t of V_h. Label this sketch V_h.

(b) Ball has mass 0.65 kg.

Calculate, for ball,

(i) maximum kinetic energy,

(ii) maximum potential energy above the ground.

Reference: Past Exam Paper – November 2014 Paper 22 Q2



Solution 527:

(a)

(i) The graph is a straight line from time t = 0.60 s to t = 1.2 s and the magnitude of V_v, |V_v| = 5.9 at time t = 1.2 s. V_v = – 5.9 at time t = 1.2 s i.e. line is for negative values of Vv

(ii)

Equation for uniformly accelerated motion: s = ut + ½ at²

{To obtain the maximum height, we need to consider the vertical component of velocity. Note that from the graph, the initial velocity is at time t = 0.6s and is zero. At this position, the ball is at its maximum height. The ball falls under an acceleration (due to gravity) of 9.81ms^-2. The time for which the ball reaches the ground is 1.2 – 0.6 = 0.6s}

EITHER Maximum height s = 0 + 0.5(9.81)(0.6)² = 1.8 (1.77) m

OR Maximum height = area under graph = 0.5 (5.9) (0.6) = 1.8 (1.77) m

(iii)

{The value of the complete velocity vector V is not given. Only a component is given in the graph. So, from the vertical component of velocity V_v, we can calculate the overall velocity V and then calculate the horizontal component from it.}

EITHER Horizontal component V_h = V cos 60° and V_v = V sin 60°

{Alternatively, the component could be calculate directly as follows}

OR V_h = 5.9 / tan 60°

OR V_h = 5.9 tan 30°

Horizontal component V_h = 3.4 ms^-1

(iv)

{The horizontal component is not affected by gravity.}

The graph is a horizontal line at 3.4 from t = 0 to t = 1.2 s [to half a small square]

(b)

(i)

{The speed used here is the overall velocity V and can be calculated as shown above.}

Kinetic energy = ½ mv² = 0.5 (0.65) (6.81)² = 15 (15.1) J

(ii) Potential energy (= mgh) = 0.65 (9.81) (1.77) = 11 (11.3) J