Thursday, April 2, 2015

Physics 9702 Doubts | Help Page 102

  • 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:
Answer: D.
Kinetic energy = ½ mv2

A graph of displacement against time for a single air particle is shown.  The gradient of the displacement-time graph gives the velocity of the air particle at that point in time. This is done by calculating the gradient of the tangent at that point.

The gradient (and hence, velocity) is found to be zero at the maximum displacement (the tangent is horizontal) and maximum when the displacement is zero (the tangent is steepest).

Thus, at time = 0, T and 2T the velocity is zero and hence kinetic energy is zero. [A and C incorrect]

But, between time = 0 and T or between time = T and 2T the displacement is zero (in case) twice. So, the velocity (and kinetic energy) reaches its maximum value 2 times in each of the 2 intervals. [B is incorrect]

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 0th order maxima is smaller and the difference in angle of diffraction of the 2nd order maxima from the 1st 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 Vv of the velocity of ball from t = 0 to t = 0.60 s is shown in Fig.

Assume air resistance is negligible.
(i) Complete Fig for the time until the ball reaches B.
(ii) Calculate maximum height reached by the ball.
(iii) Calculate horizontal component Vh of the velocity of the ball at time t = 0.
(iv) On Fig, sketch variation with t of Vh. Label this sketch Vh.

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

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

{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 Vv, we can calculate the overall velocity V and then calculate the horizontal component from it.}
EITHER Horizontal component Vh = V cos 60° and Vv = V sin 60°
{Alternatively, the component could be calculate directly as follows}
OR Vh = 5.9 / tan 60°
OR Vh = 5.9 tan 30°
Horizontal component Vh = 3.4 ms-1

{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]

{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


  1. In qu526, why are we finding the angles between the different orders. How does that relate to the intensity, and based on that, how do we arrive to the conclusion that C is the answer,

    1. First of all, maximum intensity here is just constructive interference. This is represented by the arrows in the diagrams.

      A and D can be eliminated as explained above. Now, to compare B and C, it can be seen in the diagrams that the difference between them is that the directions of the arrows and their separation.

      So, by analyzing the angles for the different orders, unlike choice B, the separation in angles does NOT DECREASE as the orders increase.

  2. For question 527 (a), how did you come up with the value for the straight line, Vh?

    1. vh is the horizontal component of the speed v and hence vh = v cos60

      now, the vertical component v_v = v sin60. from the graph, v_v = 5.9. So, from this equation V can be made the subject of formula (without evaluation the number) and it is replaced in the equation for vh.

  3. could you please explain me in more details, how did you find the velocity to be 6.81? and how do we know that the graph is a straight line?

    1. Vh = 5.9 tan 30° and Vh = V cos 60°. So, V = 5.9 tan 30° / cos 60° = 6.81 ms-1.

      The gradient of the graph gives the acceleration which is g, the acceleration of free fall here. So, g is constant, the gradient needs to be constant, and thus the graph is a straight line.

    2. how s.9 tan30 .could you explain??

    3. it's better that you use that cos or sin one

  4. For Q527, does it matter if we take the reading from the graph as 6.0 instead of 5.9 for the value of Vv?

    1. it can be read quite readily at 5.9. Using 6.0 may affect the other calculations

  5. For solution 527,can we use initial vertical component as 5.9 at t=0 untill ball reaches max height at t=0.6 to calculate max height.

    1. If we use the first part of the motion, i.e. the ball moving up towards the maximum height, we need to carefully consider the signs for the vector quantities involved:

      s = ut + ½ at^2 = (5.9×0.6) + ½ (-9.81) (0.6^2) = 1.77 m


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