# 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 mv

^{2}

D ½ mv

^{2}sin

^{2}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 mv

^{2}
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]__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, (θ

_{1}– θ

_{0}= θ

_{1}) < (θ

_{2}– θ

_{1})

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 at

^{2}
{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 mv

^{2}= 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

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,

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

DeleteA 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.

Alright thank you

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

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

Deletenow, 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.

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?

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

DeleteThe 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.

how s.9 tan30 .could you explain??

Deleteit's better that you use that cos or sin one

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

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

DeleteFor 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.

ReplyDeleteIf 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:

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

how to look tan theta over here??

ReplyDelete