Wednesday, December 10, 2014

Physics 9702 Doubts | Help Page 34

  • Physics 9702 Doubts | Help Page 34

Question 210: [Kinematics > Linear motion]
A linear accelerator sends a charged particle along the axis of a set of coaxial hollow metal cylinders as shown in the diagram.
The particle travels at constant speed inside each cylinder. The particle crosses the gaps between the cylinders at equal time intervals, and at each gap its kinetic energy increases by a fixed amount. Which of the graphs best represents the way in which v, the velocity of the particle varies with d, the distance along its tracks?

Reference: Past Exam Paper – J90 / I / 6

Solution 210:
Answer: C.
The particle travels at constant speed inside each cylinder and at each gap its kinetic energy increases by a fixed amount. Then, as it travels inside the next cylinder its speed is constant.
In the v-d graph, a constant velocity is represented by a horizontal line. Since, from the diagram, the lengths of the gaps are relatively smaller than the lengths of the cylinders, the increase in velocity should only correspond to a smaller interval of d. [D and E are incorrect]

The sizes of the gaps are identical.
Let the fixed amount by which the kinetic energy increases = x
Kinetic energy = ½ mv2. So, the speed, v depends on the square root of the kinetic energy.

Let the initial kinetic energy = y. The initial velocity is proportional to √y.
At 1st gap,
Kinetic energy = y + x
Speed v is proportional to √(y+x)

At 2nd gap,
Kinetic energy = y+x + x = y + 2x  
Speed v is proportional to √(y+2x)

At 3rd gap,
Kinetic energy = y+2x + x = y + 3x
Speed v is proportional to √(y+3x)

Note that at each gap, the speed increases while the amount x remains the same. The change in velocity (represented by the diagonal lines in the graph) is given by the difference in velocity after the gap to that before the gap.
After 1st gap, change in velocity = √(y+x) – √y
After 2nd gap, change in velocity = √(y+2x) – √(y+x)
After 3rd gap, change in velocity = √(y+3x) – √(y+2x)

Therefore, as the particle travels through the gaps, its speed increases while the amount x, by which the kinetic energy increases, remains the same. So, at the later gaps, the percentage change in velocity of the particle is less than at the earlier gaps (because v is becoming greater while x remains the same). {For example, consider x = 1 and initial y = 1. After 1 gap, y = 1+1 = 2. After 10 gaps, y = 10+1 = 11 where 10 was the KE previously. So, it can be said that after the 1st gap, y increased by 100% but after the 10th gap, y increased by only 10%}

Thus, at the later gaps, the diagonal lines in the graph decreases. [A is incorrect since the diagonal lines remain constant and D is incorrect since the diagonal lines increases at later gaps]

Question 211: [Kinematics > Projectile motion]
A body is moving with constant speed in the y-direction. For positive values of y, it experiences a uniform acceleration in the x-direction. Which one of the paths A to E does it follow?

Reference: Past Exam Paper – J81 / II / 4

Solution 211:
Answer: B.
Acceleration is the rate of change (increase) of velocity.
For a body experiencing a uniform acceleration, its speed increases as the time t increases.

The graph cannot be a straight line as this indicates that the change in the x-direction is similar to that in the y-direction. This is incorrect since the speed is constant in the y-direction while there is a uniform acceleration in the x-direction. [C is incorrect]

The speed in the x-direction keeps on increasing while that in the y-direction remains constant. So, further in the path, the value of speed in the x-direction would always be greater than that of the speed in the y-direction. That is, as we go further on the path, the value of (y / x) decreases. This is represented by the gradient on the graph. [D and E show graphs with increasing gradient and are therefore incorrect]

As the body moves further, the speed in the x-direction is much greater than at the earlier stages. So, the gradient at the final stage should be much less than the initial gradient. [A is incorrect as the gradient only changes by a small amount]

Question 212: [Kinematics > Projectile motion]
A ball is suspended from an electromagnet attached to a trolley which is travelling at a constant of 10ms-1 to the left. The trolley is illuminated by a stroboscope flashing at a constant rate. The diagram represents the viewpoint of a stationary camera.

The ball is released and a series of stroboscopic images of the ball are recorded on a single photographic plate.
Which diagram best represents what is seen on the photographic plates?

Reference: Past Exam Paper – N91 / I / 4

Solution 212:
Answer: C.
The trolley is travelling to the left at a constant speed of 10ms-1. When released, the horizontal velocity of the ball is also 10ms-1 to the left. Therefore, the image (dot) representing the initial stage is the one at the right-most on the photographic plate. Images are recorded at a regular time interval (at a constant rate).

The ball has a constant horizontal speed of 10ms-1. So, the images of the ball should move towards the left on the plate. [E is incorrect]

Additionally, the horizontal speed of the ball is constant {it is assumed that there are no resistive forces such as air resistance that would cause the horizontal velocity to change}. That is, at a regular time interval, the distance travelled to the left is constant. This is represented by the horizontal displacement towards the left of the images. [B is incorrect since the horizontal displacement is increasing and D is incorrect since this displacement is decreasing]

As the ball is released, it experiences a uniform acceleration (due to gravity) downwards. Acceleration is the rate of change (increase) of velocity, so the speed keeps on increasing every second. That is, the distance travelled vertically downwards increasing as the further images are taken.  [A is incorrect since the vertical displacement is constant]

Question 213: [Kinematics > Linear motion]
(i) Define velocity
(ii) Explain how the displacement and the acceleration of an object may be found from a velocity-time graph of its motion.

(b) A train has mass 450 000kg and a normal operating speed of 50ms-1. It can be considered to be accelerated by a constant force of 180kN and braked by a constant force of 330kN. The train starts from rest. Calculate its acceleration and the time it takes to reach its operating speed.

(c) The inhabitants of a certain town would like trains to make an additional stop at their station. The train would stand for two minutes to allow passengers to get on and off and further delay would be caused by having to slow down and speed up.
(i) Sketch speed-time graphs for a train which does stop at a town and for a train which does not.
(ii) Calculate the total delay caused by making the additional stop.
(iii) Why could this delay be reduced by having fewer carriages?

(d) Explain why, in practice, trains do not have a constant acceleration.

Reference: Past Exam Paper – November 1990 / III / 1

Solution 213:
(i) Velocity is defined as the rate of change of displacement

(ii) From the velocity-time graph of the motion of an object, the displacement of the object in a specific time interval is obtained by the area under the graph in that interval. The instantaneous acceleration at time t is given by the gradient of the graph at that time.

(Constant) Force causing the acceleration = 180kN = 180 000N
Force, F = ma
Acceleration, a = F / m = 180000 / 450000 = 0.40ms-2

Initial speed, u = 0. Final speed, v = 50ms-1. Acceleration, a = 0.40ms-2. Time t???
Acceleration, a = (v – u) / t
Time, t = (50 – 0) / 0.4 = 125s


The difference in the journey of the 2 trains is that one train stops at the town while the other continues its constant motion without at the town.

First consider the train that stops at the town.
Braking force causing the deceleration = 330kN = 330 000N
Acceleration = F / m = 330000 / 450000 = (–) (11 / 15) ms-2
Final speed (stop at town), v = 0. Initial speed, u = 50ms-1.
Acceleration, a = (v – u) / t
Time, t = 50 / (11/15) = 68.18s

The train stands for 2 minutes (= 120s) at the town.

After the 2 minutes, the train needs to accelerate again to reach to constant operating speed of 50ms-1. As calculated in (b), the time it takes to achieve this speed is 125s.

The total time to stop and reach constant speed again = 68.18 + 120 + 125 = 313.18s

But during the deceleration to stop at the town, the train travels a distance. Let the distance travelled during deceleration = sD.
sD = ut + ½ at2 = 50(68.18) + 0.5(- 11/15)(68.18)2 = 1704.55m

Also, when the train accelerates again to reach the constant speed, it travels a certain distance. Let this distance = sA.
sA = ut + ½ at2 = 0(125) + 0.5(0.4)(125)2 = 3125m

Total distance travelled = 1704.55 + 3125 = 4829.55m

Now consider the train that does not stop at the town.
For a proper comparison, we need to consider the original time taken by the train (that does not stop at the town) to travel this distance at its constant speed of 50ms-1. {This can be identified by comparing the 2 graphs}

Speed = Distance / Time
Time = 4829.55 / 50 = 96.591s

Time delay = 313.18 – 96.591 = 216.589 = 216.6s

By having fewer carriages, the mass of the train would be smaller. This causes the deceleration to be greater (a = F / m) resulting in the train to stop in a smaller time. (The time for accelerating again is irrelevant here because this would also have the same effect on the train that does not stop at the town.)

In practice, trains do not have a constant acceleration. This is due to the total mass of the passengers getting on and off. This mass has not been accounted in the calculations above. The total mass of the passengers will always vary depending on the size and number of passengers present on the train at a specific time.
(We are not considering the resistive forces which are present in real cases. These forces would also cause the trains to not have a constant acceleration)

Question 214: [Waves > Graph]
Displacement-time graph is shown for particular wave.

Second wave of similar type has twice the intensity and half the frequency.
When drawn on same axes, what would the second wave look like?

Reference: Past Exam Paper – June 2006 Paper 1 Q24

Solution 214:
Answer: B.
The original graph contains 2 periods of the wave. If the frequency is halved, only 1 period can be displayed on the same axis. [C and D are incorrect]

Intensity, I is proportional to (amplitude, A)2. So, the amplitude is proportional to I.
If the intensity is doubled, the new amplitude becomes (2I) = √2 √I. So, the new amplitude is greater by a factor of √2 (≈1.4) than the original amplitude. [A is incorrect]

It is incorrect to assume that twice the intensity of a wave would be the result of twice the amplitude.

No comments:

Post a Comment

If it's a past exam question, do not include links to the paper. Only the reference.
Comments will only be published after moderation

Currently Viewing: Physics Reference | Physics 9702 Doubts | Help Page 34