# Physics 9702 Doubts | Help Page 125

__Question 630: [Elastic and Plastic Behaviour]__**(a)**Tensile forces are applied to opposite ends of a copper rod so that rod is stretched. Variation with stress of the strain of the rod is shown in Fig.1.

(i) Use Fig.1 to determine Young
modulus of copper

(ii) On Fig.1, sketch a line to show
variation with stress of the strain of the rod as the stress is reduced from
2.5x10

^{6}Pa to zero. No further calculations expected.**(b)**Walls of the tyres on a car are made of a rubber compound.

Variation with stress of the strain
of a specimen of this rubber compound is shown in Fig.2.

As the car moves, walls of the tyres
bend and straighten continuously.

Use Fig.2 to explain why walls of
the tyres become warm

**Reference:**

*Past Exam Paper – June 2010 Paper 22 Q5*

__Solution 630:__**(a)**

(i)

Young modulus = stress / strain (=
gradient in linear region)

Data is chosen using point in the linear
region of the graph.

Young modulus (= gradient) = (2.1x10

^{8}) / (1.9x10^{-3}) = 1.1x10^{11}Pa
(ii) {Note that this mark was
removed from the assessment, owing to power-of-ten inconsistency in printed
question paper (should be 2.5x10

^{8}instead of 2.5x10^{6})}
Graph (should have been a straight
line that) does not reach zero strain at stress = 0. (Deformation occurs – it
has a greater value of strain at stress = 0 than initial curve)

**(b)**

.Either

The area between the lines (in the
graph) represents the energy

when the rubber is stretched and
then released.

This energy is seen as thermal
energy / heating.

Or

Area under curve represents energy.

The two areas are different.

Difference represents the energy
released as heat.

__Question 631: [Waves > Superposition]__**(a)**Use principle of superposition to explain the formation of a stationary wave

**(b)**Describe an experiment to determine wavelength of sound in air using stationary waves. Include a diagram of the apparatus.

**(c)**Variation with distance x of the intensity I of a stationary sound wave is shown in Fig.1.

(i) On x-axis of Fig.1, indicate
positions of all the nodes and antinodes of the stationary wave. Label the
nodes

**N**and the antinodes**A**.
(ii) Speed of sound in air is 340ms

^{-1}. Use Fig.1 to determine frequency of the sound wave.**Reference:**

*Past Exam Paper – June 2012 Paper 22 Q6*

__Solution 631:__**(a)**When two waves travelling (along the same line) in opposite directions overlap/meet,

if they have the same
wavelength/frequency, the resultant displacement is the sum of displacements if
each wave / produces nodes and antinodes.

**(b)**

Apparatus: source of sound +
detector + reflection system

Adjustment to apparatus to set up
standing waves – how recognized

Measurements made to obtain wavelength**(c)**

(i) At least 2 nodes (zero
intensity) and 2 antinodes (maximum intensity)

(ii)

Distance node to node = Î» / 2 = 34cm
(allow 33 to 35cm)

Speed c = f Î»

Frequency f = 340 / 0.68 = 500Hz
(allow 490 to 520)

__Question 632: [Simple harmonic motion]__**(a)**State what is meant by

*simple harmonic motion*.

**(b)**A trolley is attached to two extended springs, as shown in Fig.1.

Trolley is displaced along the line
joining the two springs and is then released. At one point in the motion, a
stopwatch is started. Variation with time t of velocity v of the trolley is
shown in Fig.2.

Motion of the trolley is simple
harmonic.

(i) State one time at which the
trolley is moving through equilibrium position and also state the next time
that it moves through this position.

(ii) Amplitude of vibration of the
trolley is 3.2 cm.

Determine

1. maximum speed v

_{0}of the trolley,
2. displacement of the trolley for a
speed of ½ v

_{0}.**(c)**Use answers in (b) to sketch, on the axes of Fig.3, a graph to show variation with displacement x of the velocity v of the trolley.

**Reference:**

*Past Exam Paper – November 2014 Paper 43 Q4*

__Solution 632:__**(a)**For simple harmonic motion, the acceleration / force is proportional to the displacement (from a fixed point) and

EITHER the acceleration and
displacement are in opposite directions

OR the acceleration is always
directed towards a fixed point

**(b)**

(i)

EITHER zero

__&__0.625 s
or 0.625 s

__&__1.25 s
or 1.25 s

__&__1.875 s
or 1.875 s

__&__2.5 s
(ii)

1.

Angular frequency Ï‰ = 2Ï€ / T

__and__maximum speed v_{0}= Ï‰x_{0}
Angular frequency Ï‰ = 2Ï€ / 1.25 =
5.03 rad s

^{–1}
Maximum velocity v

_{0}= 5.03 × 3.2 = 16.1 cm s^{–1}(allow 2 s.f.)
2.

(At displacement x,) Speed v = Ï‰ √(x

_{0}^{2}– x^{2})
EITHER

{when speed v = ½ v

_{0}= ½ (Ï‰x_{0}),}
½ Ï‰x

_{0}= Ï‰ √(x_{0}^{2}– x^{2})
x

_{0}^{2}/ 4 = x_{0}^{2}– x^{2}
Displacement x (= √[x

_{0}^{2}– x_{0}^{2}/4] = √[3x_{0}^{2}/4] = √[¾ (3.2)^{2}] ) = 2.8cm
OR

{when speed v = ½ v

_{0}= ½ (16.1),}
½ (16.1) = 5.03 √(3.2

^{2}– x^{2})
2.58 = 3.2

^{2}– x^{2}
Displacement x = 2.8cm

**(c)**The sketch consists of a loop with origin at its centre. It should have the correct intercepts & shape based on (b)(ii).

{At equilibrium position,
the velocity is maximum. This is point (0, 16.1) and (0, -16.1) on the displacement-velocity
graph. [The velocity can be taken as 16 since this is only a sketch].

At the maximum
displacement (amplitude), the velocity is zero for simple harmonic motion. This
is point (3.2, 0) and (-3.2, 0).

Also, as calculated, when
velocity is half the maximum value (v = 16/2 = 8), the displacement is 2.8cm.
This is point (2.8, 8), (2.8, -8), (-2.8, 8) and (-2.8, -8)}

__Question 633: [Kinematics]__
A cannon fires a cannonball with
initial speed v at angle Î± to the horizontal.

Which equation is correct for the
maximum height H reached?

A H = v sinÎ± / 2g B H = g sinÎ± / 2v C H = (v sinÎ±)

^{2}/ 2g D H = g^{2}sinÎ± / 2v**Reference:**

*Past Exam Paper – June 2013 Paper 11 Q3*

__Solution 633:__**Answer: C.**

Consider the vertical motion of the
canon ball.

Vertical component of initial
velocity = v sinÎ±

Acceleration of free fall (which is
downwards) = - g

Maximum height = H

Equation or uniformly accelerated
motion: v

^{2}= u^{2}+ 2as
0 = (v sinÎ±)

^{2}+ 2 (-g) H
H = (v sinÎ±)

^{2}/ 2g
This comment has been removed by the author.

ReplyDeleteFor 23/M/J/10 Q.7(b), check solution 635 at

Deletehttp://physics-ref.blogspot.com/2015/04/physics-9702-doubts-help-page-126.html

For 22/M/J/10 Q.7(b)(ii)2.,

Deletehttp://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-22-worked.html

In question 631(b), the question says to describe. Can you please give the written description?

ReplyDeleteThe speaker produces the sound wave, which gets reflected at the closed end of the tube.

DeleteThe reflected sound wave from the closed end superimposes with the incoming sound wave from the speaker to produce a standing wave in the tube.

A node appears as an amplitude minimum on the CRO. When the first node (minimum) is located, the position is recorded on a metre rule fixed next to the tube. The microphone is moved again until the next node is located and its position recorded. The same is done for a third node.

The wavelength is the distance between the 3 successive nodes.

thank you! :)

DeleteCan u explain when g should be taken negatively?

ReplyDeleteIt depends on how we define the directions. If the upward direction is taken as positive, then g is -ve.

Deleteand if the downward direction is taken as +ve, then g is +ve since it is downward