• 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 2.5x10⁶Pa 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) = (2.1x10⁸) / (1.9x10^-3) = 1.1x10¹¹Pa  

(ii) {Note that this mark was removed from the assessment, owing to power-of-ten inconsistency in printed question paper (should be 2.5x10⁸ instead of 2.5x10⁶)}

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 = λ / 2 = 34cm (allow 33 to 35cm)

Speed c = f λ  

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 v₀ of the trolley,

2. displacement of the trolley for a speed of ½ v₀.

(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 ω = 2π / T and maximum speed v₀ = ωx₀

Angular frequency ω = 2π / 1.25 = 5.03 rad s^–1

Maximum velocity v₀ = 5.03 × 3.2 = 16.1 cm s^–1 (allow 2 s.f.)

2.

(At displacement x,) Speed v = ω √(x₀² – x²)

EITHER

{when speed v = ½ v₀ = ½ (ωx₀),}

½ ωx₀ = ω √(x₀² – x²)

x₀² / 4 = x₀² – x²

Displacement x (= √[x₀² – x₀²/4] = √[3x₀²/4] = √[¾ (3.2)²] ) = 2.8cm

OR

{when speed v = ½ v₀ = ½ (16.1),}

½ (16.1) = 5.03 √(3.2² – x²)

2.58 = 3.2² – x²

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 α to the horizontal.

Which equation is correct for the maximum height H reached?

A H = v sinα / 2g        B H = g sinα / 2v         C H = (v sinα)² / 2g     D H = g² sinα / 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 sinα

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

Maximum height = H

Equation or uniformly accelerated motion: v² = u² + 2as

0 = (v sinα)² + 2 (-g) H

H = (v sinα)² / 2g