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Question 626: [Young modulus]

(a) Define the Young modulus.

(b) Load F is suspended from a fixed point by a steel wire. Variation with extension x of F for the wire is shown in Fig.1.

(i) State two quantities, other than gradient of the graph in Fig.1, that are required in order to determine the Young modulus of steel.

(ii) Describe how the quantities mentioned in (i) may be measured

(iii) A load of 3.0 N is applied to the wire. Use Fig.1 to calculate the energy stored in the wire.

(c) A copper wire has same original dimensions as the steel wire. Young modulus for steel is 2.2 × 10¹¹ N m^–2 and for copper is 1.1 × 10¹¹ N m^–2.

On Fig.1, sketch variation with x of F for the copper wire for extensions up to 0.25 mm. The copper wire is not extended beyond its limit of proportionality.

Reference: Past Exam Paper – June 2012 Paper 22 Q5

Solution 626:

(a)

(Young modulus (E) is defined as the ratio of stress / strain.)

E = stress/strain

(b)

(i)

1. diameter / cross sectional area / radius                               

2. original length                    

(ii)

The original length can be measured with a metre ruler / tape

The diameter can be measured with a micrometer (screw gauge)

(iii)

Energy = ½ Fe            or area under graph     or ½ kx²

Energy = ½ (3)(0.25x10^-3) = 3.8x10^-4J

(c)

{Young modulus E = stress/ strain = (Force/Area) / strain.

Young modulus E = Force / (Area x strain)

Area is the same for both.

For the same extension, strain should also be the same for both (strain = e / L where e is the extension and L is the original length).

So, the force is proportional to the Young modulus E.

Since for copper, E is half than for steel, for the same extension, the force required for the case of copper is half. So, at x = 0.25mm, the force = 3.0 / 2 = 1.5N for copper.}

Straight line through origin below the original line

Line passes through (0.25, 1.5)

Question 627: [Hooke’s law, Momentum]

(a) Variation with extension x of tension F in a spring is shown in Fig.1. 

Use Fig.1 to calculate energy stored in the spring for an extension of 4.0 cm. Explain your working.

(b) Spring in (a) is used to join together two frictionless trolleys A and B of mass M₁ and M₂ respectively, as shown in Fig.2. 

Trolleys rest on a horizontal surface and are held apart so that the spring is extended.

Trolleys are then released.

(i) Explain why, as extension of the spring is reduced, momentum of trolley A is equal in magnitude but opposite in direction to momentum of trolley B

(ii) At the instant when the extension of spring is zero, trolley A has speed V₁ and trolley B has speed V₂.

Write down

1. equation, based on momentum, to relate V₁ and V₂

2. equation to relate initial energy E stored in the spring to final energies of trolleys

(iii)

1. Show that kinetic energy E_K of an object of mass m is related to its momentum p by the expression

E_K = p² / 2m

2. Trolley A has larger mass than trolley B.

Use answer in (ii) part 1 to deduce which trolley, A or B, has the larger kinetic energy at instant when extension of spring is zero.

Reference: Past Exam Paper – June 2010 Paper 21 Q3

Solution 627:

(a)

EITHER The energy stored / work done is represented by the area under the graph OR Energy = average force x extension.

Energy stored = ½ (180) (4.0x10^-2) = 3.6J

(b)

(i)

EITHER

The total momentum before the release is zero {since trolleys are at rest}, so {from conservation of momentum} the sum of momenta (of the trolleys) after the release is also zero.

OR
The force is equal to the rate of change of momentum. The forces on the trolleys are equal and opposite.

OR

Impulse is equal to the change in momentum. The impulse on each trolley is equal and opposite.

(ii)

1. M₁V₁ = M₂V₂

2.

{Initial energy = Sum of (final) kinetic energies}

E = ½ M₁V₁² + ½ M₂V₂²

(iii)

1.

Kinetic energy, E_K = ½mv²     and momentum, p = mv          combined to give

{mv² = (mv)v = pv = p (p/m) = p²/m}

E_K = p² / 2m 

2.

The mass m of trolley B is smaller than that of trolley A. The kinetic energy of trolley B {E_K = p² / 2m} is, however, larger because the momentum of trolley A is equal to the momentum of trolley B / constant {Momentum p is equal for both [M₁V₁ = M₂V₂]. Since E_K = p² / 2m, when mass is smaller, EK is larger as m is in the denominator}.

So, trolley B has a larger kinetic energy.

{This could also be understood in terms of the velocity. Momentum p = mv is the same for both. Velocity, v = p/m and E_K depends on v² (E_K = ½mv²). Since the mass of B is smaller, its velocity is larger}

Question 628: [Electromagnetism]

(a) Define the tesla.

(b) A long solenoid has area of cross-section of 28 cm², as shown in Fig.1.

A coil C consisting of 160 turns of insulated wire is wound tightly around centre of the solenoid.

The magnetic flux density B at the centre of the solenoid is given by expression

B = μ₀n I

where I is the current in the solenoid, n is a constant equal to 1.5 × 10³ m^–1 and μ₀ is the permeability of free space.

Calculate, for a current of 3.5 A in the solenoid,

(i) magnetic flux density at the centre of the solenoid,

(ii) the flux linkage in the coil C.

(c)

(i) State Faraday’s law of electromagnetic induction.

(ii) Current in the solenoid in (b) is reversed in direction in a time of 0.80 s. Calculate the average e.m.f. induced in coil C.

Reference: Past Exam Paper – June 2013 Paper 41 & 43 Q5

Solution 628:

Go to
A long solenoid has an area of cross-section of 28 cm2, as shown in Fig. 5.1.

Question 629: [Kinematics]

Variation with time t of the displacement s for a car is shown in Fig.1.

(a) Determine magnitude of average velocity between times 5.0s and 35.0s

(b) On Fig.2, sketch variation with time t of the velocity v for the car.

Reference: Past Exam Paper – November 2011 Paper 22 Q1

Solution 629:

(a)

{Average velocity = gradient (which is constant here)}

Average velocity = 540 / 30 = 18ms^-1

(b)

Velocity of car is zero at time t = 0s  

{The graph is a horizontal line there – that is, gradient = 0 at t = 0s.}

Positive value and a horizontal line for time t = 5s to t =35s

[Gradient is constant and equal to 18ms^-1]

Line / curve through v = 0 to a negative velocity

[v = 0 (gradient = 0) at about t = 45s]

A negative horizontal line from t = 53s with magnitude less than positive value and horizontal line to time t =100s

[The car is moving in the opposite direction.

Displacement s = 590m at t = 53s. So, (considering points (53, 590) and (100, 0)) gradient = (0-590) / (100-53) = -12.6ms^-1]