Monday, April 27, 2015

Physics 9702 Doubts | Help Page 124

  • Physics 9702 Doubts | Help Page 124



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 EK of an object of mass m is related to its momentum p by the expression
EK = 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, EK = ½mv2     and momentum, p = mv          combined to give
{mv2 = (mv)v = pv = p (p/m) = p2/m}
EK = p2 / 2m 

2.
The mass m of trolley B is smaller than that of trolley A. The kinetic energy of trolley B {EK = 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 EK = 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 EK depends on v2 (EK = 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:
(a) The tesla is the unit used when a (uniform magnetic) flux is normal to a long (straight) wire carrying a current of 1 A, creates a force per unit length of 1 N m–1.

(b)
(i) Magnetic flux density B (= μ0n I) = (4π × 10–7) × (1.5 × 103) × 3.5 = 6.6 × 10–3 T

(ii) Flux linkage (= BA × n) = (6.6 × 10–3) × (28 × 10–4) × 160 = 3.0 × 10–3 Wb

(c)
(i) Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to rate of change of (magnetic) flux (linkage).

(ii)
{To reverse the current in direction, the current must first be reduced to zero, and then increased to the same value in the opposite direction. So, the flux linkage calculated above should be considered twice.}
Average e.m.f. = (2 × 3.0 × 10–3) / 0.80 = 7.4 × 10–3 V










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]




2 comments:

  1. Here are the remaining questions for now:

    22/M/J/10 Q.5(a)(ii), Q.7(b)(ii)2.

    23/M/J/10 Q.7(b)

    21/O/N/10 Q.3(c)(ii),Q.7(b)(ii)2.

    22/O/N/10 Q.4(b)(ii)

    21/O/N/11 Q.5(b),Q.7(b)(ii),(c)

    ReplyDelete
    Replies
    1. For 22/M/J/10 Q.5(a)(ii), go to
      http://physics-ref.blogspot.com/2014/10/9702-june-2010-paper-22-worked.html

      Delete

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