# 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 × 10

^{11}N m

^{–2}and for copper is 1.1 × 10

^{11}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 kx

^{2}
Energy = ½ (3)(0.25x10

^{-3}) = 3.8x10^{-4}J**(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 M

_{1}and M

_{2}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 V

_{1}and trolley B has speed V_{2}.
Write down

1. equation, based on momentum, to
relate V

_{1}and V_{2}
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}= p^{2}/ 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.

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. M

_{1}V_{1}= M_{2}V_{2}
2.

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

__E__= ½ M

_{1}V

_{1}

^{2}+ ½ M

_{2}V

_{2}

^{2}

(iii)

1.

Kinetic energy, E

_{K}= ½mv^{2}and momentum, p = mv combined to give
{mv

^{2}= (mv)v = pv = p (p/m) = p^{2}/m}
E

_{K}= p^{2}/ 2m
2.

The mass m of trolley B is smaller
than that of trolley A. The kinetic energy of trolley B {E

_{K}= p^{2}/ 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 [M_{1}V_{1}= M_{2}V_{2}]. Since E_{K}= p^{2}/ 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 v^{2}(E_{K}= ½mv^{2}). 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 cm

^{2}, 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 = Î¼

_{0}n I
where I is the current in the
solenoid, n is a constant equal to 1.5 × 10

^{3}m^{–1}and Î¼_{0}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 (= Î¼

_{0}n I) = (4Ï€ × 10^{–7}) × (1.5 × 10^{3}) × 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}]
Here are the remaining questions for now:

ReplyDelete22/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)

For 22/M/J/10 Q.5(a)(ii), go to

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