# Physics 9702 Doubts | Help Page 5

__Question 29: [Gravitation]__**(a)**Define gravitational field strength.

**(b)**Spherical planet has diameter 1.2 × 10

^{4}km. Gravitational field strength at surface of planet is 8.6 N kg

^{–1}. Planet may be assumed to be isolated in space and to have its mass concentrated at its centre. Calculate mass of planet.

**(c)**Gravitational potential at point X above surface of planet in (b) is – 5.3 × 10

^{7}J kg

^{–1}. For point Y above surface of planet, gravitational potential is – 6.8 × 10

^{7}J kg

^{–1}.

(i) State, with reason, whether point
X or point Y is nearer to planet.

(ii) Rock falls radially from rest
towards planet from one point to other. Calculate final speed of rock.

**Reference:**

*Past Exam Paper – June 2009 Paper 4 Q1*

__Solution 29:__**(a)**

Gravitational field strength is
defined as the (gravitational) force per unit mass

**(b)**

Gravitational field strength, g = GM
/ R

^{2}
{Radius = diameter / 2.
Also, distance should be converted to metre}

8.6 x (0.6x10

^{7})^{2}= M x (6.67x10^{-11})
Mass of planet, M = 4.6x10

^{24}kg**(c)**

(i)

EITHER The potential decreases as
the distance from the planet decreases {This is similar
to potential energy (mgh). Here the planet is the surface. So, as the distance (height)
from the planet increases, the potential energy increases and vice versa.}

OR The potential is zero at infinity
and point X is closer to zero {hence close to infinity}

OR The potential is proportional to –
1/r and point Y is more negative {the more negative the
potential, the smaller is r}

So, the point Y is closer to the
planet

(ii)

{Change in potential
energy = change in kinetic energy. Change in potential energy = mΔϕ where m is the mass of the rock.
Since the rock is at rest, initial velocity = 0. So, change in kinetic energy =
½mv

^{2}. m can be cancelled on both sides, leading to}
The idea of (change in potential,) Δϕ
= ½ v

^{2}
(6.8 – 5.3)x10

^{7}= ½ v^{2}
Final speed, v = 5.5x10

^{3}ms^{-1}

__Question 30: [Circular motion & Simple Harmonic motion]__
Vertical peg is attached to edge of
horizontal disc of radius r, as shown.

Disc rotates at constant angular
speed ω. Horizontal beam of parallel light produces shadow of peg on screen, as
shown.

At time zero, peg is at P, producing
shadow on screen at S. At time t, disc has rotated through angle θ. Peg is now at
R, producing shadow at Q.

**(a)**Determine,

(i) in terms of ω and t, the angle θ

(ii) in terms of ω, t and r, the
distance SQ.

**(b)**Use answer to (a)(ii) to show that shadow on screen performs simple harmonic motion:

**(c)**Disc has radius r of 12 cm and is rotating with angular speed ω of 4.7 rad s

^{–1}. Determine, for shadow on screen,

(i) Frequency of oscillation

(ii) Its maximum speed

**Reference:**

*Past Exam Paper – June 2009 Paper 4 Q4*

__Solution 30:__**(a)**

(i) Angle,
θ = ωt

(ii) Distance SQ = r sin ωt

**(b)**

This (r sin ωt) is
the solution of the equation a = – ω

^{2}x. a = – ω^{2}x is the (defining) equation of simple harmonic motion**(c)**

(i) Frequency, f = ω / 2π = 4.7 / 2π = 0.75Hz

(ii) Speed, v = rω = 4.7 x 12 = 56cms

^{-1}

__Question 31: [Hooke’s law]__**(a)**Metal wire has unstretched length L and area of cross-section A. When wire supports load F, wire extends by amount ΔL. Wire obeys Hooke’s law. Write down expressions, in terms of L, A, F and ΔL, for

(i) Applied stress.

(ii) Tensile strain in wire.

(iii) Young modulus of material of
wire.

**(b)**Steel wire of uniform cross-sectional area 7.9 × 10

^{–7}m

^{2}is heated to temperature of 650 K. It is then clamped between two rigid supports, as shown.

Wire is straight but not under
tension and length between supports is 0.62 m. Wire is then allowed to cool to
300 K. When wire is allowed to contract freely, a 1.00 m length of wire
decreases in length by 0.012 mm for every 1 K decrease in temperature.

(i) Show that change in length of
wire, if it were allowed to contract as it cools from 650 K to 300 K, would be
2.6 mm.

(ii) Young modulus of steel is 2.0 ×
10

^{11}Pa. Calculate tension in wire at 300 K, assuming that wire obeys Hooke’s law.
(iii) Ultimate tensile stress of
steel is 250MPa. Use this information and answer in (ii) to suggest whether
wire will, in practice, break as it cools.

**Reference:**

*Past Exam Paper – November 2004 Paper 2 Q5*

__Solution 31:__**(a)**

(i) Applied stress = F / A

(ii) Tensile strain in wire = ΔL / L

(iii) Young modulus (= stress /
strain) = FL / AΔL

**(b)**

(i)

{Decrease in temperature =
650 – 300 = 350K

For a 1.00m length of
wire, (since the wire decreases in length by 0.012mm for every 1K decrease in
temperature) this would correspond to a decrease of 0.012 x 350.

But this is not the same
for a 0.62m wire. This can be worked out by proportion. For a 1.oom, wire, the
overall decrease = 0.012 x 350. So, for a 0.62m wire, the corresponding
decrease = 0.62 (0.012 x 350)}

Change in length, ΔL = 0.012 x 0.62
x 350 = 2.6mm

(ii)

{Young modulus = stress /
strain = FL / AΔL}

2.0x10

^{11}= F (0.062) / [(7.9x10^{-7}) (2.6x10^{-3})]
Tension in the wire, F = 660N

(iii)

{This question can be
answered in different ways. Below, if you are considering the EITHER method,
then you should consider it on each line and if you are considering the OR
method, then it continues after the OR on each line

In the EITHER case, we try
to determine the stress when the wire is cold at 300K (assuming it does not
snap) and compare it with the ultimate tensile stress. If it is larger, then
the wire snaps since the ultimate tensile stress is the maximum stress of the
wire before it breaks. In the OR case, we try to calculate the equivalent tension
at ultimate tensile stress and compare it with the tension obtained above. If
it is less, then the wire snaps since the maximum tension that can occur in the
wire is that at ultimate tensile stress}

{Force = tension. EITHER:
Stress = tension / area = 660 / (7.9 x 10

^{-7}). OR: Tension = stress x area = 250MPa x (7.9x10^{-7})}
EITHER when cold, stress = 660 / (7.9
x 10

^{-7}) = 840 MPa OR Tension at ultimate tensile stress = 198 N
EITHER This (the stress when cold) is
greater than the ultimate tensile stress of steel OR The tension at ultimate tensile stress is less then
tension in the wire (F = 660N) given in part b(ii)

So, the wire will snap.

__Question 32: [Current of Electricity]__
Fig shows variation with applied
potential difference V of current I in electrical component C.

**(a)**

(i) State, with reason, whether
resistance of component C increases or decreases with increasing potential
difference.

(ii) Determine resistance of
component C at potential difference of 4.0 V.

**(b)**Component C is connected in parallel with resistor R of resistance 1500 Ω and battery of e.m.f. E and negligible internal resistance, as shown.

(i) On Fig, draw line to show
variation with potential difference V of current I in resistor R.

(ii) Hence, or otherwise, use Fig to
determine current in battery for e.m.f. of 2.0 V.

**(c)**Resistor R of resistance 1500 Ω and component C are now connected in series across a supply of e.m.f. 7.0 V and negligible internal resistance. Using information from Fig, state and explain which component, R or C, will dissipate thermal energy at greater rate.

**Reference:**

*Past Exam Paper – November 2004 Paper Q6*

__Solution 32:__**(a)**

(i)

The resistance is given by the ratio
of V / I (at a point)

EITHER The gradient increases OR
Current, I increases more rapidly than V

{So, the resistance decreases.

Note that the resistance
is not equal to the gradient of the line, but to the co-ordinates of the point
on the line}

(ii)

{From graph, when p.d. =
4.0V} Current = 2.00mA

Resistance of component C (= V / I)
= 2000Ω

**(b)**

(i)

{As deduced from the
graph, the resistance of component C decreases as p.d. increases. This causes
the equivalent resistance of the parallel combination to decrease as the p.d.
increases. So, current from battery increases. From Kirchhoff’s law, there is
the same p.d. across both C and R. So, as p.d. increases, current across R increases
linearly. The ratio of V / I of points on the line should always give R = 1500Ω. So, at p.d. = 6.0, I = 4.0mA}

The line drawn is a straight line
starting from the origin and passing through point (6.0V, 4.0mA)

(ii)

{Component C and resistor
R are connected in parallel, so the currents flowing through them are
different. Consider the currents on the resistor R and component C at p.d. =
2.0V from the graph}

The individual currents are 0.75mA {in component C} and 1.33mA {in
resistor R}

Current in the battery (= 0.75 +
1.33) = 2.1mA

**(c)**

{Rate of dissipation of
thermal energy = power}

The same current would flow in both
resistor R and in component C. The potential difference across component C is
larger than that across resistor R {For series
connection, the p.d. across a component is proportional to its resistance}.
Since power = VI, the power is greater in C.

__Question 33: [Dynamics > Momentum]__
A steel ball of mass 250kg is
suspended from the jib of a crane, as illustrated in Fig 3.1.

In order to demolish a wall, the
ball is pulled away from the wall and then released. The ball swings down and
hits the wall.

The variation with time t of the
speed v of the ball is shown in Fig 3.2.

**(a)**Using Fig 3.2, determine

(i)
the magnitude of the acceleration of the ball at time t = 0.8 s.

(ii) the distance moved by the ball
before it hits the wall, that is, from time t = 0 s to t = 1.6 s.

**(b)**Calculate the magnitude of

(i) the change in momentum of the
ball during its collision with the wall

(ii) the average force exerted on
the wall during the collision.

**(c)**When the ball hits the wall, 15% of the kinetic energy of the ball is converted to thermal energy in the ball. Calculate the mean temperature rise of the ball. The specific heat capacity of steel is 450 Jkg

^{-1}K

^{-1}.

**Reference:**

*D02/P2/Q3*

__Solution 33:__**(a)**

(i)

{First,
let’s try interpreting the graph. The graph is that of velocity against time.
So, gradient represent acceleration and area under graph represents the
distance travelled. The ball starts from rest (speed = 0) and its speed
increases until time t = 1.6s. The gradient decreases as time increases, so the
acceleration decreases until it is zero at t = 1.6s. At t = 1.6s to t = 1.8s,
there is a sharp decrease in velocity until it becomes zero. This is the time
of the collision. The ball comes to rest after hitting the wall}

Magnitude of acceleration of ball
(at time = 0.8s) = gradient of graph at t = 0.8s

{First draw a tangent at t
= 0.8s. The line obtained should also pass (depending on how the line is drawn)
through points (0.35, 2.6) and (1.20, 5.2). You can use these points to calculate
the gradient}

Gradient = (5.2 – 2.6) / (1.20 –
0.35) = 3.06ms

^{-2}[= 3.1ms^{-2}]
(ii)

Distance moved by ball before it
hits wall = Area under graph between t = 0.0s and t = 1.6s

{We need to count the
number of squares. There are different methods to do this. You could actually
count the number of squares and multiply it by the area of the smallest square.
On x-axis, the smallest division corresponds to 0.05s. On y-axis, the smallest
division corresponds to 0.2ms

^{-1}. So, the area of the smallest square = 0.05 x 0.2 = 0.01m.
Alternatively, you could
use trapezium rule to obtain the area directly. Choose the method which is best
and quickest for you.}

The distance should be about 5.56m

[note
that others obtained the distance as 4.9m. I obtained it close to 5.56m]

(b)

(i)

Mass of ball = 250kg

Speed of ball before collision, v

_{i}= 5.2ms^{-1}
Speed of ball after collision, v

_{f}= 0ms^{-1}
Change in momentum = mΔv = m(v

_{f}– v_{i}) = 250 (0 – 5.2) = (–) 1300kgsm^{-1}
{The magnitude may be
given alone}

(ii)

{From Newton’s 2

^{nd}law: Force is equal to the rate of change of momentum}
Magnitude of average force on

__wall__during collision = magnitude of average force on__ball__
{This is from Newton’s 3

^{rd}law. The 2 forces are equal in magnitude and opposite in direction. The momentum calculated above is actually for the ball}
Average force = Δp / Δt = 1300 /
(1.80 – 1.60) = 6500N

**(c)**

Kinetic energy when ball hits the
wall = ½mv

^{2}= ½ (250) (5.2^{2}) J
Thermal energy = mcΔθ

where m is the mass of the steel ball,
c is the specific capacity of steel and Δθ is the change in temperature

So, mcΔθ = 0.15 (½mv

^{2})
Δθ = 0.15 (½ v

^{2}) / c = 0.15 [0.5 x 5.2^{2}] / 450 = 4.51x10^{-3}K
Mean temperature rise = 4.51x10

^{-3}K

__Question 34: [Electromagnetism]__**(a)**Define magnetic flux density.

**(b)**

Flat coil consists of N turns of wire and has area A. Coil is placed so
that its plane is at angle θ to uniform magnetic field of flux density B, as
shown.

Using symbols A, B, N and θ and making reference to magnetic flux in
coil, derive expression for magnetic flux linkage through coil.

**(c)**

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

(ii) Magnetic flux density B in coil is now made to vary with time t as
shown. Sketch variation with time t of e.m.f. E induced in coil.

**Reference:**

*Past Exam Paper – November 2004 Paper 4 Q6*

__Solution 34:__**(a)**

Magnetic flux density is defined as being (numerically equal to) the
force per unit length on a straight conductor carrying unit current normal to
the field

**(b)**

Magnetic flux through the coil = BAsinθ

Magnetic flux linkage through the coil = BANsinθ

**(c)**

(i)

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

(ii)

{As stated by Faraday’s law of electromagnetic induction,
the (induced) e.m.f. is proportional to the rate of

**change**of flux (linkage). For the first section, from t = 0 to t = T, B is**constant**. So, no e.m.f. is induced.
From t = T to T = 2T, B increases at a constant rate (gradient
is constant). So, the e.m.f. induced is constant. (The direction of e.m.f. induced
can be taken arbitrarily.)

From t = 2T to t = 3T, B changes at a constant rate (gradient
is constant) but in the opposite direction. So, the e.m.f. is constant and is
induced in the opposite direction. Additionally, the change is greater than
from t= T to t = 2T, so the e.m.f. induced is greater}

The graph consists of 2 square sections in the correct positions {as explained above}, and is zero elsewhere. The pulses
are in opposite directions. The amplitude of the second pulse is about twice
the amplitude of the first.

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