# 9702 June 2014 Paper 22 Worked Solutions | A-Level Physics

## Paper 22

__Question 1__**(a)**

Show SІ base units of power are kg m

^{2}s^{–3}:
Power = work done / time OR energy / time

EITHER Units of force: kg ms

^{-2}
OR Units of kinetic energy (½mv

^{2}): kg (ms^{-1})^{2}
(Unit distance: m and (time)

^{-1}: s^{-1})
And hence Unit of power: kgms

^{-2}m s^{-1}= kgm^{2}s^{-3}**(b)**

Rate of flow of thermal energy Q /t
in material is given by Q / t = CAT / x where A is cross-sectional area of
material, T is temperature difference across thickness of material, x is
thickness of material, C is a constant.

SI base units of C:

Units of Q / t: kgm

^{2}s^{-3}
Units of area, A: m

^{2}distance, x: m and temperature, T: K
Correct substitution into C = Qx /
tAT

So, units of C: kg ms

^{-3}K^{-1}

__Question 2__
Coin is made in shape of thin
cylinder, as shown. Fig. shows measurements made in order to determine density
ρ of material used to make coin.

Quantity measurement Uncertainty

Mass 9.6g ±0.5g

Thickness 2.00mm ±0.01mm

Diameter 22.1mm ±0.1mm

**(a)**

Density ρ in
kgm

^{-3}:
ρ = m / V

Volume, V = (πd

^{2}/4) t = 7.67x10^{-7}m^{3}
ρ = (9.6x10

^{-3}) / [π({22.1/2}x10^{-3})^{2}(2.00x10^{-3})] = 12513kgm^{-3}**(b)**

(i)

Percentage uncertainty in ρ:

Δρ/ρ x 100% = [Δm/m + Δt/t + 2(Δd/d)]
x 100%

Δρ/ρ x 100% = 5.21% + 0.50% + 0.905%
= 6.6% (6.61%)

(ii)

ρ with its actual uncertainty:

ρ = 12500 ± 800 kgm

^{-3}

__Question 3__**(a)**

Newton’s first law of motion states
that a body / mass / object continues (at rest or) at constant / uniform
velocity unless acted on by a

__resultant__force.**(b)**

Box slides down slope, as shown.
Angle of slope to horizontal is 20°. Box has mass of 65 kg. Total resistive
force R acting on box is constant as it slides down slope.

(i)

Names and directions of other 2
forces acting on box:

The weight:

__vertically__down
The normal / reaction / contact
(force): perpendicular / normal

__to the slope__
(ii)

Variation with time t of velocity v
of box as it moves down slope is shown.

1.

Use data from Fig to show that
acceleration of box is 2.6ms

^{-2}:
Acceleration of box = gradient OR (v
– u) / t OR Δv/t

Acceleration of box = (6.0 – 0.8) /
(2.0 – 0.0) = 2.6ms

^{-2}
2.

Resultant force on box:

Resultant force, F = ma = 65 x 2.6 =
169N (allow to 2 or 3 sf)

3.

Resistive force R on box:

Component of weight along slope =
mgsinθ (= 218N)

218 – R = 169

Resistive force, R = 49N

__Question 4__**(a)**

Gravitational potential energy is
the energy of a

__mass__due to its position in a__gravitational field__
Kinetic energy is the energy (a mass
has) due to its motion / speed / velocity

**(b)**

Ball of mass 400g is thrown with initial
velocity of 30.0 m s

^{–1}at angle of 45.0° to horizontal, as shown. Air resistance negligible. Ball reaches maximum height H after a time of 2.16 s.
(i)

1.

Initial kinetic energy of ball:

Initial Kinetic energy = ½ mv

^{2}= ½ (0.4)(30)^{2}= 180J
2.

Maximum height H of ball:

s = (30sin45)(2.16) + ½ (-9.81)(2.16)

^{2}= 22.94 (22.9) m
OR

s = (30sin45)

^{2}/ (2 x 9.81) = 22.94 (22.9) m
3.

Gravitational potential energy of
ball at height H:

Gravitational potential energy, GPE
= mgh = 0.4(9.81)(22.88) = 89.8 (90) J

(ii)

1.

Kinetic energy of ball at maximum
height:

KE at maximum height = initial KE –
GPE = 180 – 90 = 90J

2.

Why kinetic energy of ball at
maximum height not zero:

The (horizontal) velocity is not
zero / (object) is still moving / answer explained in terms of conservation of
energy

__Question 5__**(a)**

Young modulus is defined as the
ratio of stress to strain

**(b)**

2 wires P and Q of same material and
same original length l

_{o}are fixed so that they hang vertically, as shown. Diameter of P is d and diameter of Q is 2d. Same force F is applied to lower end of each wire.
(i)

Ratio of stress in P

**to**stress in Q:
Stress = Force / Area

In P, stress = F / (πd

^{2}/4)
In Q, stress = F / (πd

^{2})
Ratio of stress in P

**to**stress in Q = 4 (or 4:1)
(ii)

Ratio of strain in P

**to**strain in Q:
Young modulus, E is the same for
both wires (as they are of the same material) [e.g. E

_{P}= E_{Q}]
Strain = stress / E

So, ratio of strain in P

**to**strain in Q = 4 (or 4:1)

__Question 6__
Battery is connected in series with
resistors X and Y, as shown. Resistance of X is constant. Resistance of Y is
6.0 Ω. Battery has electromotive force (e.m.f.) 24 V and zero internal
resistance. Variable resistor of resistance R is connected in parallel with X.

Current І from battery is changed by
varying R from 5.0 Ω to 20 Ω. Variation with R of І is shown.

**(a)**

Why potential difference (p.d.)
between points A and C is 24 V for all values of R:

There are no lost volts / energy in
the battery OR There are no lost volts / energy lost in the internal
resistance.

**(b)**

Use Fig to explain variation of p.d.
across resistor Y as R is increased:

The current / I decreases (as R
increases). So, the p.d. across resistor Y decreases (as R increases).

OR

The parallel resistance (of X and R)
increases. So, the p.d. across the parallel resistors increases and so, the
p.d. (across Y) decreases.

**(c)**

For R = 6.0 Ω,

(i)

Show that p.d. between points A and
B is 9.6V:

(from graph) For R = 6.0 Ω, Current
= 2.4A

So, p.d. across points A and B = 24 –
(2.4 x 6) = 9.6V

OR

Total resistance = (24V / 2.4A =)10 Ω

(parallel resistance = (10 – 6 =) 4Ω),
p.d. = 2.4 x (4/10) = 9.6V

(ii)

Resistance of X:

Resistance through AB, R = 9.6 / 2.4
= 4.0Ω

1/6 + 1/X = 1/4

So, resistance of X = 12Ω

OR

Current through R, I

_{R}= 9.6 / 6.0 = 1.6A
Current through X, I

_{X}= 2.4 – 1.6 = 0.8A
Resistance of X = (9.6 / 0.8 =) 12Ω

(iii)

Power provided by battery:

Power provided by battery = VI or EI
or V

^{2}/R or E^{2}/R or I^{2}/R
(total resistance through circuit = 6
+ [1/12 + 1/6]

^{-1}= 10)
Power = (24 x 2.4) or (24)

^{2}/10 or (2.4)^{2}(10) = 57.6W**(d)**

Explain qualitatively how power
provided by battery changes as resistance R is increased:

The power provided by the battery
decreases since the

__e.m.f.__is constant or power = 24 x current, and the current decreases OR the__e.m.f.__is constant or power = 24^{2}/ resistance, and the resistance increases.

__Question 7__**{Detailed explanations for this question is available as Solution 679 at Physics 9702 Doubts | Help Page 137 -**

*http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-137.html*}
please explain Q7(c)

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