# 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)

ReplyDeleteExplanations updated

DeleteQuestion 4 b ii kinetic energy should be zero at maximum height!!! Plz explain it

ReplyDeleteIt is not zero because the KE is due to both the horizontal velocity and the vertical velocity.

DeleteAt maximum height, the vertical velocity is zero, but NOT the horizontal velocity since it still continues to move. Thus, the KE is not zero