# 9702 June 2014 Paper 21 Worked Solutions | A-Level Physics

## Paper 21

__Question 1__**(a)**

(i)

Velocity is defined as EITHER the
rate of change of displacement OR the (change in) displacement / time (taken)

(ii)

Speed and velocity:

Speed has magnitude only while
velocity has both a magnitude and a direction

**(b)**

Car of mass 1500 kg moves along
straight, horizontal road. Variation with time t of velocity v for car is shown.
Brakes of car are applied from t = 1.0 s to t = 3.5 s. For time when brakes are
applied,

(i)

Distance moved by car:

Idea of (distance =) area under the
graph / use of s = [(u+ v) / 2] t

Distance, s = [(18 + 32) / 2] x 2.5
= 62.5m

(ii)

Magnitude of resultant force on car:

Acceleration, a = (gradient =) (18 –
32) / 2.5 [= - 5.6]

Resultant force, F = ma = 1500 x
(–)5.6 = (–) 8400N

**(c)**

Direction of motion of car in (b) at
time t = 2.0 s is shown. On Fig, show with arrows the directions of
acceleration (label arrow A) and resultant force (label arrow F):

The arrow labeled A and the arrow
labeled F are both to the left

{The velocity of the car
is decreasing, so the acceleration opposes the motion. [If acceleration was in
the direction of motion, the velocity would increase] The resultant force and
the acceleration are always in the same direction: F = ma.}

__Question 2__**(a)**

(i)

Power is defined as the work (done)
/ time (taken)

(ii)

Use definition in (i) to show that
power may also be expressed as product of force and velocity:

Work done = Force x Displacement (in
direction of the force)

So, power = Force x Displacement /
time (taken) = Force x Velocity

**(b)**

Lorry moves up road that is inclined
at 9.0° to horizontal, as shown. Lorry has mass 2500 kg and is travelling at
constant speed of 8.5 m s

^{−1}. Force due to air resistance is negligible.
(i)

Useful power from engine to move
lorry up the road:

Weight = mg

Power, P = Fv = [2500(9.81)(sin 9)]
x 8.5 = 33 (32.6kW)

(ii)

2 reasons why rate of change of
potential energy of lorry is equal to power calculated in (i):

There is no gain or loss of kinetic
energy

There is no work (done) against air
resistance

__Question 3__
Uniform plank AB of length 5.0 m and
weight 200 N is placed across stream, as shown. Man of weight 880 N stands
distance x from end A. Ground exerts vertical force F

_{A}on plank at end A and vertical force F_{B}on plank at end B. As man moves along plank, plank is always in equilibrium.**(a)**

(i)

Why sum of forces F

_{A}and F_{B}is constant no matter where man stands on plank:
Since the resultant force is zero,
weight of the plank + weight of the man = F

_{A}+ F_{B}OR 200 (N) + 880 (N) = F_{A}+ F_{B}
(ii)

Man stands distance x = 0.50 m from
end A. Use principle of moments to calculate magnitude of F

_{B}:
Principle of moments: Clockwise
moments = Anticlockwise moments

(Anticlockwise moments: ) F

_{B}x 5.0
(Clockwise moments: ) 880(0.5) +
200(2.5)

So, F

_{B}= (440 + 500) / 5.0 = 188N**(b)**

Variation with distance x of force F

_{A}is shown. On axes of Fig, sketch graph to show variation with x of force F_{B}:
Graph is a straight line with
positive gradient with the start point (0, 100) and
finish point (5, 980)

{From (a)(i), it was seen
that 200 (N) + 880 (N) = F

_{A}+ F_{B}. So, F_{A}+ F_{B}= 1080N.
From graph in (b), at x =
0m, F

_{A}= 980N. So, at x = 0m, F_{B}= 1080 – 980 = 100N. At x = 5.0m, F_{A}= 100N, so F_{B}= 1080 – 100 = 980N.
Additionally, since F

_{A}and F_{B}are related by a linear equation (the equation does not contain any power on F_{A}or F_{B}), the graph is a line}

__Question 4__
Metal ball of mass 40 g falls
vertically onto spring, as shown. Spring is supported and stands vertically.
Ball has speed of 2.8 m s

^{−1}as it makes contact with spring. Ball is brought to rest as spring is compressed.**(a)**

Show that kinetic energy of ball as
it makes contact with spring is 0.16 J:

Kinetic energy of ball = ½ mv

^{2}= ½ (0.040) (2.8)^{2}= 0.157J or 0.16J**(b)**

Variation of force F acting on
spring with compression x of spring is shown. Ball produces maximum compression
X

_{B}when it comes to rest. Spring has spring constant of 800 N m^{−1}. Use Fig to
(i)

Compression X

_{B}:
Spring constant, k = F / x or Force, F = kx

Compression, X

_{B}= 14 / 800 = 0.0175m
(ii)

Show that not all kinetic energy in
(a) is converted into elastic potential energy in spring:

Area under the graph = elastic
potential energy stored OR ½kx

^{2}OR ½Fx
The energy stored = 0.1225J, which
is less than the kinetic energy (of 0.16J)

__Question 5__**(a)**

For a wave on surface of water:

(i)

Displacement and amplitude:

The displacement is the distance
from the equilibrium position / undisturbed position / midpoint / rest position

The amplitude is the maximum
displacement

(ii)

Frequency and time period:

The frequency is the number of
wavefronts / crests passing a point per unit time / number of oscillations per
unit time

The time period is the time between
adjacent wavefronts OR time for one oscillation

**(b)**

Fig represents waves on surface of
water in ripple tank at 1 particular instant of time. Vibrator moves surface of
water to produce waves of frequency f. Speed of waves is 7.5 cm s

^{−1}. Where waves travel on water surface, maximum depth of water is 15 mm and minimum depth is 12 mm.
(i)

For waves,

1.

Amplitude:

Amplitude = ([15 – 12] / 2 =) 1.5mm

2.

Wavelength:

Wavelength = 25 / 6 = 4.2cm or
4.2x10

^{-2}m
(ii)

Time period of oscillations of
vibrator:

Speed, v = Î» / T or
v = f Î»

__and__T = 1 / f
Time period, T = 4.2 / 7.5 = 0.56s

**(c)**

Explain whether waves on surface of
water shown are

(i)

Progressive or stationary:

The waves on the surface of the
water are progressive since wavefront / crest moving / energy is transferred by
the waves

(ii)

Transverse or longitudinal:

The waves on the surface of the
water are transverse since the vibration is perpendicular to the direction of
energy transfer / wave velocity OR travel

__of the wave / wavefronts__

__Question 6__**(a)**

Difference between electromotive
force (e.m.f.) and potential difference (p.d.):

The electromotive force (e.m.f.) is
the energy converted from chemical / other forms to electrical per unit charge

The potential difference (p.d.) is
the energy converted from electrical to other forms per unit charge

**(b)**

Battery of e.m.f. 12 V and internal
resistance 0.50 Î© connected to 2 identical lamps, as shown. Each lamp has
constant resistance. Power rating of each lamp is 48 W when connected across a
p.d. of 12 V.

(i)

Why power dissipated in each lamp
not 48 W when connected as shown in Fig:

The potential difference across the
lamp is

__less than__12V OR there are lost volts / power / energy in the battery / internal resistance
(ii)

Resistance of 1 lamp:

Resistance, R = V

^{2 }/ P (or V = IR__and__P = VI) = 12^{2}/ 48 = 3.0Î©
(iii)

Current in battery:

(Resistance of parallel combination
of lamps, R

_{T}= [1/3 + 1/3]^{-1}= 1.5Î©)
Current, I = E / (R

_{T}+ r) = 12 / 2.0 = 6.0A
(iv)

Power dissipated in 1 lamp:

(Current in one lamp = 6/2 = 3A)

Power of each lamp = I

^{2}R = (3.0)^{2}x 3.0 = 27W**(c)**

A 3

^{rd}identical lamp is placed in parallel with battery in circuit of Fig. Explain effect on terminal p.d. of battery:
There is less resistance (in the
circuit) / more current. So, there is more lost volts / less potential difference
across the battery.

__Question 7__**(a)**

Î±-particle: a helium nucleus

Î²-particle: an electron

Î³-radiation: an

__electromagnetic__radiation / wave / ray or photon**(b)**

Changes to proton number and nucleon
number of nucleus when emission occurs of

(i)

Î±-particle:

Atomic number / proton number: Z – 2

Nucleon / mass number: A – 4

(ii)

Î²-particle:

Atomic number / proton number: Z + 1

Nucleon / mass number: no change in
A

(iii)

Î³-radiation:

Atomic number / proton number: no
change

Nucleon / mass number: no change

This comment has been removed by the author.

ReplyDeleteQuestion 2 B

ReplyDeleteWhy is road friction ignored?

No information has been given about friction in the question. So, it has be neglected.

DeleteSo instead of saying air resistance is negligible if they said that it was a smooth plane.

DeleteNo information about air resistance is given. We would have neglected air resistance then?

Is that how we go about?

Yes. We cannot account for it if we don't have enough data.

DeleteHowever, there can be a question where you need to explain why ....

and the answer could be that air resistance needs to be taken into account ...

So either data about friction/air resistance will be given in the question or it will be calculated in the question in one of the parts of the question.

DeleteIf its not asked to be calculated or given in question we neglect it.

Correct?

Pleasee answer. Thankss :)

Yes.

DeleteThanks a bunch.

ReplyDeleteA random question.

ReplyDeleteWill be glad if youll help. Thanks. :)

If they ask to answer qualitatively about energy conversions of a ball moving down a ramp.

Will they always mention if friction is present or not?

Because if it is GPE will be converted to KE and heat

If its not GPE will only be converted to KE.

Will they always make clear if its with friction or without friction?

Pleasee replyy. This question really has been confusing me.

It depends. Sometimes they may mentioned the slope to be frictionless, so you don't include the heat due to friction.

DeleteBut if not, I believe you should account for friction.

Just one LAST thing

ReplyDeleteFrictionless means no road friction but there might be air resistance.

So if they state frictionless

I should TOTALLY ignore energy loss as heat?

THANKYOU SO MUCH!!!! ^.^

Assuming heat loss is only due to friction, then yes.

DeleteBut if air resistance is being considered, you need to account for the work done against it.

This comment has been removed by the author.

DeleteThats my question

DeleteWhen do we know air resistance is being considered in a QUALITATIVE question.

If a car moves down a frictionless surface . Write the energy conversions.

How are we supposed to guess it is or is not??

Pleaseeee replyyy. Im anxiously waiting. Thankyou:)

In a qualitative question, air resistance is always considered, unless stated otherwise in the question.

DeleteCar moving down a frictionless surface:

Chemical --> Kinetic + Work against air resistance

This comment has been removed by the author.

ReplyDeleteIn question 2b i should not we use fvcos(9)??

ReplyDeleteNo, we should consider the component along the slope.

DeleteFor Question 6c) How does a lower resistance in circuit mean there are more less volts? Please help :)

ReplyDeletea lower resistance means that the current flowing in the circuit is greater.

DeleteLoss volt = Ir

So, this increases