Wednesday, October 29, 2014

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

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

Paper 21

Question 1
Velocity is defined as EITHER the rate of change of displacement OR the (change in) displacement / time (taken)

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

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,
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

Magnitude of resultant force on car:
Acceleration, a = (gradient =) (18 – 32) / 2.5 [= - 5.6]
Resultant force, F = ma = 1500 x (–)5.6 = (–) 8400N

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

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

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.
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)

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 FA on plank at end A and vertical force FB on plank at end B. As man moves along plank, plank is always in equilibrium.
Why sum of forces FA and FB is constant no matter where man stands on plank:
Since the resultant force is zero, weight of the plank + weight of the man = FA + FB           OR 200 (N) + 880 (N) = FA + FB

Man stands distance x = 0.50 m from end A. Use principle of moments to calculate magnitude of FB:
Principle of moments: Clockwise moments = Anticlockwise moments
(Anticlockwise moments: ) FB x 5.0
(Clockwise moments: ) 880(0.5) + 200(2.5)
So, FB = (440 + 500) / 5.0 = 188N

Variation with distance x of force FA is shown. On axes of Fig, sketch graph to show variation with x of force FB:
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) = FA + FB. So, FA + FB = 1080N.

From graph in (b), at x = 0m, FA = 980N. So, at x = 0m, FB = 1080 – 980 = 100N. At x = 5.0m, FA = 100N, so FB = 1080 – 100 = 980N.

Additionally, since FA and FB are related by a linear equation (the equation does not contain any power on FA or FB), 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.
Show that kinetic energy of ball as it makes contact with spring is 0.16 J:
Kinetic energy of ball = ½ mv2 = ½ (0.040) (2.8)2 = 0.157J or 0.16J

Variation of force F acting on spring with compression x of spring is shown. Ball produces maximum compression XB when it comes to rest. Spring has spring constant of 800 N m−1. Use Fig to
Compression XB:
Spring constant, k = F / x        or Force, F = kx
Compression, XB = 14 / 800 = 0.0175m

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 ½kx2      OR ½Fx
The energy stored = 0.1225J, which is less than the kinetic energy (of 0.16J)

Question 5
For a wave on surface of water:
Displacement and amplitude:
The displacement is the distance from the equilibrium position / undisturbed position / midpoint / rest position
The amplitude is the maximum displacement

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

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.
For waves,
Amplitude = ([15 – 12] / 2 =) 1.5mm

Wavelength = 25 / 6 = 4.2cm or 4.2x10-2m

Time period of oscillations of vibrator:
Speed, v = λ / T           or v = f λ         and T = 1 / f
Time period, T = 4.2 / 7.5 = 0.56s

Explain whether waves on surface of water shown are
Progressive or stationary:
The waves on the surface of the water are progressive since wavefront / crest moving / energy is transferred by the waves

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
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

Battery of e.m.f. 12 V and internal resistance 0.50 Ω 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.
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

Resistance of 1 lamp:
Resistance, R = V2 / P (or V = IR and P = VI) = 122 / 48 = 3.0Ω

Current in battery:
(Resistance of parallel combination of lamps, RT = [1/3 + 1/3]-1 = 1.5Ω)
Current, I = E / (RT + r) = 12 / 2.0 = 6.0A

Power dissipated in 1 lamp:
(Current in one lamp = 6/2 = 3A)
Power of each lamp = I2R = (3.0)2 x 3.0 = 27W

A 3rd 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
α-particle: a helium nucleus

β-particle: an electron

γ-radiation: an electromagnetic radiation / wave / ray or photon

Changes to proton number and nucleon number of nucleus when emission occurs of
Atomic number / proton number: Z – 2
Nucleon / mass number: A – 4

Atomic number / proton number: Z + 1
Nucleon / mass number: no change in A

Atomic number / proton number: no change
Nucleon / mass number: no change


  1. This comment has been removed by the author.

  2. Question 2 B
    Why is road friction ignored?

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

    2. So instead of saying air resistance is negligible if they said that it was a smooth plane.
      No information about air resistance is given. We would have neglected air resistance then?
      Is that how we go about?

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

      However, 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 ...

    4. 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.

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

      Pleasee answer. Thankss :)

  3. A random question.
    Will 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.

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

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

  4. Just one LAST thing

    Frictionless 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!!!! ^.^

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

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

    2. This comment has been removed by the author.

    3. Thats my question
      When 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:)

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

      Car moving down a frictionless surface:
      Chemical --> Kinetic + Work against air resistance

  5. This comment has been removed by the author.


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