• 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 lorry moves up a road that is inclined at 9.0° to the horizontal, as shown in Fig. 2.1.

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

A metal ball of mass 40 g falls vertically onto a spring, as shown in Fig. 4.1.

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^-2m

(ii)

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

(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 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.

(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 = V² / P (or V = IR and P = VI) = 12² / 48 = 3.0Ω

(iii)

Current in battery:

(Resistance of parallel combination of lamps, R_T = [1/3 + 1/3]^-1 = 1.5Ω)

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 = I²R = (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)

α-particle: a helium nucleus

β-particle: an electron

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

(b)

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

(i)

α-particle:

Atomic number / proton number: Z – 2

Nucleon / mass number: A – 4

(ii)

β-particle:

Atomic number / proton number: Z + 1

Nucleon / mass number: no change in A

(iii)

γ-radiation:

Atomic number / proton number: no change

Nucleon / mass number: no change