A Level Physics 9702 Paper 21, May/June 2023

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9702/21 · May/June 2023

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a(i)

Define power.

[1]

a(ii)

Use the definition of power to show that the SI base units of power are $kg m^{2} s^{- 3}$ .

[1]

The intensity $I$ of a sound wave moving through a gas is given by $I = f^{2} A^{2} v k$ where $f$ is the frequency of the wave, $A$ is the amplitude of the wave, $v$ is the speed of the wave and $k$ is a constant that depends on the gas. Determine the SI base units of $k$ .

[3]

A rigid uniform beam of weight $W$ is connected to a fixed support by a hinge, as shown in Fig. 2.1. A compressed spring exerts a total force of $8.2 N$ vertically upwards on the horizontal beam. A block of weight $0.30 N$ rests on the beam. The right-hand end of the beam is connected to the ground by a string at an angle of $30^{\circ}$ to the horizontal. The tension in the string is $4.8 N$ . The distances along the beam are shown in Fig. 2.1. The beam is in equilibrium. Assume that the hinge is frictionless.

a(i)

Show that the vertical component of the tension in the string is $2.4 N$ .

[1]

a(ii)

By taking moments about the hinge, determine the weight $W$ of the beam.

[3]

a(iii)

Calculate the horizontal component of the force exerted on the beam by the hinge.

[1]

The spring obeys Hooke’s law and has an elastic potential energy of $0.32 J$ . Calculate the compression of the spring.

[2]

c(i)

The string is cut so that the spring extends upwards. This causes the beam to rotate and launch the block into the air. The block reaches its maximum height and then falls back to the ground. Fig. 2.2 shows part of the path of the block in the air shortly before it hits the horizontal ground. The block is at a height of $0.090 m$ above the ground when it passes through point A. The block has a kinetic energy of $0.044 J$ when it hits the ground at point B. Air resistance is negligible. Calculate the decrease in the gravitational potential energy of the block for its movement from A to B.

[2]

c(ii)

Use your answer in (c)(i) and conservation of energy to determine the speed of the block at point A.

[3]

c(iii)

By reference to the force on the block, explain why the horizontal component of the velocity of the block remains constant as it moves from A to B.

[1]

c(iv)

The block passes through point A at time $t_{A}$ and arrives at point B at time $t_{B}$ . On Fig. 2.3, sketch a graph to show the variation of the magnitude of the vertical component $v_{Y}$ of the velocity of the block with time $t$ from $t = t_{A}$ to $t = t_{B}$ . Numerical values of $v_{Y}$ are not required.

[1]

A block is pulled in a straight line along a rough horizontal surface by a varying force $X$ , as shown in Fig. 3.1. Air resistance is negligible. Assume that the frictional force exerted on the block by the surface is constant and has magnitude $2.0 N$ . The variation with time $t$ of the momentum $p$ of the block is shown in Fig. 3.2.

State Newton’s second law of motion.

[1]

b(i)

Use Fig. 3.2 to determine, for the block at time $t = 2.0 s$ , the magnitude of: the resultant force on the block

[1]

b(ii)

Use Fig. 3.2 to determine, for the block at time $t = 2.0 s$ , the magnitude of: the force $X$ .

[1]

On Fig. 3.3, sketch a graph to show the variation of force $X$ with time $t$ from $t = 0$ to $t = 6.0 s$ .

[3]

A beaker in air contains a liquid. The base of the beaker is in contact with the liquid and has area $A$ , as shown in Fig. 4.1. The liquid has density $ρ$ and fills the beaker to a depth $h$ .

By using the definitions of pressure and density, show that $p = ρ g h$ where $p$ is the pressure due to the liquid that is exerted on the base of the beaker and $g$ is the acceleration of free fall.

[3]

Suggest why the equation in (a) does not give the total pressure on the base of the beaker.

[1]

Fig. 4.2 shows the variation of the total pressure inside the liquid with depth $x$ below the surface. Determine the density of the liquid.

[2]

A solid cylinder is held stationary by a wire so that the base of the cylinder is level with the surface of the liquid, as shown in Fig. 4.3. The cylinder has length $4.0 \times 10^{- 2} m$ and cross-sectional area $3.7 \times 10^{- 4} m^{2}$ . The tension in the wire is $0.53 N$ . The cylinder is now lowered and then held stationary by the wire so that the top of the cylinder is level with the surface of the liquid. Calculate the new tension in the wire.

[2]

a(i)

An electromagnetic wave in a vacuum has a wavelength of $8.4 \times 10^{- 6} m$ . State the name of the principal region of the electromagnetic spectrum for the wave.

[1]

a(ii)

Calculate the frequency, in $THz$ , of the wave.

[2]

b(i)

An arrangement that uses a double slit to demonstrate the interference of light from a laser is shown in Fig. 5.1. The light from the laser has a wavelength of $6.2 \times 10^{- 7} m$ and is incident normally on the slits. The separation of the two slits is $a$ . The slits and screen are parallel and separated by a distance of $2.8 m$ . An interference pattern of bright fringes and dark fringes is formed on the screen. The distance on the screen across 8 bright fringes is $22 mm$ , as illustrated in Fig. 5.2. The light waves emerging from the two slits are coherent. State what is meant by coherent.

[1]

b(ii)

Calculate the separation $a$ of the slits.

[3]

c(i)

Fringe P is the central bright fringe of the interference pattern in (b). Fringe Q and fringe R are the nearest dark fringe and the nearest bright fringe respectively to the right of fringe P, as shown in Fig. 5.2. Calculate the difference in the distances (the path difference) from each slit to the centre of fringe Q.

[1]

c(ii)

State the phase difference between the light waves meeting at the centre of fringe R.

[1]

A metal wire in a circuit has a length of $1.8 m$ and a cross-sectional area of $1.5 \times 10^{- 6} m^{2}$ . The total number of free electrons (charge carriers) in the wire is $2.3 \times 10^{23}$ . There is a current in the wire so that a charge of $172 C$ moves past a fixed point in the wire in a time of $2.5 minutes$ .

Show that the number density of the free electrons in the wire is $8.5 \times 10^{28} m^{- 3}$ .

[1]

Calculate the average drift speed of the free electrons.

[3]

A battery of electromotive force (e.m.f.) $9.6 V$ and negligible internal resistance is connected in series with two fixed resistors and a thermistor, as shown in Fig. 7.1. The fixed resistors have resistances of $3400 Ω$ and $5800 Ω$ . The reading on the voltmeter in the circuit is $6.0 V$ .

Calculate the current in the resistor of resistance $5800 Ω$ .

[2]

Calculate the resistance of the thermistor.

[2]

The initial energy stored in the battery is $2.6 \times 10^{4} J$ . Assume that the e.m.f. of the battery is constant. Determine the final energy stored in the battery after a charge of $330 C$ has moved through it.

[2]

d(i)

The environmental conditions change causing an increase in the resistance of the thermistor. State whether there is a decrease, increase or no change to: the temperature of the thermistor

[1]

d(ii)

the current in the thermistor

[1]

d(iii)

the potential difference across the thermistor.

[1]

An isolated stationary nucleus X decays by emitting an $α$ -particle to form a nucleus Y. Nucleus Y and nucleus Z are isotopes of the same element.

a(i)

By comparing the number of protons in each nucleus, state and explain whether the charge of nucleus Y is less than, greater than or the same as the charge of: nucleus Z

[1]

a(ii)

nucleus X.

[2]

Use the principle of conservation of momentum to explain why nucleus Y cannot be stationary immediately after the decay of nucleus X.

[2]

Mark scheme

9702/21 · May/June 2023

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1a(i) · 1 mark

B1
work done per unit time

1a(ii) · 1 mark

B1
( $P = W / t$ gives) units: $kg m^{2} s^{- 2} / s = kg m^{2} s^{- 3}$

1b · 3 marks

C1
( $I = P / A$ so) units of $I$ : $kg m^{2} s^{- 3} / m^{2}$ or $kg s^{- 3}$
C1
units of $f$ : $s^{- 1}$ and units of $A$ : $m$ and units of $v$ : $m s^{- 1}$
A1
units of $k$ : $kg s^{- 3} / [(s^{- 1})^{2} m^{2} m s^{- 1}]$ $= kg m^{- 3}$

2a(i) · 1 mark

A1
(component = ) $4.8 \sin 30^{\circ} = 2.4 (N)$

2a(ii) · 3 marks

C1
$(8.2 \times 0.50)$ or $(W \times 0.60)$ or $(0.30 \times 0.80)$ or $(2.4 \times 1.2)$
C1
$(8.2 \times 0.50) = (W \times 0.60) + (0.30 \times 0.80) + (2.4 \times 1.2)$
A1
$W = 1.6 N$

2a(iii) · 1 mark

A1
force $= 4.8 \cos 30^{\circ}$ $= 4.2 N$

2b · 2 marks

C1
$E = \frac{1}{2} F x$
A1
$0.32 = \frac{1}{2} \times 8.2 \times x$ $x = 0.078 m$

2c(i) · 2 marks

C1
$((Δ) E) = m g (Δ) h$ or $W (Δ) h$
A1
$= 0.30 \times 0.090$ $= 0.027 J$

2c(ii) · 3 marks

C1
$E = \frac{1}{2} m v^{2}$
C1
$E = 0.044 - 0.027 (= 0.017)$
A1
$v^{2} = (2 \times 0.017) / (0.30 / 9.81) (where v = speed at A)$ $v = 1.1 {m s}^{- 1}$

2c(iii) · 1 mark

B1
(gravitational / resultant) force / weight is vertical

2c(iv) · 1 mark

B1
straight line with positive gradient starting from non-zero value of $v_{Y}$ at time $t_{A}$ to a time $t_{B}$

3a · 1 mark

B1
(resultant) force (on an object) is proportional to / equal to the rate of change of momentum

3b(i) · 1 mark

A1
resultant force = e.g. $6.0 / 4.0$ $= 1.5 N$

3b(ii) · 1 mark

A1
force $X = 1.5 + 2.0$ $= 3.5 N$

3c · 3 marks

B1
from $t = 0$ to $t = 4.0 s$ : horizontal line at any non-zero value of $X$
B1
from $t = 0$ to $t = 4.0 s$ : horizontal line at $X = 3.5 N$
B1
from $t = 4.0 s$ to $t = 6.0 s$ : horizontal line at $X = 2.0 N$

4a · 3 marks

B1
$ρ = m / V$ or $ρ = m / A h$
B1
$p = F / A$ or $p = W / A$
B1
appropriate algebra leading to $p = ρ g h$ e.g. $p = ρ A h g / A$ or $ρ V g / A$ or $ρ V g / (V / h)$ and (so) $p = ρ g h$

4b · 1 mark

B1
there is atmospheric / air pressure

4c · 2 marks

C1
$Δ p = ρ g Δ h$ e.g. $(9.66 - 9.60) \times 10^{4} / 8.0 \times 10^{- 2} = ρ \times 9.81$
A1
$ρ = 760 - 770 {kg m}^{- 3}$

4d · 2 marks

C1
$F = ρ g V$ $= 760 \times 9.81 \times 3.7 \times 10^{- 4} \times 4.0 \times 10^{- 2} (= 0.11 N)$
A1
tension $= 0.53 - 0.11$ $= 0.42 N$

5a(i) · 1 mark

B1
infrared

5a(ii) · 2 marks

C1
$v = f λ$ or $c = f λ$
A1
$f = 3.0 \times 10^{8} / 8.4 \times 10^{- 6}$ $= 3.6 \times 10^{13} (Hz)$ $= 36 THz$

5b(i) · 1 mark

B1
constant phase difference (between the waves) (with time)

5b(ii) · 3 marks

C1
$λ = a x / D$
C1
$x = 22 / 8$ or $2.75 (mm)$ or $22 \times 10^{- 3} / 8$ or $2.75 \times 10^{- 3} (m)$
A1
$a = (6.2 \times 10^{- 7} \times 2.8) / (22 \times 10^{- 3} / 8)$ $= 6.3 \times 10^{- 4} m$

5c(i) · 1 mark

A1
difference in distances $= 6.2 \times 10^{- 7} / 2$ $= 3.1 \times 10^{- 7} m$

5c(ii) · 1 mark

A1
phase difference $= 360^{\circ}$

6a · 1 mark

A1
(number density =) $2.3 \times 10^{23} / (1.5 \times 10^{- 6} \times 1.8) = 8.5 \times 10^{28} {(m}^{- 3})$

6b · 3 marks

C1
$I = Q / t$ or $I = 172 / 2.5 \times 60$ or $I = 1.1 (5)$
C1
$I = n A v q$ $172 / (2.5 \times 60) = 1.5 \times 10^{- 6} \times 8.5 \times 10^{28} \times v \times 1.6 \times 10^{- 19}$
A1
$v = 5.6 \times 10^{- 5} {m s}^{- 1}$

7a · 2 marks

C1
$9.6 = 6.0 + (I \times 5800)$ or $3.6 = I \times 5800$
A1
$I = 6.2 \times 10^{- 4} A$

7b · 2 marks

C1
$9.6 = 6.2 \times 10^{- 4} \times (3400 + 5800 + R)$ or $6.0 = 6.2 \times 10^{- 4} \times (3400 + R)$
A1
$R = 6.3 \times 10^{3} Ω$

7c · 2 marks

C1
$(Δ E =) 9.6 \times 330 (= 3170 J)$
A1
final stored energy $= 2.6 \times 10^{4} - 3170$ $= 2.3 \times 10^{4} J$

7d(i) · 1 mark

B1
decrease

7d(ii) · 1 mark

B1
decrease

7d(iii) · 1 mark

B1
increase

8a(i) · 1 mark

B1
Y and Z have equal numbers of protons and (so) they have the same charge

8a(ii) · 2 marks

M1
Y has (two) fewer protons (than X)
A1
(so) Y has less charge (than X)

8b · 2 marks

B1
(total) momentum before decay is zero or X has zero / no momentum
B1
(total momentum after decay must be zero so) Y must have equal (and opposite) momentum to $α$ -particle (so cannot be stationary / must have speed/velocity)

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A Level Physics 9702 Paper 21, May/June 2023

Solve it on ExamPal

Practise

Exam

Question paper

Mark scheme

1a(i) · 1 mark

1a(ii) · 1 mark

1b · 3 marks

2a(i) · 1 mark

2a(ii) · 3 marks

2a(iii) · 1 mark

2b · 2 marks

2c(i) · 2 marks

2c(ii) · 3 marks

2c(iii) · 1 mark

2c(iv) · 1 mark

3a · 1 mark

3b(i) · 1 mark

3b(ii) · 1 mark

3c · 3 marks

4a · 3 marks

4b · 1 mark

4c · 2 marks

4d · 2 marks

5a(i) · 1 mark

5a(ii) · 2 marks

5b(i) · 1 mark

5b(ii) · 3 marks

5c(i) · 1 mark

5c(ii) · 1 mark

6a · 1 mark

6b · 3 marks

7a · 2 marks

7b · 2 marks

7c · 2 marks

7d(i) · 1 mark

7d(ii) · 1 mark

7d(iii) · 1 mark

8a(i) · 1 mark

8a(ii) · 2 marks

8b · 2 marks

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