Physics 9203 | Paper 2 Mock 01 | Written Paper 2
Question 1 Report
The image shows a magnetic resonance imaging scanner in a hospital. The patient lies on a bed that moves into the circular scanner. Powerful electromagnets create a magnetic field around the patient. The scanner is used by doctors to form images of soft tissues without using ionising radiation. Before a scan, staff ask whether the patient has a metal implant, jewellery or a magnetic card. A steel object could experience a large force in the field.
(a) What type of magnet is used to create the main field in the scanner? [1]
(b) Which object is most dangerous near the scanner: a steel key or a wooden ruler? [1]
(c) When a current in an electromagnet is increased, what happens to its magnetic field strength? [1]
(d) Sketch why a loose steel object can move towards the scanner. [1]
(e) What safety action should staff take before moving a patient into the scanner? [1]
A science club builds a crane to sort scrap metal. Fig. 1 shows the circular coil at the end of the crane cable. The coil is connected to a battery through a switch and is wrapped around an iron cylinder. The operator closes the switch only while lifting steel cans. Aluminium drinks cans must not be picked up. After each lift, the operator opens the switch so the load falls into a collection box. The diagram is not drawn to scale.
(a) What material should be used for the cylinder inside the coil? [1]
(b) Which type of can will the crane lift: aluminium or steel? [1]
(c) When the operator opens the switch, what happens to the magnetic field of the coil? [1]
(d) Sketch the field pattern around one end of the electromagnet, including arrows. [1]
MRI scanner
- (a) The main field is produced by an electromagnet. [1]
- (b) A steel key is most dangerous because steel can be magnetised and attracted strongly; a wooden ruler is not magnetic. [1]
- (c) Increasing current makes an electromagnet's magnetic field stronger. [1]
- (d) A loose steel object becomes magnetised in the scanner field and is attracted towards the strong-field region. [1]
- (e) Staff should check for and remove metal objects and check for implants before the patient enters the scanner. [1]
Scrap-metal crane
- (a) The cylinder should be soft iron. [1]
- (b) The crane lifts steel cans, not aluminium cans. [1]
- (c) Opening the switch stops the current, so the coil's magnetic field disappears or becomes zero. [1]
- (d) Field lines curve out of one end of an electromagnet and enter the other, with arrows in a consistent direction. [1]
Answer Details
MRI scanner
- (a) The main field is produced by an electromagnet. [1]
- (b) A steel key is most dangerous because steel can be magnetised and attracted strongly; a wooden ruler is not magnetic. [1]
- (c) Increasing current makes an electromagnet's magnetic field stronger. [1]
- (d) A loose steel object becomes magnetised in the scanner field and is attracted towards the strong-field region. [1]
- (e) Staff should check for and remove metal objects and check for implants before the patient enters the scanner. [1]
Scrap-metal crane
- (a) The cylinder should be soft iron. [1]
- (b) The crane lifts steel cans, not aluminium cans. [1]
- (c) Opening the switch stops the current, so the coil's magnetic field disappears or becomes zero. [1]
- (d) Field lines curve out of one end of an electromagnet and enter the other, with arrows in a consistent direction. [1]
Question 2 Report
The table below shows measurements from a jewellery workshop. A jeweller shines a narrow light beam into three polished stones to compare the number of bright internal flashes seen from above. Each stone has the same cut and is surrounded by air. The critical angle is measured for light travelling from the stone to air. The flashes are caused by rays reflecting inside the stone before they emerge. The jeweller uses the information when choosing a material for a decorative pendant.
| Stone | Critical angle / degrees | Bright flashes counted |
|---|---|---|
| quartz | 40 | 5 |
| zircon | 33 | 9 |
| synthetic spinel | 36 | 7 |
(a) Which stone produces the most bright flashes? [1]
(b) What relationship is suggested between critical angle and number of flashes? [1]
(c) Which process produces the flashes inside a stone? [1]
(d) When a ray strikes inside zircon at 35 degrees, what happens? [1]
(e) What is the difference between the critical angles of quartz and zircon? [1]
(f) Which stone would be chosen for the brightest design using this evidence? [1]
(a) Zircon produces the most bright flashes, with 9 flashes. [1]
(b) The results suggest that a lower critical angle gives more flashes. A lower critical angle means more rays meet an internal surface above the critical angle and so undergo total internal reflection. [1]
(c) The flashes are produced by total internal reflection. [1]
(d) In zircon, \(35^\circ\) is greater than its critical angle of \(33^\circ\), so total internal reflection occurs. [1]
(e) \[40^\circ-33^\circ=7^\circ\] The difference is \(7^\circ\). [1]
(f) The evidence supports choosing zircon, because it gives the highest number of bright flashes. [1]
Answer Details
(a) Zircon produces the most bright flashes, with 9 flashes. [1]
(b) The results suggest that a lower critical angle gives more flashes. A lower critical angle means more rays meet an internal surface above the critical angle and so undergo total internal reflection. [1]
(c) The flashes are produced by total internal reflection. [1]
(d) In zircon, \(35^\circ\) is greater than its critical angle of \(33^\circ\), so total internal reflection occurs. [1]
(e) \[40^\circ-33^\circ=7^\circ\] The difference is \(7^\circ\). [1]
(f) The evidence supports choosing zircon, because it gives the highest number of bright flashes. [1]
Question 3 Report
The diagram shows a diver looking up through calm water at the sky. The surface is horizontal. Rays from objects above the water reach the diver only within a circular region often called Snell's window. A ray from an object at the edge of this region refracts at the water-air boundary and reaches the diver's eye. Another ray inside the water meets the surface at a larger angle and is totally internally reflected. The normal is drawn at the point where the edge ray meets the surface.
(a) What is the refracted ray at the edge of Snell's window travelling along? [1]
(b) When does a ray at a water-air surface undergo total internal reflection? [2]
(c) Which ray shown has a larger angle of incidence, the edge ray or the reflected ray? [1]
(d) What change in speed occurs when light passes from air into water? [1]
(e) Sketch an arrow to show the direction of the reflected ray after it meets the water surface. [1]
(a) At the edge of Snell’s window, the refracted ray travels along the water surface, so it is at \(90^\circ\) to the normal. [1]
(b) Total internal reflection requires both conditions: light must travel from water to air, and its angle of incidence must be greater than the critical angle. [2]
(c) The reflected ray has the larger angle of incidence. It is the ray that reaches the surface at an angle greater than the critical angle. [1]
(d) Light slows down when it passes from air into water. [1]
(e) The reflected ray remains in the water and obeys \(\text{angle of reflection}=\text{angle of incidence}\). The arrow must point away from the surface, back into the water. [1]
Remember: at the critical angle the refracted ray is along the boundary; above it, no refracted ray emerges and total internal reflection occurs.
Answer Details
(a) At the edge of Snell’s window, the refracted ray travels along the water surface, so it is at \(90^\circ\) to the normal. [1]
(b) Total internal reflection requires both conditions: light must travel from water to air, and its angle of incidence must be greater than the critical angle. [2]
(c) The reflected ray has the larger angle of incidence. It is the ray that reaches the surface at an angle greater than the critical angle. [1]
(d) Light slows down when it passes from air into water. [1]
(e) The reflected ray remains in the water and obeys \(\text{angle of reflection}=\text{angle of incidence}\). The arrow must point away from the surface, back into the water. [1]
Remember: at the critical angle the refracted ray is along the boundary; above it, no refracted ray emerges and total internal reflection occurs.
Question 4 Report
Fig. 1 shows a model railway controller used at a community exhibition. The controller has a transformer that reduces the 230 V mains potential difference to a safer low potential difference for the track. A train motor receives electrical energy through the rails. A child notices that the train travels faster when the controller setting is raised. The organiser explains that a larger potential difference can cause a larger current in the motor circuit. The controller is kept dry, and the mains plug has a fuse. The diagram shows the mains input, transformer, control knob and rails.
(a) What is the purpose of the transformer in the controller? [1]
(b) Which quantity is measured in volts? [1]
(c) What happens to the motor current when the controller potential difference is increased, assuming its resistance is constant? [2]
A science club investigates how the resistance of a copper wire changes when its length changes. Fig. 1 shows the apparatus: a low-voltage power supply, ammeter, voltmeter, crocodile clips and a metre ruler beside the wire. For each test, the student moves one clip to increase the length of wire in the circuit. The potential difference is kept at 2.0 V. A longer wire gives a smaller current because charge has further to travel through the metal. The student uses the same thickness of copper wire throughout, then draws a graph of resistance against length. The power supply is switched off before the clips are moved.
(a) Which meter is connected in parallel with the test wire? [1]
(b) What equation is used to calculate resistance? [1]
(c) What happens to resistance when the length of the copper wire is increased? [2]
Model railway controller
(a) The transformer steps down the 230 V mains potential difference to a safer low potential difference. [1]
(b) Potential difference, also called voltage, is measured in volts. [1]
(c) The current increases. With resistance constant, \(I=V/R\), so a larger potential difference produces a larger current. [2]
Copper-wire investigation
(a) A voltmeter is connected in parallel with the test wire. [1]
(b) \(R=V/I\). [1]
(c) Resistance increases as wire length increases because charge travels through more metal and has more collisions, giving greater opposition to charge flow. [2]
Answer Details
Model railway controller
(a) The transformer steps down the 230 V mains potential difference to a safer low potential difference. [1]
(b) Potential difference, also called voltage, is measured in volts. [1]
(c) The current increases. With resistance constant, \(I=V/R\), so a larger potential difference produces a larger current. [2]
Copper-wire investigation
(a) A voltmeter is connected in parallel with the test wire. [1]
(b) \(R=V/I\). [1]
(c) Resistance increases as wire length increases because charge travels through more metal and has more collisions, giving greater opposition to charge flow. [2]
Question 5 Report
A survey team uses a laser rangefinder while checking the position of a floating marker near a harbour wall. Fig. 1 shows how the instrument works. The rangefinder sends a very short pulse of red laser light towards a reflector on the marker. Some light is reflected back to the instrument. Its electronic circuit measures the total time from emission to detection. For one reading, the displayed time is 160 ns. Use a speed of light in air of 3.0 x 108 m/s.
(a) What process returns some of the laser light to the rangefinder? [1]
(b) Calculate the total distance travelled by the light pulse in 160 ns. [3]
(c) Explain why the distance from the rangefinder to the marker is half the answer to part (b). [2]
(d) Suggest three reasons why a reflector gives a more dependable distance reading than the surface of water. [3]
(e) Explain how an uncertainty of 4 ns in the measured time affects the calculated distance to the marker. [5]
(a) Some of the laser light is returned by reflection from the reflector. [1]
(b) Convert nanoseconds to seconds:
\[160\text{ ns}=160\times10^{-9}\text{ s}\]
Use \(d=vt\):
\[d=(3.0\times10^8)(160\times10^{-9})=48\text{ m}\]
The total distance travelled by the pulse is \(48\text{ m}\). [3]
(c) The 48 m includes both the outgoing path from the rangefinder to the marker and the reflected path back to the rangefinder. The one-way distance to the marker is therefore:
\[48\div2=24\text{ m}\]
[2]
(d) A reflector gives a more dependable reading because it returns more light to the detector, producing a clearer and stronger signal. Water may absorb some light, ripples can direct reflected light away from the detector, and the water surface position is not fixed. Any three of these points gain credit. [3]
(e) First convert the time uncertainty:
\[4\text{ ns}=4\times10^{-9}\text{ s}\]
The uncertainty in the total light path is:
\[\Delta d=(3.0\times10^8)(4\times10^{-9})=1.2\text{ m}\]
This is for the outward and return journey, so the uncertainty in the one-way marker distance is:
\[\frac{1.2}{2}=0.60\text{ m}\]
The calculated distance to the marker has an uncertainty of \(\pm0.60\text{ m}\). [5]
Answer Details
(a) Some of the laser light is returned by reflection from the reflector. [1]
(b) Convert nanoseconds to seconds:
\[160\text{ ns}=160\times10^{-9}\text{ s}\]
Use \(d=vt\):
\[d=(3.0\times10^8)(160\times10^{-9})=48\text{ m}\]
The total distance travelled by the pulse is \(48\text{ m}\). [3]
(c) The 48 m includes both the outgoing path from the rangefinder to the marker and the reflected path back to the rangefinder. The one-way distance to the marker is therefore:
\[48\div2=24\text{ m}\]
[2]
(d) A reflector gives a more dependable reading because it returns more light to the detector, producing a clearer and stronger signal. Water may absorb some light, ripples can direct reflected light away from the detector, and the water surface position is not fixed. Any three of these points gain credit. [3]
(e) First convert the time uncertainty:
\[4\text{ ns}=4\times10^{-9}\text{ s}\]
The uncertainty in the total light path is:
\[\Delta d=(3.0\times10^8)(4\times10^{-9})=1.2\text{ m}\]
This is for the outward and return journey, so the uncertainty in the one-way marker distance is:
\[\frac{1.2}{2}=0.60\text{ m}\]
The calculated distance to the marker has an uncertainty of \(\pm0.60\text{ m}\). [5]
Question 6 Report
The diagram shows a coil launcher in a factory that moves small steel caps along a conveyor. A short pulse of current is sent through a coil near the cap. The coil becomes an electromagnet and produces a magnetic force on the cap. A sensor controls the pulse length so that each cap reaches the next station without hitting the glass guard. The current must be switched off before the next cap arrives. The engineer is investigating how changing the current affects the force on each cap.
(a) What makes the coil an electromagnet? [1]
(b) Which change increases the magnetic field of the coil? [2]
(c) What happens when the control system switches off the current, and why is this useful? [3]
(a) Current flowing through the coil makes it an electromagnet [1].
(b) Increasing the current increases the magnetic field [1]. Increasing the number of turns, or adding a soft iron core, also increases it [1]. [2 marks]
(c) When the current is switched off, the coil loses most of its magnetism [1]. The magnetic force on the cap decreases or stops [1], preventing continued acceleration and allowing controlled positioning [1]. [3 marks]
Answer Details
(a) Current flowing through the coil makes it an electromagnet [1].
(b) Increasing the current increases the magnetic field [1]. Increasing the number of turns, or adding a soft iron core, also increases it [1]. [2 marks]
(c) When the current is switched off, the coil loses most of its magnetism [1]. The magnetic force on the cap decreases or stops [1], preventing continued acceleration and allowing controlled positioning [1]. [3 marks]
Question 7 Report
A bicycle mechanic checks a pedal crank before a race. Fig. 1 shows the crank viewed from the side. The rider pushes down with a force of 250 N. At position A the crank is horizontal, so the force is at right angles to the crank. At position B the crank is vertical. The distance from the centre of the axle to the pedal is 0.17 m. The mechanic explains that the turning effect changes as the pedal moves.
The same force is used at both positions.
(a) Calculate the moment at position A. [2]
(b) Which position, A or B, produces the greater moment? [1]
(c) What is the moment at position B? [1]
(a) At position A, the force is perpendicular to the crank:
\[250\times0.17=42.5\text{ N m}\]
The moment is 42.5 N m. [2]
(b) Position A produces the greater moment because the perpendicular distance from the axle to the force's line of action is greatest. [1]
(c) At position B, the force acts through the axle, so the perpendicular distance is zero. The moment is 0 N m. [1]
Answer Details
(a) At position A, the force is perpendicular to the crank:
\[250\times0.17=42.5\text{ N m}\]
The moment is 42.5 N m. [2]
(b) Position A produces the greater moment because the perpendicular distance from the axle to the force's line of action is greatest. [1]
(c) At position B, the force acts through the axle, so the perpendicular distance is zero. The moment is 0 N m. [1]
Question 8 Report
During a science-club demonstration, students send a narrow light beam through three transparent blocks and towards air. They rotate each block until no ray can be seen leaving its upper face. At this point the light is reflected inside the block, so the angle inside the material is its critical angle. Table 1 records the results. The students use the same laser, a clean dry surface and the same angle scale for each block. A fibre-optic cable uses repeated internal reflection to keep light energy inside its glass core. The table can help the students decide which material would be suitable for a cable that needs light to remain trapped easily.
| Material | Critical angle / degrees |
|---|---|
| Acrylic | 42 |
| Dense glass | 38 |
| Flint glass | 35 |
Table 1
(a) Which material has the greatest refractive index? [1]
(b) Calculate the refractive index of flint glass. Use n = 1 ÷ sin c, where c is the critical angle. [2]
(c) Explain why a ray in flint glass at an angle of incidence of 36 degrees can undergo total internal reflection at a flint glass-air boundary. [1]
Fig. 1 shows a laboratory setup used to compare the apparent depth of a coin in a beaker of water. The coin is fixed to the base. A pin is moved vertically beside the beaker until its tip appears to be at the same position as the image of the coin. The observer keeps one eye at the marked position and does not move the beaker. The glass wall is thin and is ignored in this model. The real depth is the distance from the water surface to the coin.
(a) What is meant by the apparent depth of the coin? [1]
(b) Which is greater, the real depth or the apparent depth? [1]
(c) What happens to the light ray as it leaves the water? [1]
(d) Sketch a dashed line backwards from the ray in air to show where the observer sees the coin. [1]
Critical angle investigation
(a) Flint glass has the greatest refractive index. For light entering air, a smaller critical angle corresponds to a greater refractive index; flint glass has the smallest critical angle, \(35^\circ\). [1]
(b) \[n=\frac{1}{\sin c}=\frac{1}{\sin 35^\circ}=1.74\]
The refractive index is 1.74 (allow about 1.7). [2]
(c) \(36^\circ\) is greater than flint glass’s critical angle of \(35^\circ\), so the ray can undergo total internal reflection at a flint glass-air boundary. [1]
Apparent depth
(a) The apparent depth is the depth at which the coin, or its image, appears to be. [1]
(b) The real depth is greater than the apparent depth. [1]
(c) As the ray leaves water and enters air, it refracts away from the normal. [1]
(d) The dashed line is the backward extension of the emergent ray. It locates a virtual image above the real coin. [1]
Answer Details
Critical angle investigation
(a) Flint glass has the greatest refractive index. For light entering air, a smaller critical angle corresponds to a greater refractive index; flint glass has the smallest critical angle, \(35^\circ\). [1]
(b) \[n=\frac{1}{\sin c}=\frac{1}{\sin 35^\circ}=1.74\]
The refractive index is 1.74 (allow about 1.7). [2]
(c) \(36^\circ\) is greater than flint glass’s critical angle of \(35^\circ\), so the ray can undergo total internal reflection at a flint glass-air boundary. [1]
Apparent depth
(a) The apparent depth is the depth at which the coin, or its image, appears to be. [1]
(b) The real depth is greater than the apparent depth. [1]
(c) As the ray leaves water and enters air, it refracts away from the normal. [1]
(d) The dashed line is the backward extension of the emergent ray. It locates a virtual image above the real coin. [1]
Question 9 Report
The diagram shows an astronaut using a tool kit outside a space station. The astronaut and suit have mass 90 kg. Initially, the astronaut and a 5.0 kg tool kit are stationary relative to the station. The astronaut throws the kit at 6.0 m/s to the right. There is no significant external horizontal force on the astronaut-tool-kit system. Engineers use this example to show why an astronaut must hold a handrail before throwing an object. The diagram shows the direction of the tool kit after release.
(a) What is the momentum of the tool kit after it is thrown? [2]
(b) What is the total momentum of the astronaut-tool-kit system before the throw? [1]
(c) Which direction does the astronaut move after the throw? [1]
(d) What is the astronaut's speed after the throw? [1]
Table 1 contains results from a fairground bumper-car trial. Car P and car Q approach each other on a smooth floor. Their rubber buffers make them stick together after impact. The mass of car P with its driver is 180 kg and the mass of car Q with its driver is 220 kg. Directions are included because velocity, unlike speed, has direction. Take movement to the right as positive. The operator uses the results to check that the cars will move slowly enough after a collision near the loading area.
| car | mass / kg | velocity before collision / m/s |
|---|---|---|
| P | 180 | +2.0 |
| Q | 220 | -1.0 |
(a) What is the momentum of car P before the collision? [2]
(b) What is the total momentum of both cars before collision? [1]
(c) Which direction do the joined cars move after collision? [1]
(d) What is their velocity after collision? [1]
Astronaut and tool kit
- (a) \[p=mv=5.0\times6.0=30\text{ kg m/s}\]30 kg m/s to the right. [2]
- (b) Before the throw both objects are stationary, so total momentum is 0 kg m/s. [1]
- (c) To keep total momentum zero, the astronaut moves to the left, opposite to the tool kit. [1]
- (d) The astronaut has momentum \(30\text{ kg m/s}\) to the left, so:
\[v=\frac{30}{90}=0.33\text{ m/s}\]0.33 m/s to the left. [1]
Bumper cars
- (a) \[p_P=180\times2.0=+360\text{ kg m/s}\]\(+360\text{ kg m/s}\). [2]
- (b) \[p_{total}=+360+(220\times-1.0)=+140\text{ kg m/s}\]\(+140\text{ kg m/s}\). [1]
- (c) The positive total momentum means the joined cars move to the right. [1]
- (d) \[v=\frac{140}{180+220}=0.35\text{ m/s}\]0.35 m/s to the right. [1]
Answer Details
Astronaut and tool kit
- (a) \[p=mv=5.0\times6.0=30\text{ kg m/s}\]30 kg m/s to the right. [2]
- (b) Before the throw both objects are stationary, so total momentum is 0 kg m/s. [1]
- (c) To keep total momentum zero, the astronaut moves to the left, opposite to the tool kit. [1]
- (d) The astronaut has momentum \(30\text{ kg m/s}\) to the left, so:
\[v=\frac{30}{90}=0.33\text{ m/s}\]0.33 m/s to the left. [1]
Bumper cars
- (a) \[p_P=180\times2.0=+360\text{ kg m/s}\]\(+360\text{ kg m/s}\). [2]
- (b) \[p_{total}=+360+(220\times-1.0)=+140\text{ kg m/s}\]\(+140\text{ kg m/s}\). [1]
- (c) The positive total momentum means the joined cars move to the right. [1]
- (d) \[v=\frac{140}{180+220}=0.35\text{ m/s}\]0.35 m/s to the right. [1]
