Pearson Edexcel Level 3 GCE

Paper 1: Core Physics I

• 1. All multiple-choice questions must be answered with a cross in the box for the correct answer from A to D. If you change your mind about an answer, put a line through the box and then mark your new answer with a cross.

• A student throws a ball vertically upwards. Which of the following shows a free‑body force diagram for the ball immediately after it leaves the student’s hand?
• 
• A
• B
• C
• D
• Solution:
• Immediately the ball escapes the hand, the hand’s upward force stops working on it.
• As the ball is travelling higher, air resistance acts downward.
• Two downward arrows signifying the force of gravity and air resistance apply should be seen in the free-body diagram.

• Comparing this to given option:
• A: Indicates both a downward and an upward force.
• B: Illustrates a large upward force and a downward force.
• C: Show two downward forces, one arrow is air resistance
•  and other arrow is the gravitational force.
• D: There is also a downward air resistance force as the ball is moving upwards
• Correct Answer: C The free-body diagram that represents the downward force of gravity and air resistance is the right on.

• 2. Which row of the table contains only scalar quantities?

• 
• Solution:
• A: Displacement is vector quantity
• B: Momentum is a vector quantity because [math]\bar{p} = m\bar{v}[/math] in this equation [math]\bar{v}[/math] is velosity and it is a vector quantity.
• D: In this option acceleration is a vector quantity which is define as rate of change of velocity and velocity is a vector quantity.
• C: All quantities are scalar quantities.
• Correct Answer: C option

• 3. A ball falls from rest through glycerine and reaches terminal velocity. Which of the following graphs shows how displacement s varies with time t for the ball?

• 
• Solution:
• A ball accelerates initially due to gravity when it falls through a viscous fluid, such as glycerine, but as its speed increases, so does the drag force.
• Eventually, the drag force balances the gravitational force, and the ball reaches a constant terminal velocity.

• A:
• Indicates that the ball stops moving, which is untrue since it achieves a
• constant velocity.
• It first shows displacement rising quickly before levelling out.
• B:
• Indicates a gradual increase in displacement followed by a faster increase in velocity, which is consistent with the first acceleration phase. After then, the curve turns linear, signifying continuous velocity, or terminal velocity.
• C: Displays a linear connection, which is inaccurate because the ball accelerates after starting at rest and has a constant velocity from the beginning.
• D: Displays an S-shaped curve, which is common for displacement-time graphs that show an initial acceleration, a period of nearly constant velocity, and perhaps another change in velocity. This curve is not immediately relevant to achieving a single terminal velocity from rest, though.
• Correct Answer:
• Graph B accurately represents the ball’s motion. It illustrates an early period when the slope (velocity) increases, followed by a period where the slope remains constant, reflecting the terminal velocity.

• 4. A piece of conducting putty is shaped into a cylinder of uniform cross‑sectional area, as shown. The length of the cylinder is l. The resistance between the two ends is 8.0Ω

• 

• The piece of putty is then rolled out until the length is 2l. Which of the following is now the value of the resistance between the two ends?

• 
• Solution:
• [math]R_1 = 8.0 \, \Omega \\
  L_1 = l \\
  R_1 = \frac{\rho L_1}{A_1} \\ \text{put value} \\ 
  8.0 = \frac{\rho l}{A_1}[/math]
• When the putty is rolled out, its length = [math]L_1 = 2l[/math]
• If the length doubles, the cross-sectional area must halve since the putty’s volume stays the same.
• So
• [math]A_2 = \frac{A_1}{2}[/math]
• For new resistance:
• [math]R_2 = \frac{\rho L_2}{A_2} \\
  R_2 = \frac{\rho \, 2l}{A_1/2} \\
  R_2 = \frac{\rho (2(2l))}{A_1} \\
  R_2 = \frac{\rho \, 4l}{A_1} \\
  R_2 = 4 \left(\frac{\rho l}{A_1}\right) \\ \text{Put the value in this equation} \\
  R_2 = 4(8) \\
  R_2 = 32 \, \Omega
  [/math]
• The new value of the resistance between the two end is 32[math] \Omega [/math]
• Correct Answer: D

• 5. A car is travelling at a constant speed in a straight line along a horizontal road. Which row of the table gives a Newton’s third law pair of forces?

• 
• Solution:
• Every action has an equal and opposite response in order to recognize the Newton’s third law pair of forces. This implies that the forces need to:
• – Act on various objects.
• – Be of the same kind (for example, both gravitational and normal forces).
• – Have opposite directions and equal magnitudes.
• A: Since they are different forces, their magnitudes may differ.
• B: These forces meet the requirements. The road applies an equal and opposite normal force to the automobile, and the car applies a normal force to the road.
• C: Normal force is a contact force, whereas weight is a gravitational force. There are several kinds of forces.
• D: These are different types of forces, much like C. The gravitational force of the automobile on Earth is the response force to the car’s weight (gravitational force of Earth on car).
• Correct Answer: B
• 
•  Newton’s Third law of motion
• 6. The diagram shows an electric motor pulling a truck of mass m along a slope. The truck moves through a vertical height h and a distance l along the slope, during a time t. There is a potential difference V across the motor and a current l in the motor.
• 
• Solution:
• The motor’s efficiency may be determined by calculating the ratio of its useable output power to its entire input power.
• Efficiency of the motor:
• ⇒ Input Power:
• The electrical power used, which is the result of multiplying the potential difference (V) across the motor by the current (I) passing through it, provides the motor’s input power.
• [math]P_{in} = VI[/math]
• Input Energy:
• [math]E_{in} = P_{in} \times t \\
  E_{in} = VI t[/math]
• Output energy:
• [math]E_{out} = mgh[/math]
• Efficiency:
• The ratio of useable output energy to total input energy is called efficiency ([math]\eta[/math]):
• [math]\eta = \frac{\text{Output Energy}}{\text{Input Energy}} \\
  \eta = \frac{mgh}{VIt}[/math]
• So:
• Correct Answer: D
• 
  

  7. A student investigated the e.m.f. and internal resistance of a battery. The student produced the following sketch graph.

• 
• Which row of the table gives the quantities plotted?
• Solution:
• The question is to determine which quantities are plotted on the x-axis and y-axis of the provided sketch graph, which depicts an examination of a battery’s internal resistance and e.m.f.
• ⇒ Explanation:
• The graph displays a negative-sloped linear relationship, which is typical of a battery’s terminal potential difference when current is extracted from it.
• The equation for terminal potential difference (V) is given by
• [math]V = E – Ir[/math]
• Where E is the e.m.f., I is the current, and r is the internal resistance.
• ⇒ Analyzing the graph:
• When the number on the x-axis rises, the y-axis shows a quantity that falls linearly. This behavior is consistent with the connection between current and terminal potential difference.
• ⇒ Evaluating the options:
• A and B (e.m.f on y-axis):
• A battery’s e.m.f. (E) remains constant regardless of circuit resistance or current, unless the battery is dead.
• Therefore, rather than a line that slopes downward, a graph with e.m.f. on the y-axis would be a horizontal line.
• C (terminal potential difference Vs. Circuit resistance):
• While circuit resistance affects the terminal potential difference, the connection is usually not a straightforward linear decline as the graph shows.
• D (Terminal potential difference Vs. Current):
• According to the equation
• [math]V = E – Ir[/math]
• The terminal potential difference (V) changes linearly with the current (I) if E and r remain constant. The negative slope shows that the voltage drop across the internal resistance (Ir) causes the terminal potential difference to decrease as current increases.
• This is exactly the same as the provided sketch graph. When the current (I) is zero, the graph’s y-intercept would reflect the e.m.f. (E), and the slope’s magnitude would indicate the internal resistance (r).
• Correct Answer: D

• 8.  A hall is thrown vertically upwards at a velocity of 6 m.s^-1. Which of the following gives the maximum height, in m, reached by the hall?

• 
• Solution:
• Finding Vertical Projectile Motion’s Maximum Height
• The following formula may be used to determine the highest height that a ball thrown vertically upwards can reach:
• By using the third equation of motion:
• [math]u^{2} = v^{2} + 2aS[/math]
• Where:
• [math]\begin{gather} \text{u is the final velocity (at maximum height, u = 0 m.s}^{-1}\text{ )} \\
  \text{v is the initial velocity (u = 6.0 m.s}^{-1}\text{ )} \\
  \text{a is the acceleration due to gravity (a = -9.81 m.s}^{-2}\text{ , negative} \\
  \text{because it acts downwards, opposing the upward motion)} \\
  \text{s is the displacement (maximum height, h)} \end{gather}[/math]
• Putting all values in this equation
• [math]\begin{gather} u^{2} = v^{2} + 2aS \\
  0^{2} = 6^{2} + 2 \times (-9.81) \times h \\
  0 = 6^{2} – 2 \times (9.81) \times h \\
  2 \times (9.81) \times h = 6^{2} \\
  h = \frac{6^{2}}{2 \times (9.81)} \end{gather}[/math]
• the correct answer: A

• 9. A student connects the circuit shown. The battery has negligible internal resistance.

• 
• The student increases the resistance of the variable resistor from 0 Ω to 40 Ω. Determine the range of readings on the voltmeter.
• Solution:
• Calculate the total resistance when the variable resistance is at its minimum (0 Ω)
• The circuit’s overall resistance is equal to the resistance of the fixed resistor when the variable resistor is set to 0 Ω:
• [math]R_{\text{total}} = 10 \, \Omega + 0 \, \Omega = 10 \, \Omega[/math]
• Calculate the current in the circuit:
• By using Ohm’s law
• [math]I = \frac{V}{R_{\text{total}}} \\
  I = \frac{6}{10} \\
  I = 0.6 \,\text{A}[/math]
• Calculate the voltage:
• [math]V_{(10 \, \Omega)} = I \times R_{(10 \, \Omega)} \\
  V_{(10 \, \Omega)} = 0.6 \times 10 \\
  V_{(10 \, \Omega)} = 6 \,\text{V}[/math]
• This is the minimum reading of the volt meter.
• Calculate the total resistance when the variable resistor is at its maximum (40 Ω)
• [math]R_{\text{total}} = 10 \, \Omega + 40 \, \Omega = 50 \, \Omega[/math]
• For the current by using Ohm’s Law
• [math]I = \frac{V}{R_{\text{total}}} \\
  I = \frac{6}{50} \\
  I = 0.12 \,\text{A}[/math]
• Calculate the voltage across the 10  resistor (voltmeter reading)
• [math]V_{(10 \, \Omega)} = I \times R_{(10 \, \Omega)} \\
  V_{(10 \, \Omega)} = 0.12 \times 10 \\
  V_{(10 \, \Omega)} = 1.2 \,\text{V}[/math]
• Final Answers:
• Maximum reading on voltmeter = 6.0 V
• Minimum reading on voltmeter = 1.2 V
• 10. A model rocket accelerates vertically upwards then decelerates due to gravity until it reaches a maximum height.

• (a) A velocity‑time graph for the rocket until it reaches maximum height is shown.

• 
• Show that the rocket reaches a maximum height of about 68m.
• Solution:
• Determine the area under the velocity-time graph to demonstrate that the rocket achieves a maximum height of around 68 meters.
• This is because the area under a velocity-time graph indicates the displacement or distance travelled.
• ⇒ Identify the shape and dimensions of the area:
• A triangle is formed by the graph. The whole time until the velocity drops to zero (maximum height), or 4.5 seconds, is the base of the triangle.
• The greatest speed attained, 30 m/s, is shown by the height of the triangle.
• Calculate the area of the triangle:
• [math]\text{Area} = \tfrac{1}{2} (\text{Base})(\text{Height}) \\
  \text{Area} = \tfrac{1}{2} (4.5)(30) \\
  \text{Area} = \tfrac{1}{2} (135) \\
  \text{Area} = 67.5 \,\text{m}^2[/math]
• The maximum height reaches by the rocket is 67.5 m, Which is approximately 68 m.

• b. When the rocket reaches the maximum height of 68m, a parachute opens. Almost instantly, the rocket reaches a terminal velocity of 2.0 m s ^–1.

• Complete the velocity‑time graph below for the motion of the rocket until it reaches the ground.
• 
• Solution:
• ⇒ Parachute opens and terminal velocity (After maximum height):
• The rocket’s velocity will be negative because it is now travelling downward.
• As a result, the velocity should rapidly shift from 0 m/s to -2.0 m/s on the graph from the point when it is 0 m/s.
• The deceleration brought on by the parachute opening and the rocket’s rapid terminal velocity is represented by this change.
• Descent at terminal Velocity:
• A horizontal line at -2.0 m/s should be visible on the graph until the rocket touches down.
• Thus, from the time at which terminal velocity is attained, the horizontal line at -2.0 m/s should continue for around 34 seconds. The graph would reach about 38 seconds if the peak occurred at 4 seconds.
• ⇒ To complete the graph:
• Draw a sharp line downward from the velocity point at 0 m/s (about 4 seconds on the x-axis) to -2.0 m/s.
• Then, assuming the apex occurred at 4 seconds and the drop was 34 seconds, draw a horizontal line at -2.0 m/s until the time axis reaches about 38 seconds.

• (c). The rocket is fired upwards a second time when the wind is blowing. The rocket falls with a vertical velocity of [math]2.0 ms^{–1}[/math] and a horizontal velocity of [math]1.5 ms^{–1}[/math].

• Determine the velocity of the rocket by drawing a scaled vector diagram.

• Solution:
• ⇒ Draw the vertical velocity vector:
• Using the scale of your choice, draw a vector with a length equal to ([math]2 \,\text{m}\,\text{s}^{-1}[/math]) that points downward (since the rocket is falling).

  

• ⇒ Draw the horizontal velocity vector:
• Using your preferred scale, draw a vector with a length equal to ([math]1.5 \,\text{m}\,\text{s}^{-1}[/math]) from the tip of the vertical velocity vector that points horizontally (in the wind’s direction).
• ⇒ Draw the resultant velocity vector:
• From the vertical velocity vector’s beginning point to the horizontal velocity vector’s tip, draw a vector. This indicates the rocket’s final velocity.
• ⇒ Measure the magnitude of velocity:
• Using the scale of your choice, measure the length of the resulting vector and convert it back to m/s. This will be the velocity’s magnitude.
• ⇒ Measure the angle to the horizontal:
• Calculate the angle that separates the horizontal velocity vector from the resulting velocity vector. The angle to the horizontal is this.
• By Pythagorean theorem to calculate velocity:
• [math]v = \sqrt{(2)^2 + (1.5)^2} \\
  v = \sqrt{4 + 2.25} \\
  v = \sqrt{6.25} \\
  v = 2.5 \,\text{m}\,\text{s}^{-1}[/math]
• For calculate the angle to the horizontal:
• [math]\theta = \tan^{-1}\!\left(\tfrac{2}{1.5}\right) \\
  \theta = \tan^{-1}(1.33) \\
  \theta \approx 53.1^{\circ}[/math]

• 
  

  11. A student investigates how the resistance of a length of nichrome wire changes with temperature.

• (a) The student takes measurements to determine the resistance of the wire at different temperatures.

• (i) Draw a diagram of the circuit the student could use.

• Solution:
• The experiment looks at how a nichrome wire’s resistance varies with temperature.
• To find the resistance at various temperatures, the student measures.
• 

• (ii) The wire has a thin electrically insulating coating so that it can be coiled up without causing a short circuit.

• The student places the coil of wire into a water bath so the temperature of the wire can be varied.

• Describe how the student could determine the temperature of the wire accurately.

• Solution:
• ⇒ Use a calibrated thermometer or temperature sensor:
• As near to the coiled wire as you can, submerge a thermometer or a calibrated temperature sensor (such as a thermocouple or RTD) completely in the water bath.
• ⇒ Allow for thermal equilibrium:
• Before obtaining the temperature reading, give the water bath’s temperature—and, by extension, the nichrome wire’s—enough time to stabilise and achieve thermal equilibrium.
• ⇒ Stir the water bath:
• The temperature of the wire may be precisely determined by gently stirring the water bath to guarantee a consistent temperature distribution across the bath. 

• (b) Explain, in terms of particle behaviour, why the resistance of the nichrome wire changes as temperature increases.

• Solution:
• According to the problem, nichrome wire’s resistance varies with temperature.
• 
• ⇒ Step 1
• Explain the effect of increased temperature on atoms.
• – The atoms in the nichrome wire acquire more thermal energy as the temperature rises.
• – The atoms vibrate more forcefully around their fixed locations as a result of this increased energy.
• ⇒ Step 2
• Describe the impact on electron flow.
• – As the temperature increases, the nichrome wire’s atoms get more thermal energy.
• – This extra energy causes the atoms to oscillate more violently around their fixed places.
• ⇒ Step 3
• Relate collision to resistance.
• – The unimpeded movement of electrons is hampered by these more collisions.
• – The nichrome wire’s electrical resistance increases as a result of this impedance.
• Solution:
• The nichrome wire’s atoms move more violently at higher temperatures, which causes them to collide with conduction electrons more frequently. This hinders electron transport and raises resistance.

• 12. A wire‑wound resistor consists of a long length of wire wound around an insulating core. A technician finds a wire‑wound constantan resistor labelled 80 Ω.

• a) Calculate the length of the constantan wire used to make the resistor. resistivity of constantan wire at room temperature [math]= 4.9 × 10^{–7} Ω m [/math] diameter of wire = 0.28 mm

• Solution:
• To calculate the length of the constantan wire used to make the resistor, we use the formula for resistance:
• [math]R = \rho \frac{L}{A}[/math]
• Where R is resistance, [math] \rho [/math] is resistivity, L is length, and 𝐴 is the cross-sectional area.
• Given Value:
• Resistance, R = 80 [math] \Omega [/math]
• Resistivity of constantan wire, [math]\rho = 4.9 \times 10^{-7} \; \Omega \,\text{m}[/math]
• Diameter of wire, d = 0.28 mm = [math]0.28 × 10^{-3} m[/math]
• Calculate the cross-section area of the wire:
•  The radius of the wire is [math]r = \frac{d}{2}[/math]
• [math]r = \frac{0.28 \times 10^{-3}}{2} \ \text{m} \\
  r = 0.14 \times 10^{-3} \ \text{m}[/math]
• Area:
• [math]\begin{gather}
  A = \pi r^2 \\
  A = (3.14)\left(0.14 \times 10^{-3}\right)^2 \\
  A = 6.1575 \times 10^{-8} \ \text{m}^2
  \end{gather}[/math]
• Calculating the resistance of the wire:
• [math]\begin{gather}
  R = \rho \frac{L}{A} \\
  L = \frac{RA}{\rho} \\
  L = \frac{(80)\left(6.1575 \times 10^{-8}\right)}{4.9 \times 10^{-7}} \\
  L \approx 10.05 \ \text{m}
  \end{gather}[/math]

• b) A potential difference of 9.8 V is applied across the resistor and the current in the resistor is 0.12A.

• Deduce whether the value labelled on the resistor is supported by these data.

• Uncertainty in the potential difference = ±0.1V

• Uncertainty in the current = ±0.01A

• Solution:

• Given Date:
• – The potential difference across the resistance is V = 9.8 V
• – The current in the resistor is I = 0.12 A
• – The uncertainty in the potential difference is [math]\Delta V = \mp 0.1 \ \text{V}[/math]
• – The uncertainty in the current is [math]\Delta I = \mp 0.01 \ \text{A}[/math]
• – The labelled resistance value is 80 [math] \Omega [/math]
• Formula:
• By Ohm’s law
• [math]R = \frac{V}{I}[/math]
• – The sum of the fractional uncertainties of the denominator and numerator determines the fractional uncertainty in a quotient.
• Solve:
• By Ohm’s law
• [math]\begin{gather}
  R = \frac{V}{I} \\
  R = \frac{9.8}{0.12} \ \Omega \\
  R \approx 81.67 \ \Omega \\
  R \approx 82 \ \Omega
  \end{gather}[/math]
• – Fraction Uncertainty in voltage:
• [math]\begin{gather}
  \frac{\Delta V}{V} = \frac{0.1}{9.8} \\
  \frac{\Delta V}{V} \approx 0.0102
  \end{gather}[/math]
• – Percentage Fraction Uncertainty in voltage:
• [math]\begin{gather}
  \frac{\Delta V}{V} \times 100\% \\
  \approx 1.02\%
  \end{gather}[/math]
• – Fraction Uncertainty in current:
• [math]\frac{\Delta I}{I} = \frac{0.01}{0.12} \approx 0.0833[/math]
• – Percentage Fraction Uncertainty in current:
• [math]\begin{gather}
  \frac{\Delta I}{I} \times 100\% \\
  \approx 8.33\%
  \end{gather}[/math]
• – Total Fraction Uncertainty in current:
• [math]\begin{gather}
  \frac{\Delta R}{R} = \frac{\Delta V}{V} + \frac{\Delta I}{I} \\
  \frac{\Delta R}{R} = 0.0102 + 0.0833 \\
  \frac{\Delta R}{R} = 0.0935
  \end{gather}[/math]
• – Total percentage Fraction Uncertainty in current:
• [math]\begin{gather}
  \frac{\Delta R}{R} \times 100\% \\
  = 9.35\%
  \end{gather}[/math]
• – Absolute Uncertainty in resistance:
• [math]\begin{gather}
  \Delta R = R \times \frac{\Delta R}{R} \\
  \Delta R \approx 81.67 \times 0.0935 \\
  \Delta R \approx 7.63 \, \Omega \\
  \Delta R \approx 8 \, \Omega
  \end{gather}[/math]
• – Determine the range of possible resistance values
• The resistance range is R+∆R
• [math]R = 82 \pm 8 \, \Omega[/math]
• The range is approximately 74 Ω to 90.

• 13. Some students used a plank to make a bridge to cross a stream. The plank rested on a rock and a wall as shown. Assume the plank is uniform.

• 
• 
  

  (i) Show that the weight of the plank is about 250N

• mass of plank = 25 kg

• Solution:
• [math]W = m \times g [/math]
• Mass of plank = 25 kg
• [math]\begin{gather}
  W = m \times g \\
  W = 25 \times 9.8 \\
  W = 245 \, N
  \end{gather}[/math]
• Putting values
• Therefore, the weight of the plank is about 245 N.

• (ii) Determine the force exerted by the wall on the plank

•  – Let L be the total length of the plank, L = 5.0 m
• – Since the plank is uniform, its weight W acts at its center of mass, which is at [math]\frac{L}{2} = 2.5 \, \text{m}[/math] from either end.
• – Let [math]R_{\text{rock}}[/math] be the upward force exerted by the rock on the plank.
• – Let [math]R_{\text{wall}}[/math] be the upward force exerted by the wall on the plank.
• – The wall is located 1.4 m from the right end of the plank, meaning its distance from the rock (left end) is 5-1.4 = 3.6 m
• ⇒ The principle of moments:
• We may remove [math]R_{\text{rock}}[/math] from the equation by taking a minute to consider the point of contact with the rock (the pivot point) in order to determine [math]R_{\text{wall}}[/math].
• For balance, the total of the clockwise and anticlockwise moments must be equal.
• Calculate moments:
• Moment due to the plank’s weight (clockwise moment about the rock):
• [math]W \times \frac{L}{2} = W \times 2.5 \, \text{m}[/math]
• Moment due to the wall’s force (anticlockwise moment about the rock):
• [math]R_{\text{wall}} \times (L – 1.4) = R_{\text{wall}} \times 3.6 \, \text{m}[/math]
• Set up the equilibrium equation:
• – Assuming the plank is in equilibrium:
• [math]\begin{gather}
  W \times 2.5 = R_{\text{wall}} \times 3.6 \, \text{m} \\
  R_{\text{wall}} = \frac{2.5W}{3.6}
  \end{gather}[/math]
• Where W is the weight of the plank = 245 N
• Put in this equation
• [math]\begin{gather}
  R_{\text{wall}} = \frac{2.5W}{3.6} \\
  R_{\text{wall}} = \frac{2.5 \times 245}{3.6} \\
  R_{\text{wall}} = 170 \, N
  \end{gather}[/math]

• b) A student of weight 550N states that the plank will tip if she walks from the rock to the other end of the plank. Justify the student’s statement

• ⇒ Analyze the tipping condition:
• When the student’s weight (together with the plank’s weight) creates a clockwise moment around the wall that is more than the greatest anticlockwise moment that the rock can produce, the plank will tilt.
• The center of gravity of the plank-student system will go past the wall, resulting in a tip, if the student moves past the plank’s center of mass and towards the end that is not supported by the wall.
• ⇒ Consider moments about the wall:
• The wall serves as the pivot point for tilting as the trainee approaches the unsupported end. The plank’s weight causes the wall to rotate anticlockwise.
• A clockwise moment is created about the wall by the student’s weight.
• ⇒ Calculate moments:
• The weight of the plank causes an anticlockwise moment about the wall:
• The plank’s center of mass is 2.5 away from the rock.
• Five meters separate the wall and the rock. Thus, the plank’s center of mass is located to the left of the wall
• [math]\begin{gather}
  5 – 2.5 = 2.5 \, \text{m} \\
  \text{Moment (Plank)} = 250 \times 2.5 \\
  \text{Moment (Plank)} = 625 \, \text{Nm} \, \, (\text{anticlockwise})
  \end{gather}[/math]
• Clockwise instant caused by student weight: The student will be 1.4 meters beyond the wall if they go to the “other end” (the unsupported end).
• [math]\begin{gather}
  \text{Moment (student)} = 550 \times 1.4 \\
  \text{Moment (student)} = 770 \, \text{Nm} \; (\text{clockwise})
  \end{gather}[/math]
• Compare moments:
• The student’s clockwise moment (770 Nm) is larger than the weight of the plank’s anticlockwise moment (625 Nm). This implies that the plank will tilt as it revolves around the wall in a clockwise direction.
• ⇒ Justification:
• The plank will tip when the student, weighing 550, approaches the end of the plank beyond the wall because the clockwise moment created by their weight about the wall (which serves as a pivot) will be greater than the anticlockwise moment created by the plank’s weight.

• 
  

  14. A student connects a filament bulb to a battery. The battery has internal resistance. The student connects an identical bulb in parallel with the first bulb, as shown.

• 

• He continues to connect identical bulbs in parallel.

• The student observes that the bulbs get dimmer as more bulbs are connected. He also observes that the temperature of the battery increases.

• Explain these observations.

•  Solution:
• ⇒ Bulbs getting dimmer as more bulbs are connected:
• The internal resistance of the battery causes the voltage across each bulb to drop as additional bulbs are connected in parallel, making the bulbs dimmer.
• Explanation:
• The external circuit’s overall equivalent resistance drops when identical bulbs are connected in parallel.
• The total current drawn from the battery rises as a result of this drop in total external resistance.
• A higher overall current causes a greater voltage drops across the battery’s internal resistance because of its internal resistance.
• [math]V_{\text{internal}} = I_{\text{total}} \times r_{\text{internal}}[/math]
• As a result, the battery’s terminal voltage—the voltage accessible to the external circuit—drops.
• [math]V_{\text{terminal}} = EMF – V_{\text{internal}}[/math]
• The lights look dimmer because they are connected in parallel, which means that the voltage across each bulb is equal to the terminal voltage. As this voltage drops, so does the power that each bulb ([math]P = V^2 IR[/math]) dissipates.
• ⇒ Temperature of the battery increasing:
• The battery’s temperature rises as a result of more heat being released within the battery due to the higher total current passing through its internal resistance.
• Explanation:
• As previously mentioned, increasing the number of parallel bulbs lowers the overall exterior resistance, which raises the total current drawing from the battery.
• Power is released as heat by the battery’s internal resistance, which is provided by
• [math]P_{\text{heat}} = I_{\text{total}}^{2} \times r_{\text{internal}}[/math]
• A larger total current causes the battery’s internal resistance to produce more heat, which raises the battery’s temperature noticeably.

• 15. A bullet of mass 12g moved at a speed of 450 m.s-1. The bullet hit a soft wooden block of mass 2.5kg which was attached to a wire, as shown. The bullet became stuck in the wooden block which swung upwards.

• 

• (a) (i) Show that the momentum of the bullet is about [math]5 kg.ms^{-1}[/math]

• Solution:
• – Mass of bullet, [math]m_b = 12 \, \text{g} = 0.012 \, \text{kg}[/math]
• – Speed of bullet, [math]v_b = 450 \, \text{m}\,\text{s}^{-1}[/math]
• Calculate the momentum of the bullet:
• [math]\begin{gather}
  \text{Momentum } p = m \times v \\
  p_b = m_b \times v_b \\
  p_b = 0.012 \times 450 \\
  p_b = 5.4 \, \text{kg}\,\text{m}\,\text{s}^{-1}
  \end{gather}[/math]

• (ii) Determine the maximum change in vertical height of the wooden block

• ⇒ Apply conservation of momentum during the inelastic collision:
• – Let M be the mass of the wooden block and V be the initial velocity of the combined bullet – block system immediately after impact.
• [math]\begin{gather}
  \text{Total momentum before collision} = \text{Total momentum after collision} \\
  m_b v_b = (m_b + M)V
  \end{gather}[/math]
• – Given M = 2.5 kg
• [math]\begin{gather}
  m_b v_b = (m_b + M)V \\
  5.4 \, \text{kg}\,\text{m}\,\text{s}^{-1} = (0.012 + 2.5)V \\
  V = \frac{5.4}{0.012 + 2.5} \\
  V = \frac{5.4}{2.512} \\
  V \approx 2.15 \, \text{m}\,\text{s}^{-1}
  \end{gather}[/math]
• ⇒ Apply conservation of momentum during the inelastic collision:
• As the combined system swings upward to its full height, the kinetic energy of the impact is transformed into gravitational potential energy.
• [math]\begin{gather}
  \text{Initial kinetic energy} = \text{Final potential energy} \\
  \frac{1}{2}(m_b + M)V^2 = (m_b + M)gh \\
  \frac{1}{2}V^2 = gh
  \end{gather}[/math]
• -[math]g = 9.8 \, \text{m}\,\text{s}^{-2} \quad \text{(standard gravitational acceleration)}[/math]
• [math]\begin{gather}
  \frac{1}{2}V^2 = gh \\
  \frac{1}{2}(2.15)^2 = (9.8)h \\
  h = \frac{\tfrac{1}{2}(2.15)^2}{9.8} \\
  h = \frac{2.31125}{9.8} \\
  h \approx 0.236 \, \text{m} \\
  h \approx 0.24 \, \text{m}
  \end{gather}[/math]
• The maximum change in vertical height of the wooden block is approximately [math]0.24 m[/math].

• b) The soft wooden block was replaced with a hard steel block of the same mass.

• Solution:
• ⇒ Change in momentum of the bullet:
• The bullet embeds itself in the wooden block, causing an inelastic collision in which the block and bullet travel in tandem.
• The bullet bounces in the steel block scenario, which means that its end velocity is opposite to its beginning velocity.
• When the bullet strikes the steel block instead of the wooden block, this shift in direction indicates a greater change in momentum for the projectile.
• ⇒ Momentum transfer to the block:
• The overall momentum of the system prior to the collision must be equal to the total momentum following the collision, as per the principle of conservation of momentum.
• The bullet sends a stronger impulse (change in momentum) to the steel block because it undergoes a greater change in momentum when it bounces off the block.
• ⇒ Initial velocity and kinetic energy of the block:
• In comparison to the wooden block, the steel block experiences a higher initial velocity immediately following the contact due to this stronger impulse.
• As a result, the steel block has additional kinetic energy just after the impact.
• ⇒ Conversion to potential energy and vertical height:
• The block’s kinetic energy is transformed into gravitational potential energy as it swings higher.
• A higher maximum vertical height may be attained because the steel block’s greater initial kinetic energy can be transformed into more potential energy.

• 16) A student released a trolley from the top of a ramp of length of about 1.5m, as shown.

• The student investigated how the speed v of the trolley at the bottom of the ramp varied as the height h of the ramp was increased.

• The student placed a light gate connected to a data logger at position A to measure v as the card passed through the light gate.

• 
• 
  

  a) Describe how the student could measure h accurately

• Solution:
• To measure h accurately, the student could:
• At the location where (h) has to be measured, place a vertical ruler or meter rule next to the ramp.
• Verify that the highest point of the ramp is in line with the measurement and that the ruler is perpendicular to the base.
• From the horizontal surface, read the value on the ruler that represents the ramp’s highest vertical height.
• As an alternative, make sure the measurement is taken vertically from the ramp’s base to its top by using a spirit level or set square.

• (b) The student derived the following equation for the motion of the trolley 

• [math]h = \frac{v^2}{2g}[/math]

• Where g is the acceleration due to gravity.

• (i) Explain why plotting a graph of h against [math]v^2[/math]  will produce a straight line.

• Solution:
•   The equation is [math]h = \frac{v^2}{2g}[/math]
• In this equation, h is the dependent variable (y-axis), [math]v^2[/math] is the independent variable (x-axis), and [math] \frac{1}{2g}[/math] is a constant.
• This equation can be written in the form of a straight-line equation
• [math]y = mx + c[/math]
• Where,
• [math]y = h, \; x = v^2, \; m = \tfrac{1}{2g} \; (\text{the gradient}), \; c = 0 \; (\text{y-intercept})[/math]
• Plotting (h) versus ([math]v^2[/math]), as the equation has a constant multiplier and no constant term, will produce a straight line that passes through the origin with a gradient of [math]\frac{1}{2g}[/math]

• ii) The student varied h and measured corresponding values of v. The results are recorded below.

• Plot a graph of h on the y-axis against [math]v^2[/math]  on the x-axis on the grid opposite use the additional column in the table for your processed data.

• Solution:
• Calculate [math]v^2[/math]  for each data point:
• The values for (h) in cm and (v) in m/s are given in the table. For each equivalent ([math]v^2[/math]) value, we must compute (v).
•  

• ⇒ Prepare the graph axes:
• – The y-axis should represent h (height in cm)
• – The x-axis should represent [math]v^2 (\text{speed squared in } (m/s)^2)[/math]
• – To accommodate the range of values determined in Step 1, select suitable scales for both axes. The range for (h) is 10.8 cm to 58.7 cm. 1.9044 (m/s)²) to 11.9716 (m/s)² are the ranges for ([math]v^2[/math]).
• Plot the points:
• Plot each ([math]v^2 , h[/math]) pair on the graph paper. For example, the first point would be approximately (1.90, 10.8), the second (3.92, 18.9), and so on.
• Draw the line of best fit:
• Since the equation [math]h = \tfrac{1}{2g} v^2[/math] is in the form of
• [math]y = mx[/math]
• Where
• [math]y = h, \; x = v^2, \; m = \tfrac{1}{2g}[/math]
• is a constant,the graph of h against [math]v^2[/math] should be a straight line passing through the origin (assuming on energy losses). Draw a straight line that best fits the plotted points.

  

• Explanation:
• In the problem, the height (h) at which a tram is discharged on a ramp is compared to its speed (v) at the bottom. A direct proportionality between (h) and (v²) is suggested by the derived equation ([math]h = \frac{v^2}{2g}
  [/math]), where the constant of proportionality is [math]\frac{1}{2g}
  [/math]. Plotting (h) versus (v2 should provide a straight line, which visually validates this connection and enables additional analysis, such as taking the graph’s gradient and calculating the value of (g).

• (iii)The student used her results to plot a graph and determine a value for g.

• She concluded that her value was consistent with the value of g given on the data sheet at the back of this paper.

• Comment on the student’s conclusion.

•  Solution:
•  Comment on the student’s conclusion:
•  The student came to the conclusion that the value of (g) on the data sheet and the amount she had calculated were consistent.
• Comment:
• Consistency:
• Her conclusion is sound if the graph’s computed value of (g) is around the accepted value of (g), which is roughly 9.8 m/s².
• ⇒ Factors Affecting Consistency:
• Ideal Condition:
• Ideal circumstances, such as a frictionless ramp and very little air resistance, are assumed by the derived equation ([math]h = \frac{v^2}{2g} [/math]). These variables might cause inconsistencies in an actual experiment.
• Precision of data:
• The apparent consistency may also be impacted by the quantity of significant figures used in the computations and measurements.
• Validity:
• The conclusion of consistency is supported if the experiment’s value falls within a reasonable range of the theoretical value (taking experimental uncertainty into account).
• Ideally, the student should compute the error or percentage difference to quantify this consistency.

• 17. An empty lift is positioned at the first floor of a building. It is suspended by 6 identical steel cables of length [math]50\,m[/math].

• a) Calculate the extension of each lift cable.

• Cross – sectional area of a cable = [math]3.1\times 10^{-4}\ m^2[/math]
  Young modulus of steel = [math]200\ GPa[/math]
  Weight of lift = [math]12\ kN[/math]
• Solution:
• Number of cables = [math]6[/math]
• Length of each cable, [math]L = 50\ m[/math]
• Cross – sectional area of a cable, [math]A = 3.1\times 10^{-4}\ m^2[/math]
• Young modulus of steel, [math]E = 200\ GPa = 200\times 10^9\ Pa[/math]
  Weight of lift, [math]W = 12\ kN = 12\times 10^3\ N[/math]
• Calculate the force:
• [math]F = \frac{W}{\text{Number of cables}}[/math]
  [math]F = \frac{12\times 10^3}{6}[/math]
  [math]F = 2\times 10^3\ N[/math]
• Appling Young’s Modulus formula to find extension:
• [math]E = \frac{\text{stress}}{\text{strain}}[/math]
  [math]E = \frac{F/A}{\Delta L/L}[/math]
  [math]E = \frac{F\times L}{\Delta L \times A}[/math]
  [math]\Delta L = \frac{F\times L}{E\times A}[/math]
  [math]\Delta L = \frac{(2\times 10^3)(50)}{(3.1\times 10^{-4})(200\times 10^9)}[/math]
  [math]\Delta L = \frac{100\times 10^3}{6.2\times 10^7}[/math]
  [math]\Delta L = 1.61\times 10^{-3}\ m[/math]
• The extension of each lift cable is approximately [math]1.61\times 10^{-3}\ m[/math]

• b) Ten people enter the lift. After 5 seconds the lift starts to move upwards and stops at the top floor of the building.

  The graph shows the total tension force in the cables during this time.

• 

• (i) Calculate the total mass of the people in the lift

  Weight of lift = 12 kN

• Solution:
• The forces operating on the lift whether it is at rest or travelling at a steady speed must be taken into account in order to determine the total mass of the passengers.
• According to the graph, the cable’s stress is [math]19.0\ kN[/math] while the lift is stationary (before 5 seconds and between 7 and 27 seconds).
• ⇒ Calculate the total weight to Newtons:
• The entire weight of the lift and its cargo equals the strain in the cable while the lift is stationary or travelling at a steady speed.
• [math]\text{Total weight} = \text{tension in cable} = 19.0\ kN[/math]
  [math]1\ kN = 1000\ N[/math]
  [math]19\ kN = 19\times 1000 = 19000\ N[/math]
  [math]W = mg[/math]
  [math]m = \frac{W}{g}[/math]
  [math]m = \frac{19000}{9.8}[/math]
  [math]m = 1938.8\ kg[/math]
• – Calculate the mass of the lift:
• [math]12\ kN = 12\times 1000 = 12000\ N[/math]
  [math]W = mg[/math]
  [math]m = \frac{W}{g}[/math]
  [math]m = \frac{12000}{9.8}[/math]
  [math]m = 1224.5\ kg[/math]
• [math]\text{Mass of people} = \text{total mass} – \text{mass of lift}[/math]
  [math]\text{Mass of people} = 1938.8 – 1224.5[/math]
  [math]\text{Mass of people} = 714.3\ kg[/math]
• The total mass of the people in the lift is approximately [math]714.3\ kg[/math].

• (ii)Explain the motion of the lift between 5 s and 29 s.

  Solution:

• The entire weight of the lift and its cargo equals the strain in the cable while the lift is stationary or travelling at a steady speed.
• 5 s to approximately 6.5 s:
• Over [math]19.0 kN[/math], the wires’ stress rises to [math]19.5\ kN[/math]. This suggests that the lift is increasing in speed.
• The upward tension force must be larger than the downward force of gravity (the weight of the lift and passengers) in order for there to be an upward acceleration.
• Approximately 6.5 s to 27 s:
• At [math]19.0 kN[/math], the wires’ tension remains constant. This indicates that the lift is travelling upward at a constant speed (or zero acceleration) as there is no net force acting on it.
• 27 s to 29 s:
• Below [math]19.0 kN[/math], the wires’ tension drops to [math]18.5 kN[/math]. This means that as it gets closer to the top level, the elevator is slowing down, or decelerating.
• When the upward tension force is less than the downward force of gravity, there is a downhill acceleration (or upward deceleration).

• (c) When the lift is empty, a technician removes one of the cables for maintenance. Assess how removing one cable would affect the extension of the remaining cables.

• Solution:
• – An empty lift is suspended by 6 identical steel cables.
• – The cables are of length [math]50\ m[/math]
• – One cable is removed for maintenance
• By using Hooke’s Law
  [math]F = kx[/math]
• – Total weight of the lift is distributed among the supporting cables.
  Determine the change in force on each of the remaining cables, then use Hooke’s Law to connect this change to the extension.
• Let [math]W[/math] be the weight of the empty lift
• Initial force per cable:
  [math]F_{\text{initial}} = \frac{W}{6}[/math]
• After removing one cable, 5 cables remain
  [math]F_{\text{final}} = \frac{W}{5}[/math]
• Now Compare the force and extension:
  [math]F_{\text{final}} > F_{\text{initial}}\\ \frac{1}{5} > \frac{1}{6}[/math]
• The extension of each remaining cable will rise since extension is related to force.
• When one cable is removed, the stress on the other cables rises, causing those cables to be extended farther.