OCR A Physics OCR Paper 1 2022
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SECTION A
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You should spend a maximum of 30 minutes on this section.
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Write your answer to each question in the box provided.
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Answer all the questions.
- 1. A student has constructed the table below of possible scalar and vector quantities.
- Answer: D
- Explain:
- – Mass is a scalar quantity because it only has magnitude, while centripetal force is a vector quantity because it has both magnitude and a specific direction (towards the center of the circular path).
- – In the other options, acceleration, displacement, and amplitude/wavelength are incorrectly categorized.
- 2. The diameter of a wire is measured in five different places along its length. The results are shown below.
- [math]1.92\ \text{mm}[/math]
- [math]1.88\ \text{mm}[/math]
- [math]1.90\ \text{mm}[/math]
- [math]1.86\ \text{mm}[/math]
- [math]1.89\ \text{mm}[/math]
- What is the absolute uncertainty in the diameter of this wire?
- A. [math]0.01\ \text{mm}[/math]
- B. [math]0.03\ \text{mm}[/math]
- C. [math]0.05\ \text{mm}[/math]
- D. [math]0.06\ \text{mm}[/math]
- Answer: B
- Explain:
- The absolute uncertainty of a repeated set of measurements can be estimated using half of the range:
- [math]\text{Absolute Uncertainty}=\frac{\text{Maximum Value}-\text{Minimum Value}}{2}[/math]
- [math]\text{Maximum Value}=1.92\ \text{mm}[/math]
- [math]\text{Minimum Value}=1.86\ \text{mm}[/math]
- [math]\text{Absolute Uncertainty}=\frac{1.92-1.86}{2}[/math]
- [math]\text{Absolute Uncertainty}=\frac{0.06}{2}[/math]
- [math]\text{Absolute Uncertainty}=0.03\ \text{mm}[/math]
- 3. A student has plotted a velocity against time graph for a trolley moving down a ramp.
- Which of the following pair of quantities can be determined from the gradient of the graph and the area under the graph?
- A. acceleration, displacement
- B. acceleration, impulse
- C. displacement, kinetic energy
- D. force, work done
- Answer: A
- Explain:
- – The gradient of a velocity-time graph represents the rate of change of velocity, which is acceleration.
- [math]a=\frac{\Delta x}{\Delta t}[/math]
- – The area under a velocity-time graph represents the product of velocity and time, which is displacement.
- [math]s=vt[/math]
- 4. The diagram below shows the directions and magnitudes of the three forces acting on an object at a specific time as it moves through water.
- The weight of the object is [math]1.20\ \text{N}[/math], the upthrust on the object is [math]0.80\ \text{N}[/math] and the drag is [math]0.60\ \text{N}[/math].
- Which statement is correct about this object at this specific time?
- A. It has reached its terminal velocity.
- B. It is accelerating.
- C. It is decelerating.
- D. It is moving upwards.
- Answer: C
- Explain:
- [math]\text{Total Upward Force}=\text{Upthrust}+\text{Drag}[/math]
- [math]\text{Total Upward Force}=0.80\ \text{N}+0.60\ \text{N}[/math]
- [math]\text{Total Upward Force}=1.40\ \text{N}[/math]
- [math]\text{Total Downward Force}=\text{Weight}[/math]
- [math]\text{Total Downward Force}=1.20\ \text{N}[/math]
- The net force is [math]1.40\ \text{N}-1.20\ \text{N}=0.20\ \text{N}[/math] directed upwards.
- – Since fluid drag always opposes the direction of motion and is pointing upwards, the object must currently be moving downwards.
- – Because the net force (upwards) opposes the direction of motion (downwards), the object is decelerating.
- 5. The object below is in equilibrium.
- A force, not shown on the diagram, also acts on the object at point P.
- Which of the following shows the correct direction and magnitude of the force acting at point P?
- A.
B. 
C. 
D. - Answer: B
- Explain:
- The two vertical 10 N forces point in opposite directions and completely balance each other out vertically. For the entire object to remain in translational equilibrium, the net force must be zero. Therefore, the unknown force acting at point P must exactly counteract the remaining 5.6 N force. Since the top force is 5.6 N pointing up and to the left at 30º to the horizontal, the balancing force at P must point in the exact opposite direction: down and to the right at 30º below the horizontal line, as shown in option B.
- 6. A particle X of mass m collides with a stationary particle Y of mass 4m.
- Immediately after the collision the particle X is moving at velocity v1 at an angle of 60° to its original direction and the particle Y is moving with velocity v2 at 90° to the velocity of particle X.
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- What is the value of the ratio [math]\frac{V_1}{V_2}[/math]?
- A. 2.3
- B.3.9
- C. 4.0
- D. 6.9
- Answer: A
- Explain:
- – Let the original direction of particle X be along the horizontal axis.
- – Since particle Y moves at 90º 𝑟𝑒𝑙𝑎𝑡𝑖𝑣𝑒 𝑡𝑜 𝑝𝑎𝑟𝑡𝑖𝑐𝑙𝑒 𝑋 (𝑤ℎ𝑖𝑐ℎ 𝑖𝑠 𝑎𝑡 + 60º), particle Y must move at an angle of
- 60º − 90º = −30º
- relative to the horizontal axis.
- – Applying the conservation of linear momentum perpendicular to the initial direction of motion (vertically)
- [math]
\begin{aligned}
0 &= mv_1\sin 60^\circ – 4mv_2\sin 30^\circ \\
mv_1\sin 60^\circ &= 4mv_2\sin 30^\circ \\
v_1\left(\frac{\sqrt{3}}{2}\right) &= 4v_2\left(\frac{1}{2}\right) \\
v_1\sqrt{3} &= 4v_2 \\
\frac{v_1}{v_2} &= \frac{4}{\sqrt{3}}
\end{aligned}
[/math] - 7. A metal block of mass m is heated by an electric heater.
- The graph of temperature θ against time t for this block is shown below.
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- The power of the heater is P. The gradient of the straight-line graph is G.
- What is the correct expression for the specific heat capacity c of the metal?
- A. 𝒄 = 𝑮
- B. [math]c = \frac{PG}{m}[/math]
- C. [math] c = \frac{mP}{G} [/math]
- D. [math] c = \frac{P}{mG}[/math]
- Answer: D
- Explain:
- – The thermal energy supplied by the heater in time t is given by
- 𝑄 = 𝑃. 𝑡
- – The heat energy absorbed by the metal block is
- 𝑄 = 𝑚𝑐∆𝜃
- – The gradient G of the temperature – time graph is
- [math]
\begin{aligned}
G = \frac{\Delta\theta}{t} \\
\text{Meaning: } \Delta\theta &= Gt \\
\text{Substituting this back into the equation:} \\
Pt &= mc(Gt) \\
P &= mcG \\
c &= \frac{P}{mG}
\end{aligned}
[/math]