A student investigates how the speed of a trolley changes as it rolls down a long ramp. The student places six light gates at equal intervals along the ramp...

Assessment: Physics 0625 | Paper 3 Mock 01 | Theory (Core) Subject: Physics - 0625

Question 1 Report

A student investigates how the speed of a trolley changes as it rolls down a long ramp. The student places six light gates at equal intervals along the ramp, each 0.50 m apart. The trolley is released from rest.

diagram

A card of length 5.0 cm is attached to the trolley. The times to pass each gate are recorded.

gate number123456
distance from start / m0.501.001.502.002.503.00
time for card to pass / s0.1000.0710.0580.0500.0450.041
speed / m/s0.501.00

(a) Complete the table by calculating the speed at gates 2, 3, 5 and 6. [2]

(b) Describe the pattern shown by the speed values. [1]

(c) Using the speed at gate 1 and gate 6, calculate the acceleration of the trolley. The time between the trolley passing gate 1 and gate 6 is 3.2 s. [2]

(d) Use the equation v² = u² + 2as to calculate the acceleration using the speed at gate 1 and gate 6 and the distance between them. [3]

(e) Compare your answers to (c) and (d) and comment on the agreement. [1]

(f) State two assumptions made in calculating the acceleration. [2]

(g) Explain why the card length should be as short as practical. [1]

Answer Details

(a) Completing the speed table [2]

Speed at each gate = card length / time for card to pass = 0.050 m / time:

GateTime / sSpeed / m/s
10.1000.050 / 0.100 = 0.50
20.0710.050 / 0.071 = 0.70
30.0580.050 / 0.058 = 0.86
40.0500.050 / 0.050 = 1.00
50.0450.050 / 0.045 = 1.11
60.0410.050 / 0.041 = 1.22

[1] for gates 2 and 3 correct; [1] for gates 5 and 6 correct.

(b) Pattern in the speed values [1]

The speed increases as the trolley moves further down the ramp. [1]

This is expected because gravity accelerates the trolley down the slope.

(c) Acceleration using v, u, and t [2]

Using the speeds at gate 1 (u = 0.50 m/s) and gate 6 (v = 1.22 m/s), with a time interval of 3.2 s:

\( a = \frac{v - u}{t} = \frac{1.22 - 0.50}{3.2} \) [1]

\( a = \frac{0.72}{3.2} = 0.23 \text{ m/s}^2 \) [1]

(d) Acceleration using v\(^2\) = u\(^2\) + 2as [3]

The distance between gate 1 and gate 6 is 3.00 - 0.50 = 2.50 m:

\( v^2 = u^2 + 2as \)

\( 1.22^2 = 0.50^2 + 2 \times a \times 2.50 \) [1]

\( 1.4884 = 0.25 + 5.0a \) [1]

\( 5.0a = 1.2384 \)

\( a = \frac{1.2384}{5.0} = 0.25 \text{ m/s}^2 \) [1]

(e) Comparison of the two values [1]

The two values (0.23 m/s\(^2\) and 0.25 m/s\(^2\)) are similar, suggesting the acceleration is approximately uniform. The small difference is due to measurement uncertainties in the timing. [1]

(f) Two assumptions [2]

  1. The acceleration is uniform (constant) along the ramp. [1]
  2. Friction is constant along the ramp. [1]

(g) Why the card should be as short as practical [1]

A shorter card passes through the gate more quickly, so the measured speed is closer to the instantaneous speed at that point on the ramp. [1]

A long card gives an average speed over the length of the card, which is less precise when the trolley is accelerating.

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