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.
A card of length 5.0 cm is attached to the trolley. The times to pass each gate are recorded.
| gate number | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| distance from start / m | 0.50 | 1.00 | 1.50 | 2.00 | 2.50 | 3.00 |
| time for card to pass / s | 0.100 | 0.071 | 0.058 | 0.050 | 0.045 | 0.041 |
| speed / m/s | 0.50 | 1.00 |
(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]
(a) Completing the speed table [2]
Speed at each gate = card length / time for card to pass = 0.050 m / time:
| Gate | Time / s | Speed / m/s |
|---|---|---|
| 1 | 0.100 | 0.050 / 0.100 = 0.50 |
| 2 | 0.071 | 0.050 / 0.071 = 0.70 |
| 3 | 0.058 | 0.050 / 0.058 = 0.86 |
| 4 | 0.050 | 0.050 / 0.050 = 1.00 |
| 5 | 0.045 | 0.050 / 0.045 = 1.11 |
| 6 | 0.041 | 0.050 / 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]
(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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