Trolley on an Inclined Runway: Interpreting Ticker-Tape Data
A student investigates the acceleration of a trolley released from rest down a runway tilted at 8.0° to the horizontal. A ticker-timer, vibrating at 50 Hz, leaves dots on a paper tape attached to the trolley. The student selects five consecutive intervals, each containing five spaces between dots, so each interval represents 0.10 s. The measured lengths of these intervals, in order, are 2.4 cm, 3.6 cm, 4.8 cm, 6.0 cm and 7.2 cm. The mass of the trolley is 0.80 kg. The student expects the acceleration to equal g sin 8.0°, taking g = 9.8 m/s², but the value obtained from the tape is smaller. The runway is horizontal across its width, and the tape runs freely. Friction is present at the trolley axles and between the tape and the timer. The student repeats the run three times and obtains accelerations within 3% of one another.
1. Calculate the acceleration of the trolley from the tape and decide whether the five intervals are consistent with uniform acceleration.
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Each interval lasts 0.10 s. The average velocity in an interval is its length divided by 0.10 s:
- v1 = 0.024/0.10 = 0.24 m/s
- v2 = 0.036/0.10 = 0.36 m/s
- v3 = 0.48 m/s
- v4 = 0.60 m/s
- v5 = 0.72 m/s
These average velocities can be treated as the instantaneous velocities at the midpoints of the intervals. The velocity rises by 0.12 m/s every 0.10 s, so:
a = Δv/Δt = 0.12/0.10 = 1.2 m/s².
Equivalently, the interval lengths increase by 1.2 cm each 0.10 s, which is a constant second difference. Since the increments are equal, the tape is consistent with uniform acceleration. A reading uncertainty of about 0.1 cm per length would give roughly ±0.2 m/s², so the constancy is within experimental uncertainty rather than exact.
2. Compare the tape acceleration with g sin 8.0° and explain the difference in terms of forces.
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The expected acceleration is:
g sin 8.0° = 9.8 × sin 8.0° = 1.36 m/s².
The measured value, 1.2 m/s², is about 0.16 m/s² smaller. Along the runway, the component of weight is:
mg sinθ = 0.80 × 9.8 × sin 8.0° = 1.09 N.
The resultant force needed for a = 1.2 m/s² is:
ma = 0.80 × 1.2 = 0.96 N.
Therefore the total resistive force is about 1.09 − 0.96 = 0.13 N. This is consistent with axle friction and the drag of the tape through the timer. The difference is systematic, not random, because all repeats give a similar low value. Repeating the run would not remove it; the friction must be measured or compensated for.
3. The student proposes increasing the tilt to 20° so that friction becomes negligible. Evaluate this proposal.
Read the worked answer
At 20°, the expected acceleration would be g sin 20° = 3.35 m/s². The same friction force of about 0.13 N would then be a smaller fraction of the driving force: about 0.13/(0.80 × 3.35) = 4.9%, compared with about 12% at 8°. That would reduce the percentage error.
However, friction is not eliminated; it still causes a systematic offset. At a steeper angle the trolley may bounce or leave the runway, the tape may snatch, and the 0.10 s intervals may become too short to time well. The better judgement is to keep a moderate angle, measure the friction by tilting the runway until the trolley moves at constant velocity, and subtract that force, or to compare the velocity-time gradient with g sinθ while explicitly accounting for friction.