Light Gates on a Low-Friction Ramp
A trolley is released from rest and rolls down a ramp inclined at 2.0° to the horizontal, measured with a digital inclinometer reading to ±0.1°. A card of width 2.00 cm is mounted on the trolley. Two light gates, separated by 0.600 m along the ramp, record the card blocking times t1 = 0.0860 s and t2 = 0.0345 s. The student calculates the average speed through each gate as w/t and then uses a = (v2² − v1²)/(2s) to obtain the acceleration. The student compares this with a = g sin θ and concludes that the trolley moves 'essentially without friction' because the two accelerations are the same order of magnitude. The mass of the trolley is 0.500 kg. The light-gate timer has a resolution of 0.0001 s.
1. Calculate v1, v2, the acceleration a, and the value of g sin θ. Hence estimate the coefficient of kinetic friction μ, stating one assumption.
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v1 = 0.0200/0.0860 = 0.233 m/s; v1² = 0.0541 m²/s².
v2 = 0.0200/0.0345 = 0.580 m/s; v2² = 0.3361 m²/s².
a = (0.3361 − 0.0541)/(2 × 0.600) = 0.2820/1.200 = 0.235 m/s².
g sin θ = 9.81 × sin 2.0° = 9.81 × 0.03490 = 0.342 m/s².
If the difference is attributed to constant kinetic friction, mg sin θ − μmg cos θ = ma, so μ = (g sin θ − a)/(g cos θ) = (0.342 − 0.235)/(9.81 × 0.999) = 0.011.
Assumption: friction is constant and air resistance is negligible; the angle is accurately known.
2. The student claims the motion is essentially frictionless. Evaluate this claim using your value of μ and the ±0.1° uncertainty in the angle. Is the discrepancy significant?
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The claim is not supported. The driving component is mg sin θ = 0.500 × 9.81 × 0.0349 = 0.171 N. The friction force implied by μ = 0.011 is μmg cos θ = 0.011 × 0.500 × 9.81 × 0.999 = 0.054 N. That is about 31% of the driving force - measurable, not negligible.
The ±0.1° angle uncertainty changes sin θ by about cos θ × (0.1° in radians) = 0.00174, so it changes g sin θ by about 0.017 m/s² (5%). The observed gap is 0.107 m/s², roughly six times that uncertainty. So the gap is significant compared with the angle uncertainty. The trolley is low-friction, but the data do not justify calling it frictionless.
3. A classmate says the acceleration is underestimated because v = w/t is only an average speed while the card passes the gate. Assess this criticism. Would a much wider card make the result worse?
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The criticism is not valid for constant acceleration. While the card passes a gate, the measured speed w/t is the average speed over the card width. For constant acceleration, that average equals the instantaneous speed at the instant the centre of the card passes the gate. The same card is used at both gates, so the card centre travels the gate separation s between those two instants. Substituting the two average speeds into (v2² − v1²)/(2s) therefore gives the correct acceleration to first order.
A much wider card would average over a longer time. If the acceleration were changing, that would smooth out the variation and could hide a real effect; a very wide card could also be affected by air resistance or flexing. But for the uniform acceleration assumed here, widening the card does not by itself create the bias the classmate suggests. The real issue is friction, not the card width.