Trolley on a Ramp: When the Ticker Tape Disagrees with Theory
A group of students set up a 2.00 m ramp inclined at 12.0° to the horizontal. They release a 1.20 kg trolley from rest and record its motion with a ticker-tape timer vibrating at 50 Hz. The tape shows four successive 0.10 s intervals of 0.80 cm, 2.40 cm, 4.00 cm and 5.60 cm. From these data the students calculate an acceleration of 1.60 m/s². Using g = 9.81 m/s², they predict a = g sin 12.0° = 2.04 m/s². The trolley’s wheels are slightly worn, one axle squeaks, and the ramp has a thin layer of dust. The teacher asks them to explain the gap and to decide whether the result supports the hypothesis that acceleration down a frictionless incline depends only on the angle. They must also state one improvement and one safety precaution.
1. Using the four tape intervals, calculate the trolley’s acceleration. Show clearly how you obtain the velocity at the midpoint of each interval and why the midpoints are used.
Read the worked answer
Each interval lasts 0.10 s. If acceleration is uniform, the average velocity over an interval equals the instantaneous velocity at the midpoint of that interval. Convert cm to m:
- Interval 1: s₁ = 0.0080 m, so v₁ = 0.0080 / 0.10 = 0.080 m/s at t = 0.05 s.
- Interval 2: s₂ = 0.0240 m, so v₂ = 0.0240 / 0.10 = 0.240 m/s at t = 0.15 s.
- Interval 3: s₃ = 0.0400 m, so v₃ = 0.0400 / 0.10 = 0.400 m/s at t = 0.25 s.
- Interval 4: s₄ = 0.0560 m, so v₄ = 0.0560 / 0.10 = 0.560 m/s at t = 0.35 s.
Using the first and last midpoints, Δt = 0.35 − 0.05 = 0.30 s, so
a = (v₄ − v₁) / Δt = (0.560 − 0.080) / 0.30 = 1.60 m/s².
Using consecutive intervals gives the same value, for example (0.240 − 0.080) / 0.10 = 1.60 m/s², so the tape is internally consistent. The measured acceleration is 1.60 m/s², which is less than the frictionless prediction of 2.04 m/s².
2. Explain the difference between 1.60 m/s² and 2.04 m/s² in terms of forces and energy. Decide whether the hypothesis is supported, and state what the students should conclude.
Read the worked answer
Along the slope, the component of weight down the ramp is mg sin θ; friction f acts up the ramp. Newton’s second law gives
ma = mg sin θ − f, so a = g sin θ − f/m.
The dusty ramp and worn axle increase f, so the measured acceleration is smaller than g sin θ. The shortfall is 2.04 − 1.60 = 0.44 m/s², so f/m = 0.44 N/kg. With m = 1.20 kg, f ≈ 0.53 N. In energy terms, the trolley’s gravitational potential energy is converted to kinetic energy plus heat and sound at the wheels and ramp surface, so less energy becomes kinetic energy and the acceleration is lower.
The hypothesis refers to a frictionless incline. This ramp is not frictionless, so the result does not support the hypothesis as stated. It does not disprove the general relationship either; it shows that friction is a significant extra factor. The students should conclude that the angle is not the only factor: surface condition and wheel friction also affect acceleration. They should reduce friction and repeat before making a general claim.
3. State one improvement to reduce the gap and one safety precaution. Justify each briefly.
Read the worked answer
Improvement: clean the ramp and use a trolley with ball bearings or an air track. This reduces f, so the measured acceleration should be closer to g sin θ. Repeating each run and averaging also reduces the effect of random timing errors.
Safety precaution: clamp the top of the ramp firmly to the bench and place a soft cushion or box at the bottom. The trolley gains speed and could fall off the bench or hit someone’s foot; clamping prevents the ramp from slipping and the cushion stops the trolley safely.