The Trolley That Accelerates Too Slowly: Where Did 0.13 N Go?
In a school laboratory a group sets up a 1.20 m dynamics runway with a pulley at the raised end. To compensate for friction they slip a 2 mm card under one leg, give the trolley a gentle push, and judge by eye that it keeps a steady speed. The trolley (0.500 kg) is then tied to a 0.100 kg hanging mass. A 0.020 m card on the trolley passes two light gates placed 0.400 m apart, which record 0.55 m/s and 1.20 m/s. The group predicts a = (0.100 × 9.81)/0.600 = 1.64 m/s² but the gate readings give about 1.42 m/s². They conclude that friction is the cause and quote a friction force of 0.13 N. A second student objects: the pulley has to turn, the string may not be parallel to the runway, and the runway was checked once, by eye, with the trolley moving. The teacher asks for a decision on what the missing 0.13 N actually is and how the experiment should be redone.
1. Calculate the acceleration implied by the two light-gate speeds and the force needed to explain the difference from 1.64 m/s². Decide whether friction between trolley and runway can be that force, given the way the runway was compensated, and state what else must contribute.
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Acceleration from the gates. For uniform acceleration, v² = u² + 2as, so
a = (1.20² − 0.55²)/(2 × 0.400) = (1.44 − 0.3025)/0.800 = 1.42 m/s².
Force bookkeeping. The only external driving force on the (trolley + hanging mass) system is the weight of the hanging mass, 0.100 × 9.81 = 0.981 N. The expected acceleration is 0.981/0.600 = 1.64 m/s². The observed acceleration needs only 0.600 × 1.42 = 0.852 N, so about 0.13 N is missing.
Can runway friction explain it? Only if the compensation is imperfect. The runway was tilted so that m₁g sinθ = f for the trolley alone, so a freely moving trolley should keep a steady speed. Adding the string barely changes the normal force, so f and m₁g sinθ still cancel, and runway friction cannot produce the shortfall in that case.
The 0.13 N must then come mainly from the pulley: bearing friction (a torque that acts as a backward force at the rim) and the pulley's rotational inertia, which behaves like extra mass m_p/2 on the string. A pulley of about 0.18 kg would explain the whole deficit by inertia alone, which is heavy for a laboratory pulley, so bearing friction is probably significant too. The string being non-parallel changes the normal force and hence the friction; and the compensation was only eyeballed, so some genuine friction cannot be ruled out. The honest conclusion is that the shortfall is not evidence for runway friction.
2. The card on the trolley is 0.020 m wide, so each gate times the card's passage and divides width by time. Does this make the recorded speeds systematically too low, too high, or unbiased? Does the 0.400 m separation need correcting?
Read the worked answer
What the gate measures. Each reading is the average velocity of the card during the passage, that is, over a 0.020 m stretch.
Bias. For motion with constant acceleration, the average velocity over a time interval equals the instantaneous velocity at the middle of that interval in time. The middle of the card's passage in time is the instant when the card's centre is level with the gate; for constant acceleration the mid-time position is also the mid-point in space, so the recorded value is the speed of the card centre exactly as it crosses the gate. To first order the reading is unbiased - it is neither systematically low nor high.
A residual second-order effect appears only if the acceleration changes markedly over the passage or if the card is long compared with the distance travelled. Here 0.020 m against 0.400 m makes it well below 0.1%, far smaller than the reading scatter.
Distance. The card centre travels from gate 1 to gate 2, a displacement of exactly 0.400 m, so the separation needs no correction.
What does matter. Alignment: the card must be vertical and the gates square to the runway, otherwise the timed width is not the true 0.020 m. The blocking times are only 36 ms and 17 ms, so a gate that clips the card edge or a card with a ragged edge produces a larger error than the card width itself.
3. Explain how to test whether the runway is really compensated, and describe a procedure that would let the group separate the pulley's effect from friction.
Read the worked answer
Testing the compensation. Remove the string. Give the trolley a gentle push and record the speed at both gates; if the runway is truly compensated the two speeds should agree within the reading uncertainty (about ±0.01 m/s here). Repeat in the opposite direction. A runway that looks level, or that was judged by eye once, is not evidence.
Separating friction from the pulley. Repeat the run with several hanging masses, for example 0.050, 0.100, 0.150 and 0.200 kg, and plot the measured acceleration a against m₂g. The model is
m₂g − f = (m₁ + m₂ + m_p/2)a,
so the graph should be a straight line with gradient 1/(m₁ + m₂ + m_p/2) and intercept −f/(m₁ + m₂ + m_p/2). A best-fit line through five or six points separates the constant force f (an intercept) from the extra inertia of the pulley (a change of gradient) - which two readings can never do. If the fitted gradient corresponds to an inertial mass greater than m₁ + m₂, the excess is the pulley's effective mass m_p/2, and its size identifies whether the pulley is the problem.
Practical fixes. Use a light, well-lubricated pulley; keep the string parallel to the runway, checked with a set square; and re-check the compensation after changing the hanging mass, because any vertical component of the tension alters the normal force and therefore the friction.