Trolley, Ticker-Tape and a Claim About Inertia
A student investigates Newton's second law with a 0.500 kg trolley on a horizontal runway, pulled by a string over a pulley by a 0.100 kg hanging mass. A ticker-tape timer vibrating at 50 Hz marks the tape, and the displacements in four successive 0.20 s intervals are 4.0 cm, 10.4 cm, 16.6 cm and 23.0 cm. Taking g = 10 m/s^2, the student predicts an acceleration of 1.67 m/s^2 and is puzzled that the tape gives a smaller value. An extra 0.500 kg is then fixed to the trolley, and the measured acceleration falls to 0.85 m/s^2. The student writes in the report: 'Acceleration halved when the mass was doubled, so a is inversely proportional to m, confirming F = ma.' The teacher comments that the conclusion is only partly justified, and that the tension in the string is not 1.0 N.
1. Use the tape to calculate the acceleration of the trolley, and explain why it is less than the 1.67 m/s^2 the student predicted.
Read the worked answer
The timer vibrates 50 times a second, so each interval is 0.02 s and ten ticks span T = 0.20 s.
For uniformly accelerated motion from the first and last displacements:
a = (s4 - s1) / (3T^2) = (0.230 - 0.040) m / (3 x 0.040 s^2) = 0.190 / 0.120 = 1.58 m/s^2
The shortfall is about 5%, and the student has ignored two effects:
- Friction in the pulley bearing and between trolley and runway opposes the motion. A net force of 0.500 x 0.09 = 0.045 N is enough to account for the gap.
- The 1.67 m/s^2 prediction assumes the string pulls with the full weight of the hanging mass (1.0 N). It does not, because the hanging mass is itself accelerating downwards (see the last task), so the actual driving force is only about 0.84 N.
A smaller third effect is the drag of the tape through the timer. The measured value is therefore not a failure of F = ma: the net force on the trolley is 0.500 x 1.58 = 0.79 N, which is consistent with the true tension minus friction.
2. Evaluate the student's conclusion: 'Acceleration halved when the mass was doubled, so a is inversely proportional to m, confirming F = ma.'
Read the worked answer
It is partly right, but for the wrong reason.
The mass that is actually being accelerated is the whole moving system: trolley plus hanging mass. That goes from 0.600 kg to 1.100 kg, a factor of 1.83, not 2. The acceleration ratio is 1.58 / 0.86 = 1.84, which matches the mass ratio closely, so a is roughly proportional to 1/m at constant driving force.
Problems with the reasoning:
- Two points prove nothing. Many curves pass through two points. A proper test uses five or six loads and plots a against 1/m, checking that the graph is a straight line through the origin.
- The driving force is not exactly constant, because the tension changes when the acceleration changes.
- Friction is a bigger fraction of the smaller net force, so the two points are not affected equally.
A prediction from F = ma with 1.0 N driving and about 0.05 N friction gives a = 0.95/1.10 = 0.86 m/s^2, which agrees well. So the data are consistent with Newton's second law, but the student's argument does not establish it. The phrase 'halved when doubled' is a coincidence of the numbers chosen.
3. Explain why the tension in the string is not equal to the weight of the 0.100 kg hanging mass, and calculate the tension.
Read the worked answer
The hanging mass is accelerating downwards at 1.58 m/s^2, so the forces on it are not balanced. Taking downwards as positive:
mg - T = ma T = m(g - a) = 0.100 x (10 - 1.58) = 0.84 N
Its weight is 1.0 N, so the string pulls the trolley with only 0.84 N.
Check on the trolley: T - friction = 0.84 - 0.05 = 0.79 N = 0.500 x 1.58. Consistent.
The physics is that the hanging mass has to be accelerated, so part of its own weight is 'used up' in giving it that acceleration. For accurate work the hanging mass should be much smaller than the trolley, so that a is small and T is close to mg. Here it is one-sixth of the trolley mass, which is marginal, and the student should say so in the evaluation.