The method and the results are on the collisions lab. Do the analysis yourself first, then check the answers below.
Quick answer: In a collision, the total momentum of the colliding objects is the same before and after, provided no outside net force acts. These answers go with the AP Physics 1 lab on collisions: the graph, the calculations, the uncertainties and the conclusion explained, for all 7 analysis questions.
Model values are calculated from the sample times. Your own results will differ slightly, but the conclusions about conserved momentum, lost kinetic energy and the likely sources of error should be the same.
Level: AP Physics 1 (algebra-based) · Curriculum: AP P1 Unit 4
p = m v. Trial 1: before 0.50 × 0.800 = 0.400 kg m/s; after 0.75 × 0.526 = 0.395 kg m/s. Trial 2: before 0.500; after 0.492 kg m/s. Trial 3: before 0.625; after 0.612 kg m/s. Values agree to within about 2 %.
Question 2 Percentage difference
Difference = (pafter − pbefore) ÷ pbefore × 100: -1.3 %, -1.6 % and -2.0 %. These are within the timing uncertainty and the small friction loss, so momentum is conserved.
Question 3 Energy retained
Before: 0.160, 0.250, 0.391 J. After: 0.104, 0.161, 0.250 J. Fraction retained: 0.65, 0.64, 0.64. Theory: 0.50 ÷ 0.75 = 0.67. The measured fractions are close to the theory and slightly lower because of friction.
Question 4 Graph of momentum
Plot pafter on the vertical axis. The points lie on a straight line through the origin with a gradient close to 1. The intercept is zero because no motion before means no motion after.
Question 5 Timing uncertainty
Percentage uncertainty in t = 0.01 ÷ 0.50 × 100 = 2 %. Since u = 0.40 ÷ t, the percentage uncertainty in u is also 2 %, and the momentum before (m × u) has about 2 % uncertainty, because the mass uncertainty is negligible.
Question 6 Why is p after slightly low?
Friction and air drag act on the carts after the collision and slow them before they pass the second gate. The effect is reduced by raising the track end very slightly to compensate for friction, using an air track, or timing a short distance just after the collision.
Question 7 Centre of mass check
vcm = (0.50 × 0.800 + 0.25 × 0) ÷ 0.75 = 0.533 m/s. The measured speed of the joined carts is 0.526 m/s. They agree to within 1.3 %, as expected because the centre of mass velocity stays constant.
