Inferring Collision Type from Spring Compression
On a frictionless horizontal track, a 0.50 kg block moving at 4.0 m/s to the right strikes a stationary 1.50 kg block that is attached to an ideal spring of constant 200 N/m fixed to a wall on the right. The moving block does not stick to the second block; after the collision the second block compresses the spring and momentarily stops. Sensors record the maximum spring compression as 0.10 m. The collision lasts a short time compared with the subsequent spring motion. Determine what the compression implies about the collision rather than assuming in advance that it is elastic or perfectly inelastic.
1. Test the perfectly inelastic hypothesis: if the blocks stick together, find the common velocity and the predicted maximum spring compression. Compare with the observed 0.10 m.
Read the worked answer
For a perfectly inelastic collision, momentum conservation gives (0.50)(4.0) = (0.50 + 1.50)v, so v = 2.0/2.0 = 1.0 m/s.
The kinetic energy after the collision is K = 0.5(2.0)(1.0)² = 1.0 J. This energy compresses the spring: 0.5 k x² = 1.0, so x² = 2.0/200 = 0.010 and x = 0.10 m.
This matches the observed compression exactly. Therefore the data are consistent with a perfectly inelastic collision in which the blocks move together after impact.
2. Test the elastic hypothesis by finding the second block’s speed and the predicted compression. Does it match the observation?
Read the worked answer
For an elastic collision with a stationary second block, v2 = [2m1/(m1 + m2)]v1 = 2(0.50)/2.0 = (0.50)(4.0) = 2.0 m/s.
The second block’s kinetic energy is K2 = 0.5(1.50)(2.0)² = 3.0 J. The predicted compression satisfies 0.5(200)x² = 3.0, so x² = 0.030 and x = 0.173 m.
The predicted 0.173 m is much larger than the observed 0.10 m. An elastic collision would also leave the first block moving at v1 = (0.50 − 1.50)/2.0 = −2.0 m/s, rebounding to the left. The observed compression therefore rules out an elastic collision at the stated incoming speed.
3. Calculate the kinetic energy lost in the collision under the consistent interpretation and explain where that energy goes.
Read the worked answer
Initial kinetic energy: K_i = 0.5(0.50)(4.0)² = 4.0 J.
If the blocks move together at 1.0 m/s, final kinetic energy is K_f = 0.5(2.0)(1.0)² = 1.0 J. The loss is 3.0 J.
This lost mechanical energy becomes internal energy of the blocks and track surface, including deformation, sound, and thermal energy, during the short collision. Momentum is still conserved because the track is frictionless and external horizontal forces on the two-block system are negligible during the brief impact. Energy is not conserved as kinetic energy in an inelastic collision, so the 3.0 J loss is expected rather than a contradiction.