A 1200 kg vehicle moving at 18 m/s strikes a barrier and rebounds at 2.0 m/s. Contact lasts 0.25 s. (a) Calculate momentum before and after, with initial direction positive. (b) Calculate impulse and average force. (c) Relate the force on the barrier to that on the vehicle. (d) A new barrier gives the same momentum change in 0.60 s. Calculate its average force, explain the safety advantage and state one model limitation.

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Accepted Answer

Momenta +21600 and −2400 kg m/s; impulse −24000 N s; average forces −96000 N and −40000 N. Initial momentum is 1200 × 18 = +21600 kg m/s. Rebound is opposite the chosen positive direction, so final momentum is 1200 × (−2.0) = −2400 kg m/s. Impulse equals final minus initial momentum: −2400 − 21600 = −24000 N s. The negative sign is physically important because the impulse points opposite the original motion. Average resultant force is Δp/Δt = −24000/0.25 = −96000 N. This average does not reveal the usually larger instantaneous peak force. Newton's third law gives equal-magnitude opposite-direction interaction forces on vehicle and barrier. They act on different objects, so they must not be cancelled on one free-body diagram. For 0.60 s, average force = −24000/0.60 = −40000 N. Greater stopping time reduces average force, but occupant safety also depends on peak acceleration, restraint, deformation distance and force history.

Vehicle rebound and collision safety

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