A 0.200 kg clay block moving at 6.00 m s⁻¹ strikes and sticks to a stationary 0.800 kg cart on a level low-friction track. The combined object then compresses a spring of constant 180 N m⁻¹ until momentarily at rest. Find the speed just after collision, the mechanical energy lost during impact, and the maximum compression. Explain why momentum is used during impact but mechanical energy is used during compression, and propose a measurement that tests the spring model.

Question

Accepted Answer

The combined speed is 1.20 m s⁻¹, collision energy loss is 2.88 J, and maximum compression is 0.0894 m. Momentum is approximately conserved over the short collision; post-collision kinetic energy becomes spring energy afterward. During the brief collision, the spring has not yet compressed appreciably and external horizontal impulse is negligible. Momentum conservation gives (0.200)(6.00) = (1.000)v, hence v = 1.20 m s⁻¹. Initial kinetic energy is ½(0.200)(6.00)² = 3.60 J. Immediately after impact it is ½(1.000)(1.20)² = 0.720 J. The collision is perfectly inelastic, so the lost mechanical energy is 3.60 − 0.720 = 2.88 J. It becomes deformation, thermal energy and sound rather than disappearing. After impact, negligible track losses let 0.720 J convert into elastic energy: ½kx² = 0.720. Thus x = √(1.44/180) = 0.0894 m. Measure force against compression slowly with a calibrated force sensor. A straight F–x graph through the origin supports constant k; the area beneath that graph should match the stored energy used here.

Collision followed by spring compression

What a complete response should do

How did this one feel?