A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.
A. Suppose the packages stick together. What is their common speed after the collision?
Answer: __ m/s
B. Suppose the collision between the packages is elastic. To what height does the package of mass rebound?
Answer: __ cm
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A.) By the law of conservation of energy,
The mechanical energy at the top = mechanical energy at the bottom
=> m*g*h = (1/2)m*v^2
=> v(bottom, before coll.) =sqrt(2*g*h) = sqrt(2*9.8*3.0) = 7.668 m/s
This solution is provided in 195 words and uses the law of conservation of energy to find mechanical energy, as well as the collision and differences if the collision were elastic or inelastic.