A package of mass 4.0 kg is released from rest at the top of a frictionless load

Question

A package of mass 4.0 kg is released from rest at the top of a frictionless loading dock. The worker at the bottom of the loading dock isn't paying attention. The 4.0 kg package slides down the loading dock and hits a 5.0 kg package wh.ch was sitting at rest at the bottom of the loading dock. The two packages move together after the collision. The top of the loading dock is 3.0 m above the bottom of the loading dock. How fast are the packages moving after the collision? Use conservation laws to answer the question.

Explanation / Answer

The kinetic energy at the bottom is equal to the potential energy at the top .
m g h = (1/2) m v2
From this the velocity of the 4 kg package at the bottom is
v = sqrt (2 g h) = sqrt (2 x 9.8 x 3)
v = 58.8 m/s
This is the velocity of the package before hitting the 5 kg package
Using the conservation of linear momentum
m v =(m + M) V
Where M is the 5 kg mass , m is 4 kg mass , V is the final velocity of the combined mass
V = m v / (m +M) = 4 x 58.8 / (4 +5 )
V = 26.13 m/s

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