A package of mass m is released from rest at a warehouse loading dock and slides

Question

A package of mass m is released from rest at a warehouse loading dock and slides down a h = 2.5 m high frictionless chute to a waiting truck. Unfortunatley, the truck driver went on a break without having removed the previuos package, of mass 2m, from the bottom of the chute.

(a) Suppose the packages stick together. What is their common speed after the collision? _______m/s

(b) Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound? _____cm

Explanation / Answer

By the law of energy conservation:-
PE(top) = KE(bottom)
mgh = 1/2mv^2
v = sqrt[2gh]
v = sqrt[2 x 9.8 x 2.5]
v = 7 m/s

a) By the law of momentum conservation:-
m1u1+m2u2 = (m1+m2) x v
m x 7 + 0 = 3m x v
v = 2.33 m/s

2) Let the velocity of the m is v1 and the velocity of the 2m is v2 after the collision,by the energy conservation:-
v1 - v2 = u2 - u1
v1 - v2 = -7 -------------(i)

BY the law of momentum conservation:-
m1u1 + m2u2 = m1v1 + m2v2
m x 7.67 + 0 = mv1 + 2mv2
v1+2v2 = 7 ---------------(ii)

By 2 x (i) + (ii)

3v1 = -7
v1 = -2.33 m/s
Thus again by the law of energy conservation,Let the package of mass m gain h meter due the velocity of 2.33 m/s
PE(top) = KE(bottom)
mgh = 1/2mv1^2
h = v1^2/2g
h = (2.33)^2/(2 x 9.8)
h = 0.277m

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