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 the h = 3.0 m-high, frictionless chute to a waiting truck. Unfortunately, the driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute. (Figure 1) Suppose the packages stick together. What is their common speed after the collision? Express, your answer to two significant figures and include the appropriate units. Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound? Express, your answer to two significant figures and include the appropriate units.

Explanation / Answer

Let the velocity of the package of m kg is v m/s as it reaches the bottom, By the law of energy conservation:-

=>PE(top) = KE(bottom)

=>mgh = 1/2mv^2

=>v = sqrt[2gh]

=>v = sqrt[2 x 9.8 x 3]

=>v = 7.67 m/s

1) By the law of momentum conservation:-

=>m1u1+m2u2 = (m1+m2) x v

=>m x 7.67 + 0 = 3m x v

=>v = 2.56 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 = -7.67 -------------(i)

BY the law of momentum conservation:-

=>m1u1 + m2u2 = m1v1 + m2v2

=>m x 7.67 + 0 = mv1 + 2mv2

=>v1+2v2 = 7.67 ---------------(ii)

By 2 x (i) + (ii):-

=>3v1 = -7.67

=>v1 = -2.56 m/s

Thus again by the law of energy conservation,Let the package of mass m gain h meter due the velocity of 2.56 m/s

=>PE(top) = KE(bottom)

=>mgh = 1/2mv1^2

=>h = v1^2/2g

=>h = (2.56)^2/(2 x 9.8)

=>h = 0.33m

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