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 = 2.4 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.

Express your answer to two significant figures and include the appropriate units

A) Suppose the packages stick together. What is their common speed after the collision?

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

Explanation / Answer

Speed of mass M just before hitting mass 2M,
Vo = ( 2 gh)1/2   ( h is height from where it slides on frictionless chute)

A) Applying momentum conservation during collision
   MVo = ( M+2M) V ( V is common velocity of two masses after collision)
   V = Vo/3 = (2gh)1/2 / 3 = 2.3 m/s

B) Applying momentum conservation during collision
MVo = MV1 + 2MV2 ( V1 and V2 are velocities of mass M and 2M respectively) ......1
   As collision is elastic , relative velocoties between masses before and after collision are equal. Hence
   Vo = V2 - V1 ......2
   Eliminating V2 from equations 1 and 2 , we get
   V1 = - Vo/3 ( -ve sign indicates mass moves back along the chute )
Height of mass M after collision = V12/2g = Vo2/18g = h/9 = 0.27 m

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