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 = 4.0 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.

Part A

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.

Part B

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

For the package , Using conservation of energy

Kinetic energy at bottom = potential energy at top

(0.5) m v2 = mgh

v = sqrt(2gh)

v = sqrt(2 x 9.8 x 4) = 8.85 m/s

using conservation of momentum

m = mass of package released

2m = mass of package at bottom

initial momentum of mass package = final momentum of combination

m v = (2 m + m) Vf

Vf = V/3 = 8.9/3 = 2.9 m/s

B)

v1 = final speed of package "m"

v2 = final speed of package "2m"

For perfectly elastic collision :

V2 - V1 = v -0

V2 = V1 + 8.85

using conservation of momentum

mv = mV1 + (2m) V2

8.85 = V1 + 2 (V1 + 8.85)

v1 = - 2.95 m/s

height gained can be given as

h = v12 /2g = 2.952 /(2 x 9.8)

h = 0.44 m

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