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 3m high frictionless chute to a waiting truck below. Unfortunately, the truck driver did not remove the previous package of mass 2m, now sitting at the bottom of the chute.?

a- Suppose the packages stick together. What is their common speed after the collision? Does the total kinetic energy stay conserved during this collision?

b- Suppose for a second scenario that the two packages collide with a perfect elastic collision. To what height does the first package rebound up again? What is the forward speed of the second package?

Explain work if possibile  

Explanation / Answer

The package of mass m will have a velocity of

v1i = sqrt(2gh)

v1i = 7.6681 m/s

Using conservation of momentum, the common speed after collision is

vf = 2.556 m/s   [ANSWER, PART A, common speed]

No, the total KE is not conserved as this is a perfectly inealistic collision. Only when the collision is elastic that the total KE is conserved.   [ANSWER, PART A]

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In the event that the collision is elastic, by conservation of momentum and energy,

v1f = 2.556 m/s back up the incline

Thus, it will go back

h = v1f^2 / 2g

h = 0.333 m [ANSWER, PART B, at what height]

v2f = 5.112 m/s [ANSWER, PART B, forward speed of the second package]

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