Question

(7 points) A package of mass 2m is released from rest at a warehouse loading dock at height h and slides down a frictionless chute. Unfortunately, the previous package (of mass m) was not removed, from the bottom of the chute. a) (1 point) What is the potential energy of the package (of mass 2m) at the top of the chute (relative to the dashed line)? 2m b) (I point) What is the speed V of the package of mass m at the bottom of the chute (just before hitting the package of mass m)? c) (2 point) Suppose the packages stick together. What is their common speed after the collision? d) (3 point) Now suppose the two packages collide elastically. What are the velocities of packages after the collision? (To do this find the velocity of center of mass, calculate initial and final velocities of the packages in center of mass, and then find the final velocities of the packages in the lab frame.)

Answer 1

a)

PE = mgh = 2m gh

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b)

using 3rd equation of motion

u^2 = 2 g h

u = sqrt ( 2gh)

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c)

using conservation of momentum

m1 u1 + m2 u2 = (m1 + m2) v

2 m sqrt(2 gh) = (3m) v

v = (2/3) sqrt ( 2gh)

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d).

final velocity of m1

v1 =( m1 - m2) u1 / ( m1 + m2)

v1 = m sqrt (2 gh) / 3m

v1 = sqrt(2gh) / 3

final velocity of m2

v2 = 2 m1 u1 / ( m1 + m2)

v2 = (4/3) sqrt(2 gh)

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