43 L A package of mass m is released from rest at a warehouse loading dock and s

Question

43 L A package of mass m is released from rest at a warehouse loading dock and slides down the 3.0-m-high, frictionless chute of FIGURE P10.43 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. a. Suppose the packages stick together. What is their commorn speed after the collision? b Suppose the collision between the packages is perfectly elas tic To what height does the mackage of mass m rebound? b. sli

Explanation / Answer

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apply the law of conservation of energy as

total mehanial eenrgy at top = total mehanial energy at bottom

m g h = 0.5 mv^2

V^2 = 2 g h

v^2 = 2* 9.81 * 3

v = 7.67 m/s

Now for Collision,

momentum before Collision = momentum after Collision

so

m * 7.67 = 2m* V2

V2 = 3.83 m/s
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part B:

if the Collision is perfectly elastic, rebound height is

m v = mv1 + 2m* V2

7.67 = V1 + 2V2 ---------------1

for elastic collison e =1

V1 - v2 = -e*(u1-U2)

V1-V2 = -1*(v-0)

V1 - V2 = -7.67----------------2

solving 1 and 2

v1 , the veloity of rebound mass = -2.55 m/s

from enery conservation

0.5 mv^2 = mg h

h = v^2/2g

h = 2.55^2/(2* 9.81)

h = 0.331 m
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