(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-hfrictionlesschute 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 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?

2. Relevant equations

KE = PE

KE = 1/2 mv^{2}

PE = mgh

3. The attempt at a solution

V_{i}= 0m/s

Ek = Ep

1/2mv^{2}= mgh

The masses cancel each other out, leaving me with

v = [tex]\sqrt{}2gh[/tex]

v = 58.8m/s, this is the velocity at which the box 1 hits box 2

From this point i decided to use momentum to find the final velocity

(2m+m)v_{f}= mv_{i}

This simplifies to 3V_{f}= V/3 = 19.6m/s

Why would using momentum be flawed?

When I use KE is conserved throughout this system I get another answer... that is why one of the methods I'm using is wrong

1/2mv^{2}= 1/2mv^{2}

1/2(2m+m)v^{2}= 1/2mv^{2}

Doing some algebra I get that V^{2}= V^{2}/3

For this answer the Vf = 33.95m/s

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# Energy Conservation or Momentum?

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