# Elastic Collision w/ Pendulum Bobs

1. Oct 16, 2011

### paola8

1. A pendulum bob of mass m is released from a height H above the lowest point. It collides at the lowest point with another pendulum of the same length but with a bob of mass 2m initially at rest.

Find the heights to which the bobs rise given that the collision is (A) Completely Inelastic, (B) Perfectly Elastic.

2. Relevant equations
(conservation of linear momentum)
m1u1 + m2u2 = m1v1 + m2v2

(conservation of energy)
.5mv^2 + mgh = .5mv^2 + mgh

3. The attempt at a solution
initial PE = mgH
= final KE at lowest point
= 0.5 mu^2
(A) Completely Inelastic
total mass = 3m
final PE
= 3m g *h

mgH = 3mg h
h = H/3
***answer is supposed to be H/9??***

(B) Perfectly Elastic
mgH= 0.5 mu^2

0.5 *m *u^2 + 0.5 * 2m *0^2 =0.5 * m * 0^2 + 0.5 * 2m * v^2
0.5 *m *u^2 = 0.5 * 2m * v^2
0.5 * 2m * v^2 = mgH

KE of 2m at lowest point = PE at height point
0.5 * 2m * v^2 = 2m g h1
mgH = 2mg h1
h1 = H/2
***answer is supposed to be H/9, and 4H/9??***

Could anyone help me as to why I've got the wrong answers?

Last edited: Oct 16, 2011
2. Oct 17, 2011

### ehild

Yes, it is a big truth that everything is identical with itself...

You supposed that the final potential energy is equal to the initial one, that is energy is conserved, but it is not true for an inelastic collision.

The final velocity of the lighter mass will not be zero: It does not collide with an equal mass.

ehild