Elastic Collision w/ Pendulum Bobs

[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.

Homework Equations

(conservation of linear momentum)
m1u1 + m2u2 = m1v1 + m2v2

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

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?

 

[ehild]

Homework Equations

(conservation of linear momentum)
m1u1 + m2u2 = m1v1 + m2v2

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

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

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

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.

(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

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