E_o=\frac{1}{2}mv_1^2Solve Elastic Collision: $\frac{\triangle E}{E_o}$

[thereddevils]

Homework Statement

A body of mass, m makes a head on perfectly elastic collision with a body of mass, M initially at rest. Show that

$$\frac{\triangle E}{E_o}=\frac{4(\frac{M}{m})}{(1+\frac{M}{m})^2}$$

Homework Equations

The Attempt at a Solution

Momentum: $$mu_1=mv_1+Mv_2$$

$$\triangle E= \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2-\frac{1}{2}mu_1^2$$

 

[Borek]

A body of mass, m makes a head on perfectly elastic collision with a body of mass, M initially at rest.

$$\triangle E= \frac{1}{2}mv_1^2+\frac{1}{2}Mv_2^2-\frac{1}{2}mu_1^2$$

If collision is ellastic, delta E as defined (final minus initial, right?) should equal zero.

Or am I missing something?

 

[thereddevils]

If collision is ellastic, delta E as defined (final minus initial, right?) should equal zero.

Or am I missing something?

Thanks Borek! Exactly, i think it's meant to be inellastic collision. I'll give it a try.