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## Homework Statement

When two balls are dropped like shown in situation 1 in the figure, the top ball shoots up after the impact, while the bottom ball loses some of its mechanical energy, as seen in situation 3. By simplifying the situation, we can assume that the bottom ball stops and bounces up before the top ball, so we get an elastic collision between the two, as shown in situations 2 and 4. If the mass of the small ball is [tex]m[/tex] and the mass of the large ball is [tex]M[/tex] and they have the same speed [tex]v[/tex] before the collision and velocities [tex]u[/tex] and [tex]U[/tex] after, what is [tex]u[/tex] is terms of [tex]m,M,v[/tex]?

Problem may be rewritten and simplifyed as:

Two spheres with masses [tex]m[/tex] and [tex]M[/tex] collide head on with the same speed [tex]|v|[/tex] and exit the collision with velocities [tex]u[/tex] and [tex]U[/tex]. The collision is elastic. Express [tex]u[/tex] in terms of [tex]m[/tex], [tex]M[/tex] and [tex]v[/tex]

http://img401.imageshack.us/img401/4555/fysikkkollisjon.png [Broken]

## Homework Equations

Definition of momentum: [tex]p=mv[/tex] (1)

Conservation of momentum: [tex]\Delta \Sigma p=0[/tex] (2)

For elastic collisions:

[tex]v_1+u_1=v_2+u_2[/tex] (3) (v=initial velocity , u=velocity after collision)

## The Attempt at a Solution

The problem is one-dimentional, so I have omitted the usual vector notation and instead defined a positive direction, to the right in situation 4 in the figure.

Inserting values into the equations, we get

(2)[tex]Mv-mv=MU+mu[/tex]

[tex]U=\frac{Mv-mv-mu}{M}[/tex]

(3)[tex]u-v=v+U[/tex]

substituting for U:

[tex]u-v=v+v-\frac{m}{M}(v-u)[/tex]

[tex]u+\frac{m}{M}u=3v-\frac{m}{M}v[/tex]

[tex]u(1+\frac{m}{M})=v(3-\frac{m}{M})[/tex]

[tex]u=v\frac{3-\frac{m}{M}}{1+\frac{m}{M}}[/tex]

So I have a solution. Is it reasonable? I think so, on the grounds that [tex]u=v[/tex] when [tex]m=M[/tex].

What do you think?

Thanks for any help.

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