Please answer Part (c)

A small smooth sphere A of mass m moving on a smooth horizontal table with velocity u collides directly with another small smooth sphere B, of equal size and mass 2m, which is at rest on the table.The coefficient of restitution is e.

(a)Show that the velocity acquired by B is (1+e)u/3, and find the impulse J between the spheres.

(b)Express the loss of kinetic energy due to the impact in the form E =J(1-e)u/2

(c)If the direction of motion of A goes reversed due to the impact,;show that e>1/2 and E<(1mu^2)/4

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From part a we have got that after collision;

Velocity of A = u' = `((2e-1)u)/3` at the opposite direction of its initial travel.

So Actually if it happens that means (2e-1)>0.

2e-1>0

2e>1

e>1/2

So to reverse A after collision `e>1/2` .

From part (a) we got that `J = (2m*u(1+e))/3 `

From part (b) we got that `E = (J(1-e)u)/2`

`E = (2m*u(1+e))/3xx(u(1-e))/2`

`E = (m*u^2(1-e^2))/3`

We have obtained that` e>1/2`

So we can say;

`E<(m*u^2(1-(1/2)^2))/3`

`E < m*u^2/4`

** So the answer is** `E<1/4m*u^2`

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