Please answer Part (a)

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)/2

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Before the collision;

`rarrV_A = u`

`V_B = 0`

After the collision;

`rarrV_A = u'`

`rarrV_B = v'`

Using momentum conservation at` rarr ` direction

`mu = mu'+2mv'`

`u = u'+2v' ---(1)`

Using Newtons law of restitution;

For sphere A,

`eu = u'-v' ----(2)`

By solving (1) and (2) we will get;

`u' = (u(1+2e))/3`

`v' = (u(1+e))/3`

So the velocity acquired by B after collision is `(u(1+e))/3` as required.

From momentum equation for sphere B;

`J = 2mv' = 2m*u((1+e))/3`

** So the impulse of the collision is** `J = (2m*u(1+e))/3`

**Sources:**

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Please note that ;

u' = (u(2e-1))/3

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