Two small smooth particles, **A** and B of mass **m** are attached to the two ends of an inelastic string of length **l** and are at rest on a smooth horizontal plane . The system now moves with speed u in the direction AB with the string taut. The particle B, after some time, collides with a small smooth elastic particle **C** of mass **M** which is at rest on the plane. If e is the coefficient of restitution between the particles **B** and **C** , show that the particle B moves with speed

(m- e M/m +M)*u after colliding with particle **C.**

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When the collision occurs both A and B travels at a speed of u. Once B collide with C the velocity of B will decrease and still A will have the previous velocity u.

Before collision;

`V_A = rarru`

`V_B = rarru`

`V_C = 0`

After collision

`V_A = rarru`

`V_B = rarr?`

`V_C = rarr?`

Using the conservation of momentum;

`mxxu+mxxu+Mxx0 = mxxu+mV_B+MV_C`

`mV_B+MV_C = mxxu -----(1)`

Using Newton’s law of restitution;

`V_B-V_C = -eu-----(2)`

`(2)xxM+(1)`

Here `MV_C` will get canceled.

`V_B(M+m) = -Meu+mxxu`

`V_B = (u(m-eM))/(M+m)`

** So velocity of particle B after collision is** `(u(m-eM))/(M+m)` .

*assumptions*

*The particles A and B travels in a same line*

*The collision happens in the same line as A,B travels and after collision the particles travel in same line.*

**Sources:**

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