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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.
`V_A = rarru`
`V_B = rarru`
`V_C = 0`
`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)`
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)` .
- 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.
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