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Two small smooth particles, A and B of mass m are attached to the two ends of an...
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.
1 Answer | add yours
- 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.
Best answer as selected by question asker.
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)` .
Posted by jeew-m on August 18, 2013 at 10:40 AM (Answer #1)
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