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Three balls A, B and C(m(A)=m(C)=4m(B)) ...

Three balls `A, B` and `C(m_(A)=m_(C)=4m_(B))` are placed onn a smooth horizontal surface. Ball `B` collides with ball `C` with an initial velocity `v` as shown in figure. Find the total number of collision betwenent the balls (all collisions are elastic).

A

one

B

two

C

three

D

four

Text Solution

Verified by Experts

The correct Answer is:
B
Let `V_1` and `V_2` velocities of B and C after first collision towards right
`m_B = m, m_A = m_C = 4m`
`implies mV = mV_1 + 4mV_2 and V_2 - V_1 = V implies V_1 = (-3V)/(5) , V_2 = (2V)/(5)`
and Let `V_3 and V_4` be velocity of A and B after 2nd collision towards left.
`implies (3 mV)/(5) = 4 mV_3 + mV_4 and V_3 - V_4 = (3V)/5 implies V_3 = (6V)/(25) and V_4 = (-9V)/(25)`
`implies B moves toward right again but it wont collide with C again as `V_4 < V_4`.
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