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Question

Two balls are fastened to two light strings of same length of l and of different masses m and 2m as shown in the figure. The other end of the strings are fixed at O. The strings are kept in the same horizontal line and the system is released from rest. The collision between the balls is perfectly elastic. Find the ratio of the speeds of the balls, just after the collision.


A
5:1
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B
2:1
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C
2:1
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D
5:1
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Solution

The correct option is A 5:1
Just before collision:


From WPE - Initial TE = Final TE
mgl=12mv2
v=2gl, so just before collision speed of both balls will be same.
No horizontal force is present so momentum along the horizontal direction will be conserved.
Just after collision:


Assumed just after collision velocity of 2m & m masses are v1 & v2 respectively.
Fom momentum conservation we have
2mvmv=2mv1+mv2
v=2v1+v2(1)
Collision is elastic, hence, e=1
e=Velocity of seperationVelocity of approach=v2v1v+v=1
2v=v2v1 ------(II)
from eq (I) & (II)
v2=53v v1=v3
(-ve means opposite in direction.)
v2v1=53vv3=51
Hence, the ratio of their velocities is 5:1

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