Two balls having masses m and 2m are fastened to two light strings of same length = 1m. The other ends of the strings are fixed at O. Both the balls are moved away such that strings become horizontal and ball are on different sides of the fixed point as shown in the figure. Now both the balls are released and they collide elastically. Find height raised by ball of mass m after collision (Assume string does not break)
Text Solution
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(2)
P i = mu – 2 mu ...(1)
when u = 
let after collision velocity of m is u 1 towards left and 2 m is u 2 towards right so.
P f = 2 mu 2 – mu 1 ....(ii)
mu – 2mu = 2 mu 2 – mu 1 – u = 2u 2 – u 1 ∴ u 1 – 2u 2 = u ...(1)
2mu 2 + 
mu 2 = 
2m × u 2 2 + 
mu 1 2 ....(2)
3 u 2 = 2u 2 2 + u 1 2 ....(2)
from (1) u 1 = u + 2u 2
on solving 3u 2 = 2u 2 2 + u 2 + 4u 2 2 + 4uu 2
6u 2 2 – 2u 2 + 4u.u 2 = 0
3u 2 2 + 2u.u 2 – u 2 = 0
on solving u 2 = – 
u 2 = – u or 
or 
∴ u 1 = u + 2u 2 = 
Since u 1 > 
it will complete the circle.
u 1 > 
∴ h 1 = 2 × λ . Mt ∴ h 2 = 
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