HC Verma , Exercise question no. 28?

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applying cons. of momentum at 1st jump then,sign conven is taking right as positive.

mu=-(M+m)v

so v=-mu/(M+m)

now the velocity of 2nd jump is u wrt this velocity so,its velo. wrt ground is

v'=-Mu/(m+M)

applying cons. of momentum on 2nd jump we get,

-mu=Mv"-mMu/(M+m)

from this we get v" as your ans