A small cube of mass `m` slides down a circular path of radius R cut into a large block of mass M, as shown in figure. M rests on a table, and both blocks move without friction. The blocks are initially at rest, and m starts from the top of the path. Find the horizontal distance from the bottom of block where cube hits the cable

Answer

B

Explanation

`mv_1=Mv_2` …(i)
`mgR=1/2mv_1^2+1/2Mv_2^2` …(ii)
`t=sqrt((2(R//2))/(g))=sqrt(R/g)` …(iii)
The desired distance is
`S=(v_1+v_2)t` …(iv)
Solving Eqs. (i) and (ii) of `v_1` and `v_2` and substituting in Eq. (iv), we get
`S=Rsqrt((2M+m)/(M))`