Correct Answer - A
Let `v^(1)` be the velocity of sphere after collision
`J cos 30 = 10v^(1) rArr J = (20 v^(1))/(sqrt(3))`
`J = (m)/(10)(10 cos 30)-(m)/(10)(v^(1) cos 30^(@))`
`rArr` speed of the sphere `v^(1) = (30)/(43) ms^(-1)`
speed of particle in tangential direction remains same
`10 sin 30 = 5ms^(-1)`
speed of partile `= sqrt((5)^(2)+[((30)/(43))cos 30])^(2)`
`=5.26 ms^(-1)`