A bullet of mass `10 g` moving horizontally with a speed of `400 m//s` a block of mass `390 g` and remains in it. The block slides `10 m` on rough surface before coming to rest. The friction coefficient is

Options

A. `(1)/(4)`
B. `(1)/(2)`
C. `(3)/(4)`
D. `1`

Answer

B

Explanation

By the momentum conservation
`m u + M xx 0 = (m + M)v`
For the combined block :
Retardation `a = (f)/((m + M)) = (mu (m + M) g)/((m + M) = mu g`
`v_(1)^(2) = v^(2) - 2ad`
`0 = (10)^(2) - 2 mu gd = (10)^(2) - 2 mu xx 10 xx 10`
`mu = (1)/(2)`