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A bullet of mass 0.02 kg travelling hori...

A bullet of mass 0.02 kg travelling horizontally with velocity 250`ms^(-1)` strikes a block of wood of mass 0.23 kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40m. The coefficient of sliding friction of the rough surface is `( g = 9.8 ms^(_2))`

A

`0.75`

B

`0.61`

C

`0.51`

D

`0.30`

Text Solution

Verified by Experts

The correct Answer is:
C
`m_1u_1 = (m_1+m_2) v`
`implies v = ((0.02)/(0.02+0.03))xx250`
`implies v=2/50 xx250 = 20m//s`
Conservation of energy `1/2 mv^2 = mu N xxS`
`implies 1/2 xx0.25 xx400 = mu xx 0.25 xx9.8 xx40`
`implies mu = 5/(9.8) = 0.51`
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