A wooden block of mass M rests on a horizontal surface. A bullet of mass m moving in the horizontal direction strikes and gets embedded in it. The combined system covers a distance x on the surface. If the coefficient of friction between wood and the surface is `mu` , the speed of the bullet at the time of striking the block is (where m is mass of the bullet)

Options

A. ` sqrt(( 2Mg)/( mum )) `
B. ` sqrt((2mu Mg)/( Mx)) `
C. ` sqrt( 2mugx)((M+m)/( m))`
D. ` sqrt((2mu mx)/( M+m))`

Answer

C

Explanation

Let speed of the bullet = v
Speed of the system after the collision = V
By conservation of momentum mv = (m+M) V
` rArr V (mv)/( M+m) `
So the initial K.E. acquired by the system
`= (1)/(2) (M+m) V^(2) =(1)/(2) (m+M) ((mv)/(M+m) )^(2) = (1)/(2) (m^(2) v^(2))/( 2(m+M))`
This kinetic energy goes against friction work done by friction
` = mu R xx x = mu (m+M) g xx x `
By the law of conservation of energy
`(1)/(2) (m^(2)v^(2))/( (m+M)) = mu (m+M) g xx x rArr v^(2) = 2 mu g x ((m+M)/( m))^(2) `
` therefore v= sqrt( 2 mu g x ) ( ( M+m) /( m))`