A bullet of mass 0.01 kg is fired horizontal into a 4 kg wooden block block at rest , on a horizontal surface. The coefficient of kinetic friction between the block and bullet is 0.25 the combination moves 20 m before coming to rest . With what speed did the bullet strike the block ?

Explanation

Here , mass of bullet , ` m_(1) = 0.01 kg`
mass of wooden block , ` m_(2) = 4 kg `
Coeff . Of kinetic friction , ` mu = 0.25 , `
distance moved by the combination ` s = 20 m `
Let velocity of bullet be `u_(1)`, initial velocity of block `u_(2) = 0`
If upsilon is the velocity with which the combination moves , then according to the principle of conservation of linear momentum ,
`(m_(1) + m_(2) ) upsilon = m_(1) u_(1) + m_(2) u_(2) `
`(0 . 01 + 4 ) upsilon = 0.1 u_(1) + 4 xx 0 `
`upsilon = (0.01u_(1))/(4.01)= (u_(1))/(401)`
Force of dynamic friction on the combination
` F = mu R = mu (m_(1) + m_(2)) g`
` = 0 .25 (0.1 + 4 ) xx 9.8 N `
`:.` Retardation produced
`a = (f ) /(m_(1)+ m_(2))= (0.25 xx 4.01 xx 9.8 )/(4.01) = 2 .45 ms^(-2)`
Using for the combination , the relation
`u^(2) - u^(2) = 2 as , 0 - upsilon^(2) = 2 (-2 .45 ) xx 20`
`upsilon^(2) = 4.9 xx20 = 98 or upsilon = sqrt(98)`
From (i) `u_(1) = 401 upsilon`
or `u_(1) = 401 sqrt(98)= 3969 .7 ms^(-1)` .