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A bullet of mass 10 gram is fired horizontally into a 5 kg wooden block at rest on a horizontal surface The coefficient of kinetic friction between the block and the surface is 0.1 Calculate the speed of the bullet striking the block if the combination moves 20 m before coming rest .

Answer» Here, `m_(1) = 10` gram `= 10^(-2) kg ,m_(2) = 5 kg `
`u_(2) = 0 , u_(1) = ? mu_(k) = 0.1 , s = 20 m`
Let be the velocity of the combination (bullet and block ) after the bullet pierces the block According to principle of conservation of linear momentum
`(m_(1) + m_(2)) upsilon = m_(1) u_(1) + m_(2) u_(2) = m_(1) u_(1)`
`upsilon = (m_(1)u_(1))/(m_1 + m_(2))=(10^(-2) u_(1))/(10^(-2) + 5) `
` upsilon = (u_(1))/(501) `
Now , for the combination
`u = upsilon,v = 0,s = 20 m`,
`a = -mu_(k) g = - 0.1 xx 9.8 = - 0.98 m//s^(2)`
From `upsilon^(2) - u^(2) = 2` as
`0 - upsilon^(2) = 2 (-0.98) xx 20 `
`upsilon= sqrt(39.2) = 6 .26 m//s `
From `u_(1) = 501_(upsilon)`
` = 501 xx 6.26 = 3136.26 m//s ` .


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