A bullet of mass 5.00 g is fired horizontally into a 2.26 kg wooden block at res

Question

A bullet of mass 5.00 g is fired horizontally into a 2.26 kg wooden block at rest on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.190. The bullet comes to rest in the block, which moves 2.08 m. (a) What is the speed of the block immediately after the bullet comes to rest within it? Correct: Your answer is correct. m/s (b) At what speed is the bullet fired? m/s

Explanation / Answer

We generally do not answer the questions of those who are having low rating. You are having a rating of "86%". Please rate all answers and improve your rating. Even if you find the answers are incorrect, first rate Lifesaver or helpful to the best answer and then rate other answers as not helpful. But do rate them for the good of both of us. Thanks!! Coming to the problem m1 = 5g = (5/1000) kg m2 = 2.26 Kg Let the bullet was fired into the block at a speed u1. Given initially the block was at rest=>u2 = 0 a)Let the speed of the block immediately after the bullet comes to rest within it = V. We know that 0^2 = V^2 + 2aS Given S = 2.08 m a = -u*g = -0.19*9.81 =>V = sqrt(2*0.19*9.81*2.08) = 2.785 m/s b)Momentum is conserved =>m1u1 + m2u2 = (m1+m2)*V =>m1u1 = (m1+m2)*V =>u1 = ((2.26+(5/1000))*2.785)/(5/1000) = 1261.41 m/s So the bullet was fired into the block at a speed u1 = 1261.41 m/s

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