15 points Mi4 10 6 XP008. A bullet of mass 0.017 kg traveling horizontally at a

Question

15 points Mi4 10 6 XP008. A bullet of mass 0.017 kg traveling horizontally at a high speed of 220 m/s embeds itself in a block of mass 3 kg that is sitting at rest on a nearly frictionless surface. (a) What is the speed of the block after the bullet embeds itself in the block? Vf = m/s (b) Calculate the total translational kinetic energy before and after the collision. trans,l Kerans, (c) Compare the two results and explain why there is a difference. Some of the momentum is lost in an inelastic collision. O The internal energy of the block-bulet system has increased. The Energy Principle isn't valid for an inelastic collision. Additional Materials Section 106

Explanation / Answer

(a) By the law of momentum conservation:-
=>m1u1+m2u2=(m1+m2)v
=>0.017x 220 = (0.017+3) x v
=>v =1.239m/s
(b) KE(before) = 1/2 x m1 x u1^2 = 1/2 x 0.017 x (220)^2 = 411.4J
KE(after) = 1/2 x (m1+m2) x v^2 = 1/2 x 3.017 x (1.239)^2 = 2.315J

c) the energy principle isn't valid for an inelastic collision

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