Question

A bullet of mass 0.017 kg traveling horizontally at a high speed of 210 m/s embeds itself in a block of mass 5 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? Vr = 42 x m/s (b) Calculate the total translational kinetic energy before and after the collision. Ktrans,i = 374.85 Ktrans,f= (c) Compare the two results and explain why there is a difference. • The internal energy of the block-bullet system has increased. Some of the momentum is lost in an inelastic collision. The Energy Principle isn't valid for an inelastic collision. Additional Materials

Answer 1

Sol:

Given
m1=0.017 kg
V1=210 m/s
m2=5 kg

(a)
By the law of momentum conservation:-
=>m1u1+m2u2=(m1+m2)v
0.017 x 210 = (0.017+5) x v
v = 0.7116 m/s

(b)
KE(before) = 0.5x m1 x u1^2
= 1/2 x 0.017x (210)^2
= 374.85 J

KE(after) = 1/2 x (m1+m2) x v^2
= 1/2 x 5.017x (0.7116)^2
= 1.270 J

C)
The internal energy of the block bullet system has increased

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