A bullet of mass 0.102 kg, traveling horizontally at a speed of 300 m/s, embeds

Question

A bullet of mass 0.102 kg, traveling horizontally at a speed of 300 m/s, embeds itself in a wooden block of mass 3.5 kg that is sitting at rest on an effectively frictionless surface.

a) What is the speed of the block after the bullet embeds itself?

b) Calculate the total kinetic energy of the bullet plus block before the collision.

c) Calculate the total kinetic energy of the bullet plus block after the collision.

d) Calculate increase in internal energy of the bullet-plus-block as a result of the collision.

Explanation / Answer

a) Conserving linear momentum

m1u1 = (m1+m2)v

v = 0.102*300/(0.102+3.5) = 8.5 m/s

b) KEi = 0.5m1u1^2 = 0.5*0.102*300^2 = 4590 J

c) KEf= 0.5(m1+m2)v^2 = 0.5(3.602)*8.5^2 = 129.98 J

d) Increase in internal energy = KEf - KEi = 4460.02 J

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