Question

A bullet of mass 0.056 kg traveling horizontally at a speed of 100 m/s embeds itself in a block of mass 1.5 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? v= m/s (b) Calculate the kinetic energy of the bullet plus the block before the collision: K; = (c) Calculate the kinetic energy of the bullet plus the block after the collision: Ks = (d) Was this collision elastic or inelastic? (e) Calculate the rise in thermal energy of the bullet plus block as a result of the collision: AEthermal, bullet + AEthermal, block = (f) What was the transfer of energy (microscopic work) from the surroundings into the block+bullet system during the collision? (Remember that represents energy transfer due to a temperature difference between a system and its surroundings.) Q=

Answer 1

(a ) Using conservation of momentum

m1v1 + m2v2 = ( m1 + m2)v

as block was at rest, v2 = 0

0.056 * 100 = ( 0.056 + 1.5)v

v = 3.598 m/s

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(b)

K_i = 1/2 * 0.056 * 100²

K_i = 280 J

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(c)

K_f = 1/2 * (m1 + m2) v²

K_f = 10 J

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(d)

this was an INELASTIC COLLISION as there is a loss of lot of energy.

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(e) thermal energy = 280 - 10

thermal energy = 270 J

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(f)

Q = 0

because there is no ample time for energy transfer