Question

A bullet of mass 0.107 kg traveling horizontally at a speed of 300 m/s embeds itself in a block of mass 3 kg that is sitting at rest on a nearly frictionless surface. 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.)

Answer 1

Answer:

165.77J

Explanation:

M₁ = 0.107kg

u₁ = 300m/s

m₂ = 3kg

u₂ = 0

v =

m₁u₁ + m₂u₂ = (m₁ + m₂)V

(0.107*300) + 0 = (0.107 + 3)V

V = 32.1 / 3.107 = 10.33m/s

kinetic energy of the system after collision =

½m1v² + ½m2v²

K.E = ½(m1 + m2)v²

K.E = ½(0.107+3) * 10.33²

K.E = 165.77J