Question

A miniature, spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 5.00 g, and the gun and puck have a combined mass of 120 g.With the system initially at rest, the radio-controlled trigger releases the bullet, causing the puck and empty gun to move with a speed of 0.500 m/s. Of the totalkinetic energy of the gun-puck-bullet system, what percentage is in the bullet?

Answer 1

Conservation of Momentum:ρ(i) = ρ(f) = ρ(total)m(1) = bullet mass = 5.00 g = .005 kgm(2) = total system mass = 120 g = .120 kgv(i) = bullet velocity =????v(f) = total system velocity = 0.500 m/sLet's find the velocity of the bullet:m(1)v(i) = m(2)v(f)(.005)(v(i)) = (.120)(0.500)v(i) = 60/5v(i) = 12 m/sAnd the percentage KE in the bullet is found with:KE(b) / KE(total)KE(b) = Kinetic energy of bulletKE(total) = Kinetic energy of total system[(1/2)m(1)v(i)^2]/ [(1/2)m(1)v(i)^2 +(1/2)m(2)v(f)^2)]By simplification ----> canceling the 1/2 and dividing topand bottom by m(1)v(i)^2, leaves:1/[1+(m(2)v(f)^2/m(1)v(i)^2)]1/[1+(0.03/0.72)]1/[1+.04167]1/1.04167 = 0.96

Answer 2

Kinetic energy = 1/2 mv²

velocity of both bullet and mass is same.

therefore the ratio of kinetic energies is same as ration of masses

percentage ratio = mass of bullet/(mass of gun + mass of bullet) x 100%

=5/120 x 100 = 4.17%