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 total kinetic energy of the gun-puck-bullet system,what percentage is in the bullet?

Question

Quizplus · Accepted Answer

The initial kinetic energy of the system is zero, since it is at rest. After the bullet is fired, the momentum of the system is conserved, so we have:
(mass of gun and puck)(velocity of gun and puck) = (mass of bullet)(velocity of bullet)
(120 g)(0.500 m/s) = (5.00 g)(v_b)
v_b = 24.0 m/s
The kinetic energy of the bullet is:
KE_b = (1/2)mv_b^2 = (1/2)(0.005 kg)(24.0 m/s)^2 = 1.44 J
The total kinetic energy of the system is:
KE_sys = (1/2)(120 g)(0.500 m/s)^2 + KE_b = 0.075 J + 1.44 J = 1.515 J
The percentage of kinetic energy in the bullet is:
(KE_b / KE_sys) x 100% = (1.44 J / 1.515 J) x 100% = 95.7%
Rounding to the nearest whole number, we get 96%, which is answer choice C.

A Miniature,spring-Loaded,radio-Controlled Gun Is Mounted on an Air Puck