Question

A ball of mass 0.40 kg is fired with velocity 200 m/s into the barrel of a spring gun of mass 1.5 kg initially at rest on a frictionless surface. The ball sticks in the barrel at the point of maximum compression of the spring. No energy is lost to friction. What fraction of the ball's initial kinetic energy is stored in the spring?

Answer #1

Given

Initial Velocity u = 200 m/s

Mass of ball m = .40 kg

Mass og the gun m’ = 1.5 kg

At the moment of maximum compression of the spring, the ball and the spring gun move with the same velocity v

According to conservation of momentum

Initial momentum = final momentum

mu + (m’ x 0) = (m+m’) v

v = mu/(m+m’)

= 0.40 x 200 /(0.40+1.5)

= 42.1 m/s

KE of ball = ½ mu^{2}

= 0.4 * 200^{2} /2

= 8000 J

KE of after collision = ½ (m+m’)v^{2}

= (0.4 +1.5) x 42.1^{2}/2

= 1679.79 J

So 1679.79 x 100 / 8000 = 20.99

So 20.99% of energy or roughly one fifth(1/5) of the ball’s energy is stored in the system

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