The spring of a spring gun is compressed a distance (d) of 3.2 cm from its relaxed state, and a ball of mass m = 12 g is put in the barrel.

Question

With what speed will the ball leave the barrel once the gun is fired? The spring constant is 7.5 N/cm. Assume no friction and a horizontal gun barrel.

Answer 1

The ball will leave with a speed that makes its kinetic energy equal to the potential energy of the compressed spring.

(1/2)kx² = (1/2)mv² ⇒

v = √(kx²/m)

k = 7.5 N/cm = 750 N/m
x = 3.2 cm = 0.032 m
m = 12 g = 0.012 kg

v = √[750(0.032)²/0.012] m/s = 8 m/s