Question

A spring-loaded toy gun is used to shoot a ball straight up in the air. (Figure 1)The ball reaches a maximum height H, measured from the equilibrium position of the spring. -The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to H.

Answer 1

With spring compressed by x , energy stored in spring = kx²/2 ( k is spring constant)
As ball reaches maximum height, this energy stored in spring is converted into gravitaional potential energy of ball, as kinetic energy of ball at maximum height is zero ( ball is rest for a moment at top most point). Potential energy of ball at height h = mgh.

In first case, let spring be compressed by x, maximum height of ball is H. Hence
kx²/2 = mgH   ......1
Second time spring is compressed by x/2. Let maximum height be H'
k(x/2)² / 2 = mgH'
1/4 ( kx²/2) = mg H'
using equation 1
1/4 ( mgH ) = mgH'
H' = H/4