In a spring gun having spring constant 100N/m, a small ball of mass 0.1kg is put in its barrel by compressing the spring through 0.05m as shown in Figure. The ball leaves the gun horizontally at a height of 2m above the ground. Find
(a) The velocity of the ball when the spring is released.
(b) Where should a box be placed on the ground so that the ball falls in it. `(g= 10 m//s^(2))`

Explanation

(a) When the spring is released its elastic potential energy `(1//2) kx^(2)` is converted into kinetic energy `(1//2) mv^(2)` of the ball, so, by conservation of mechanical energy
`(1)/(2) mv^(2)= (1)/(2) kx^(2) rArr v= x sqrt((k)/(m))`
So `v= 0.05 sqrt((100)/(0.1)) m//s = sqrt((5)/(2)) m//s`
(b) As initial vertical component of velocity of ball is zero, time taken by the ball to reach the ground.

`t= sqrt((2h)/(g))= sqrt((2 xx 2)/(10))= sqrt((2)/(5)) s`, So the horizontal distance travelled by the ball in this time `d= vt = sqrt((5)/(2)) xx sqrt((2)/(5))= 1m`