In a spring gun having spring constant 100 N/m, a small ball of mass 0.1 kg is put in its barrel by compressing the spring though 0.05 m as shown in Figure. The ball leaves the gun horizontally at a height of 2 m 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 that 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)rArrv=xsqrt((k)/(m))`
So `v=0.05sqrt((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((2xx2)/(10))=sqrt((2)/(sqrt(5)))s,` So the horizontal distance travelled by the ball in this time `d=vt=sqrt((5)/(2))xxsqrt((2)/(5))=1m`