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 through 0.05 mas shown in Figure. The ball leaves the gun horizontally at a height of 2 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))

In a spring gun having spring constant 100 N/m, a small ball of mass 0

1.	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 through 0.05 mas shown in Figure. The ball leaves the gun horizontally at a height of 2 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))
Answer» Solution :(a) When the spring isreleased 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`