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))

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 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))
Answer» Solution :(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((2xx2)/(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`