Q. A spring gun of spring constant $90\ N/cm$ is compressed $12\ cm$ by a ball of mass $16\ g$. If the trigger is pulled, the velocity of the ball is
A) $50\ ms^{- 1}$
B) $9\ ms^{- 1}$
C) $40\ ms^{- 1}$
D) $90ms^{- 1}$
Explanation
(d) Loss in $PE$ in spring = gain in $KE$ of ball
$\frac{1}{2}Kx^{2} = \frac{1}{2}mv^{2}$
$\frac{90}{10^{- 2}} \times \left( 12 \times 10^{- 2} \right)^{2} = 16 \times 10^{- 3}v^{2} \Rightarrow v = 90\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.