Q. A spring gun of spring constant 90 $Ncm^{- 1}$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) $90{ms}^{- 1}$
Explanation
(d) Kinetic energy of ball=potential energy of spring
$i.e.,\ \ B\ \ \frac{1}{2}mv^{2} = \frac{1}{2}kx^{2}$
$\therefore\ \ \ \ \ 16 \times 10^{- 3} \times v^{2} = \frac{90}{10^{- 2}} \times \left( 12 \times 10^{- 2} \right)^{2}$
Or $v^{2} = \frac{90 \times 144 \times 10^{- 4}}{10^{- 2} \times 16 \times 10^{- 3}}$
Or $v = 90ms^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.