Question

A spring gun has a force constant 1000 N/m. When a ball of 10 gm is shot from this gun, its spring is compressed by 10 cm. Find the maximum horizontal distance that can be achieved by the ball: (g = 10 m/s 2 )

Answer 1

Given:

A spring gun has a force constant 1000 N/m. When a ball of 10 gm is shot from this gun, its spring is compressed by 10 cm.

To find:

Maximum horizontal distance travelled by ball.

Calculation:

The ball will follow projectile motion after release from spring gun.

The potential energy of spring will be converted to kinetic energy of ball;

$\therefore \: \dfrac{1}{2} k {x}^{2} = \dfrac{1}{2} m {v}^{2}$

$= > \: \dfrac{1}{2} \times 1000 \times {(0.1)}^{2} = \dfrac{1}{2} \times 0.01 \times { v }^{2}$

$= > \: \dfrac{1}{2} \times 1000 \times {10}^{ - 2} = \dfrac{1}{2} \times {10}^{ - 2} \times { v }^{2}$

$= > {v}^{2} = 1000$

Maximum range of projectile takes place at 45° angle of Projection ;

$\therefore \: range = \dfrac{ {v}^{2} \sin(2 \theta) }{g}$

$= > \: range = \dfrac{ {v}^{2} \sin(2 \times 45 \degree) }{g}$

$= > \: range = \dfrac{ {v}^{2} \sin(90 \degree) }{g}$

$= > \: range = \dfrac{ {v}^{2} }{g}$

$= > \: range = \dfrac{ 1000 }{g}$

$= > \: range = \dfrac{ 1000 }{10}$

$= > \: range = 100 \: m$

So, final answer is :

$\boxed{ \red{ \large{ \bold{ \: range = 100 \: m}}}}$