Question

A ball of mass 100g lands at 2.5m on a
horizontal surface when it is shot from the
level of the surface at an angle of 15°
with horizontal. The ball is shot from a
fixed gun of spring constant 500N/m.
Find the compression of the spring of the
gun (in cm) : (g = 10m/s²)​

Answer 1

Given:

A ball of mass 100g lands at 2.5m on a horizontal surface when it is shot from the level of the surface at an angle of 15° with horizontal. The ball is shot from a fixed gun of spring constant 500N/m.

To find:

Compression of spring.

Calculation:

First of all , let's find the velocity with which the ball was projected :

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

$= > \: 2.5 = \dfrac{ {u}^{2} \sin(2 \times {15}^{ \circ} ) }{10}$

$= > \: 2.5 = \dfrac{ {u}^{2} \sin( {30}^{ \circ} ) }{10}$

$= > \: 25 = {u}^{2} \sin( {30}^{ \circ} )$

$= > \: 25 = {u}^{2} \times \dfrac{1}{2}$

$= > \: {u}^{2} = 50$

Now , we know that the whole potential energy of spring was converted to kinetic energy of ball during Projection:

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

$= > \: k {x}^{2} = m {v}^{2}$

$= > \: 500 {x}^{2} = 0.1 \times 50$

$= > \: 500 {x}^{2} = 5$

$= > \: {x}^{2} = \dfrac{5}{500}$

$= > \: {x}^{2} = \dfrac{1}{100}$

$= > x = \dfrac{1}{10}$

$= > x = 0.1 \: m$

$= > x = 10 \: cm$

So, final answer is:

$\boxed{ \bf{ \red{compression = x = 10\: cm}}}$