# A ball with a mass of 120 g is projected vertically by a spring loaded contraption. The spring in the contraption has a spring constant of 15 (kg)/s^2 and was compressed by 3/5 m when the ball was released. How high will the ball go?

Jul 10, 2018

The height reached by the ball is $= 2.30 m$

#### Explanation:

The spring constant is $k = 15 k g {s}^{-} 2$

The compression of the spring is $x = \frac{3}{5} m$

The potential energy stored in the spring is

$P E = \frac{1}{2} k {x}^{2} = \frac{1}{2} \cdot 15 \cdot {\left(\frac{3}{5}\right)}^{2} = 2.7 J$

This potential energy will be converted to kinetic energy when the spring is released and to potential energy of the ball

$K {E}_{b a l l} = \frac{1}{2} m {u}^{2}$

Let the height of the ball be $= h$

The acceleration due to gravity is $g = 9.8 m {s}^{-} 2$

Then ,

The potential energy of the ball is $P {E}_{b a l l} = m g h$

Mass of the ball is $m = 0.120 k g$

$P {E}_{b a l l} = 2.7 = 0.120 \cdot 9.8 \cdot h$

$h = \frac{2.7}{0.120 \cdot 9.8}$

$= 2.30 m$

The height reached by the ball is $= 2.30 m$

$2.293 \setminus m$

#### Explanation:

Assuming spring to be perfectly elastic, when the ball is released, the entire elastic energy of spring $\frac{1}{2} k {x}^{2}$ is transferred to the ball in form of kinetic energy $\frac{1}{2} m {v}^{2}$
Where, $k$ is spring constant,

$x$ initial compression of spring

$m$ is mass of ball

$u$ is the velocity of ball when it is thrown vertical upward

hence, we have

$\setminus \textrm{K . E . o f b a l l} = \setminus \textrm{e l \ast i c e \ne r g y o f s p r \in g}$

$\frac{1}{2} \left(0.120\right) {u}^{2} = \frac{1}{2} \left(15\right) {\left(\frac{3}{5}\right)}^{2}$

${u}^{2} = \frac{135}{3}$

$u = \setminus \sqrt{\frac{135}{3}}$

Now, using third equation of motion:

${v}^{2} = {u}^{2} + 2 a s$

For the motion against gravity, we have $a = - g = - 9.81 \setminus \setminus {\textrm{\frac{m}{s}}}^{2}$, initial velocity $u = \setminus \sqrt{\frac{135}{3}}$ & final velocity $v$ at the maximum height $s = h$ becomes zero i.e. $v = 0$

hence, we have

${0}^{2} = \left(\setminus \sqrt{\frac{135}{3}}\right) - 2 \left(9.81\right) h$

$h = \setminus \frac{135}{3 \setminus \cdot 19.62}$

$= 2.293 \setminus m$