A spring with a constant of 1 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 2/5 kg and speed of 1/4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

Apr 30, 2016

$0.16 m$

Explanation:

According to the law of conservation of energy, the total energy before and after a reaction must be equal.

Kinetic energy before this reaction is $\frac{1}{2} m {v}^{2}$, where $m$ is mass and $v$ is velocity. Potential energy stored in the spring is $\frac{1}{2} k {x}^{2}$, where $k$ is the spring constant and $x$ is the distance it compresses.

Therefore, according to conservation,

$\frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2}$

$m {v}^{2} = k {x}^{2}$

Now we can put in what we know of the values and rearrange to find $x$,

$\frac{2}{5} \cdot {\left(\frac{1}{4}\right)}^{2} = 1 \cdot {x}^{2}$

$\frac{1}{40} = {x}^{2}$

$x = \sqrt{\frac{1}{40}} \approx 0.16 m$