# 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 8 kg and speed of 9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

May 3, 2016

$25.5 m$

#### Explanation:

Conservation of energy tells us that there must be exactly the same amount of energy before a reaction as after.

Before the reaction energy is kinetic, since the object is moving, and energy is given by

$E = \frac{1}{2} m {v}^{2}$,

where $m$ is mass and $v$ is velocity.

After the reaction, the energy is potential, stored in the spring, and given by

$E = \frac{1}{2} k {x}^{2}$,

where $x$ is the distance the spring compresses and $k$ is the spring constant.

Since these two equations must be equal (conservation of energy above),

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

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

The question is asking you about how much the spring compresses, so you rearrange to make $x$ the subject.

$\sqrt{m {v}^{2} / k} = x$.

Put in the values you know and solve

$\sqrt{\frac{8 \cdot {9}^{2}}{1}} = x$

$\sqrt{648} = x = 25.5 m$