A spring with a constant of 12 (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 3 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?

May 13, 2016

$\sqrt{6} m$

Explanation:

Consider the inital and final conditions of the two objects (namely, spring and mass):

• Initially:
Spring is at lying at rest, potential energy = $0$
Mass is moving, kinetic energy = $\frac{1}{2} m {v}^{2}$

• Finally:
Spring is compressed, potential energy = $\frac{1}{2} k {x}^{2}$
Mass is stopped, kinetic energy = 0

Using conservation of energy (if no energy is dissipated into the surroundings), we have:

$0 + \frac{1}{2} m {v}^{2} = \frac{1}{2} k {x}^{2} + 0$

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

$\implies {x}^{2} = \left(\frac{m}{k}\right) {v}^{2}$

$\therefore x = \sqrt{\frac{m}{k}} v = \sqrt{\frac{8 k g}{12 k g {s}^{-} 2}} \times 3 m {s}^{-} 1 = \sqrt{6} m$