91Ó°ÊÓ

The magnitude of momentum is equal to the mass times the speed, and the distinct components of momentum along the x, y, and z axes are preserved.

The kinetic energy is given by$\frac{1}{2}m{v}^2$ where mis mass of the object and vis the velocity of the object.

The kinetic energy of spring is given by$\frac{1}{2}k{x}^2$ where k is spring constant and xis the distance from the equilibrium point.

• A horizontal spring with stiffness 0.5 N/m has a relaxed length of 15 cm,
• A mass of 20 g and the spring to a total length of 25 cm.
• To find the speed of the mass at the moment when the spring returns to its relaxed length of 15 cm.

Assume that the system is free of non-conservative forces, therefore energy is conserved between the starting and end states.

Write equation of conservation of energy and use the formula of energy.

$$\begin{gathered} E_1=E_f \\ \frac{1}{2}{k}_s{s}^2=\frac{1}{2}m{v}^2 \end{gathered}$$

Solve the equation for v.

$v=s\sqrt{\frac{k_s}{m}}$

Substitute 0.25-0.15 for s, 0.5for kand 0.020 for minto the obtained equation.

$$\begin{gathered} v=(0.25-0.15)\sqrt{\frac{0.5}{0.020}} \\ =0.5m{s}^{-1} \end{gathered}$$

Therefore, the speed of the mass when the spring returns to its relaxed length is $0.5{\mathrm{ms}}^{-1}$.