A rifle bullet with mass 8.00 g and initial horizontal velocity 280 m/s strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless surface and is attached to one end of an ideal spring. The other end of the spring is attached to the wall. The impact compresses the spring a maximum distance of 15.0 cm. After the impact, the block moves in SHM. Calculate the period of this motion.

\(\begin{aligned}{}{m_1} = 0.008\;{\rm{kg}}\\{m_2} = 0.992\;{\rm{kg}}\\{v_{1i}} = 280\;{\rm{m/s}}\\{v_{2i}} = 0\\x = 0.15\;{\rm{m}}\end{aligned}\)

The time required to complete one cycle is known as the period.

Using conservation of linear momentum,

\(\begin{aligned}{}{m_1}{v_{1i}} + {m_2}{v_{2i}} = {m_1}{v_{1f}} + {m_2}{v_{2f}}\\\left( {0.008 \times 280} \right) + \left( {0.992 \times 0} \right) = \left( {0.008{v_{1f}} + 0.992{v_{2f}}} \right)\\{v_{1f}} = {v_{2f}} = 2.24\;{\rm{m/s}}\end{aligned}\)

To find the spring constant,

\(\begin{aligned}{}\frac{1}{2}\left( {{m_1} + {m_2}} \right){v^2} = \frac{1}{2}k{x^2}\\\frac{1}{2}\left( {0.008 + 0.992} \right){2.24^2} = \frac{1}{2}k \times {0.15^2}\\2.51 = 0.01k\\k = 223.11\;{\rm{N/m}}\end{aligned}\)

Angular frequency,

\(\begin{aligned}{}k = \left( {{m_1} + {m_2}} \right){\omega ^2}\\223.11 = \left( {0.008 + 0.992} \right){\omega ^2}\\\omega = 14.94\;{\rm{rad/s}}\end{aligned}\)

Time period,

\(\begin{aligned}{}T = \frac{{2\pi }}{\omega }\\ = \frac{{2\pi }}{{14.94}}\\ = 0.421\;{\rm{s}}\end{aligned}\)

Hence the period of this motion is 0.421s.