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(B) $1$. Initial momentum of the system: $P_i = m v_0$. $2$. At maximum compression,the ball and the gun move together with a common velocity $V$. By

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$0.8$

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(B) $1$. Initial momentum of the system: $P_i = m v_0$. $2$. At maximum compression,the ball and the gun move together with a common velocity $V$. By conservation of linear momentum: $m v_0 = (m + M) V$. $3$. Common velocity $V = \frac{m v_0}{m + M} = \frac{60 	imes 22}{60 + 240} = \frac{60 	imes 22}{300} = \frac{22}{5} = 4.4 	ext{ m/s}$. $4$. Initial kinetic energy of the ball: $K_i = \frac{1}{2} m v_0^2$. $5$. Kinetic energy of the system at maximum compression: $K_f = \frac{1}{2} (m + M) V^2$. $6$. Energy stored in the spring $(U)$ is the loss in kinetic energy: $U = K_i - K_f = \frac{1}{2} m v_0^2 - \frac{1}{2} (m + M) V^2$. $7$. Substituting $V = \frac{m v_0}{m + M}$: $U = \frac{1}{2} m v_0^2 - \frac{1}{2} (m + M) \left( \frac{m v_0}{m + M} ight)^2 = \frac{1}{2} m v_0^2 \left( 1 - \frac{m}{m + M} ight) = \frac{1}{2} m v_0^2 \left( \frac{M}{m + M} ight)$. $8$. The fraction of initial kinetic energy stored in the spring is $\frac{U}{K_i} = \frac{M}{m + M} = \frac{240}{60 + 240} = \frac{240}{300} = 0.8$.

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