A bullet of mass \( m \) leaves the barrel of a gun of mass \( M \) with a velocity \( v \). The gun is known to recoil with a velocity \( V \). If \( k \) and \( K \) respectively denote the kinetic energies of the bullet and the gun respectively; then
\( K = \left(\frac{m}{M}\right)^2 k \)
\( K = \sqrt{\frac{m}{M}} k \)
\( K = \left(\frac{m}{M}\right) k \)
\( K = \left(\frac{M}{m}\right) k \)
Solution:
By conservation of momentum, the bullet and gun have equal momentum magnitude, \( p \). Since kinetic energy is \( K_{\text{E}} = \frac{p^2}{2\text{mass}} \), we have \( k = \frac{p^2}{2m} \) and \( K = \frac{p^2}{2M} \). Thus, \( K = \left(\frac{m}{M}\right) k \).
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