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Question

(A) Initial kinetic energy of the bullet is $K_i = \frac{1}{2} m v_0^2$.After the collision,the bullet gets lodged in the block,and they move together

Vedclass · Accepted Answer

$\frac{1}{2} m v_0^2 \left( \frac{M}{m+M} \right)$

Vedclass · Answer

(A) Initial kinetic energy of the bullet is $K_i = \frac{1}{2} m v_0^2$.After the collision,the bullet gets lodged in the block,and they move together with a common velocity $v$. By the law of conservation of linear momentum:$m v_0 = (m + M) v \implies v = \frac{m v_0}{m + M}$.The kinetic energy of the combined system immediately after the collision is $K_f = \frac{1}{2} (m + M) v^2$.Substituting the value of $v$:$K_f = \frac{1}{2} (m + M) \left( \frac{m v_0}{m + M} \right)^2 = \frac{1}{2} (m + M) \frac{m^2 v_0^2}{(m + M)^2} = \frac{1}{2} \frac{m^2 v_0^2}{m + M}$.The energy lost in the collision is $\Delta K = K_i - K_f$:$\Delta K = \frac{1}{2} m v_0^2 - \frac{1}{2} \frac{m^2 v_0^2}{m + M} = \frac{1}{2} m v_0^2 \left( 1 - \frac{m}{m + M} \right) = \frac{1}{2} m v_0^2 \left( \frac{m + M - m}{m + M} \right) = \frac{1}{2} m v_0^2 \left( \frac{M}{m + M} \right)$.

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