A ball of mass m = 60 text{ g} is shot with s | vedclass

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(B) Since there are no external horizontal forces acting on the system (ball + spring gun),the linear momentum of the system is conserved.Let $V$ be t

Vedclass · Accepted Answer

$4.4 	ext{ m/s}$

Vedclass · Answer

(B) Since there are no external horizontal forces acting on the system (ball + spring gun),the linear momentum of the system is conserved.Let $V$ be the final velocity of the combined system (ball + spring gun) after the ball stops relative to the barrel.According to the law of conservation of linear momentum:$m v_0 + M(0) = (m + M) V$Given:$m = 60 	ext{ g} = 0.06 	ext{ kg}$$M = 240 	ext{ g} = 0.24 	ext{ kg}$$v_0 = 22 	ext{ m/s}$Substituting the values:$0.06 	imes 22 + 0 = (0.06 + 0.24) V$$1.32 = 0.30 V$$V = \frac{1.32}{0.30} = 4.4 	ext{ m/s}$Thus,the speed of the spring gun after the ball stops relative to the barrel is $4.4 	ext{ m/s}$.

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