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(A) Let the mass of the shell be $2m$. At the highest point,the velocity of the shell is $v_x = v \cos 	heta$ in the horizontal direction.According t

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$3v \cos 	heta$

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(A) Let the mass of the shell be $2m$. At the highest point,the velocity of the shell is $v_x = v \cos 	heta$ in the horizontal direction.According to the law of conservation of linear momentum,the total momentum before the explosion must equal the total momentum after the explosion.Initial momentum $P_i = (2m)(v \cos 	heta)$.After the explosion,one piece of mass $m$ retraces its path,meaning its velocity is $-v \cos 	heta$.Let the velocity of the other piece of mass $m$ be $v'$.Applying conservation of momentum: $(2m)(v \cos 	heta) = m(-v \cos 	heta) + m(v')$.Dividing by $m$: $2v \cos 	heta = -v \cos 	heta + v'$.Therefore,$v' = 3v \cos 	heta$.

$A$ shell is fired from a cannon with a velocity $v$ at an angle $\theta$ with the horizontal direction. At the highest point of its path,it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is

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