A shell is fired from a cannon with velocity | vedclass

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Question

(A) Let the mass of the shell be $M$. When fired with velocity $v$ at an angle $	heta$,its velocity at the highest point is $v_x = v \cos 	heta$ in

Vedclass · Accepted Answer

$3v \cos 	heta$

Vedclass · Answer

(A) Let the mass of the shell be $M$. When fired with velocity $v$ at an angle $	heta$,its velocity at the highest point is $v_x = v \cos 	heta$ in the horizontal direction.The momentum of the shell just before the explosion is $P_i = Mv \cos 	heta$.After the explosion,the shell splits into two equal pieces of mass $m = M/2$.One piece retraces its path,meaning its velocity becomes $-v \cos 	heta$ (opposite to the original horizontal direction).Let the velocity of the other piece be $V$. By the law of conservation of linear momentum:$P_i = P_f$$Mv \cos 	heta = m(-v \cos 	heta) + mV$Since $m = M/2$,we have:$Mv \cos 	heta = \frac{M}{2}(-v \cos 	heta) + \frac{M}{2}V$Dividing by $M/2$:$2v \cos 	heta = -v \cos 	heta + V$$V = 3v \cos 	heta$Thus,the speed of the other piece is $3v \cos 	heta$.

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