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Question

(D) Let the mass of the shell be $2m$. At the highest point,the vertical component of velocity is zero,and the horizontal velocity is $u \cos 	heta$.

Vedclass · Accepted Answer

$E_2=9 E_1$

Vedclass · Answer

(D) Let the mass of the shell be $2m$. At the highest point,the vertical component of velocity is zero,and the horizontal velocity is $u \cos 	heta$. Thus,the momentum of the shell just before the explosion is $P = (2m)(u \cos 	heta)$. After the explosion,the shell breaks into two equal parts of mass $m$ each. One part retraces its path,meaning its velocity is $v_1 = -u \cos 	heta$. Let the velocity of the second part be $v_2$. By the law of conservation of linear momentum: $2m u \cos 	heta = m v_1 + m v_2$ $2m u \cos 	heta = m(-u \cos 	heta) + m v_2$ $2m u \cos 	heta = -m u \cos 	heta + m v_2$ $m v_2 = 3m u \cos 	heta \Rightarrow v_2 = 3u \cos 	heta$. The kinetic energy of the first part is $E_1 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (-u \cos 	heta)^2 = \frac{1}{2} m u^2 \cos^2 	heta$. The kinetic energy of the second part is $E_2 = \frac{1}{2} m v_2^2 = \frac{1}{2} m (3u \cos 	heta)^2 = \frac{1}{2} m (9 u^2 \cos^2 	heta) = 9 \left( \frac{1}{2} m u^2 \cos^2 	heta ight)$. Therefore,$E_2 = 9 E_1$.

$A$ cannon shell fired breaks into two equal parts at its highest point. If one part retraces its path to the cannon with kinetic energy $E_1$ and the kinetic energy of the second part is $E_2$,then

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