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(D) Let the mass of the shell be $M$. At the highest point,the velocity of the shell is $v_x = u \cos 	heta$. The shell breaks into two equal parts o

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$E_2 = 9 E_1$

Vedclass · Answer

(D) Let the mass of the shell be $M$. At the highest point,the velocity of the shell is $v_x = u \cos 	heta$. The shell breaks into two equal parts of mass $m = M/2$.Let the velocity of the first part (which retraces its path) be $v_1 = -u \cos 	heta$. By the law of conservation of linear momentum:$M(u \cos 	heta) = m v_1 + m v_2$$M(u \cos 	heta) = (M/2)(-u \cos 	heta) + (M/2)v_2$$u \cos 	heta = -0.5 u \cos 	heta + 0.5 v_2$$1.5 u \cos 	heta = 0.5 v_2$$v_2 = 3 u \cos 	heta$The kinetic energy of the first part is $E_1 = \frac{1}{2} (M/2) (u \cos 	heta)^2 = \frac{1}{4} M u^2 \cos^2 	heta$.The kinetic energy of the second part is $E_2 = \frac{1}{2} (M/2) (3 u \cos 	heta)^2 = \frac{1}{2} (M/2) (9 u^2 \cos^2 	heta) = \frac{9}{4} M u^2 \cos^2 	heta$.Comparing $E_1$ and $E_2$:$E_2 = 9 	imes (\frac{1}{4} M u^2 \cos^2 	heta) = 9 E_1$.

$A$ cannon shell fired breaks into two equal parts at its highest point. One part retraces its path to the cannon with kinetic energy $E_1$,and the kinetic energy of the second part is $E_2$. The relation between $E_1$ and $E_2$ is:

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