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(D) Let the mass of the shell be $M$ and its velocity at the highest point be $v_h = u \cos 	heta$. The momentum at the highest point is $P = M u \co

Vedclass · Accepted Answer

$E_2 = 9 E_1$

Vedclass · Answer

(D) Let the mass of the shell be $M$ and its velocity at the highest point be $v_h = u \cos 	heta$. The momentum at the highest point is $P = M u \cos 	heta$.After breaking into two equal parts of mass $M/2$, one part retraces its path, meaning its velocity is $-u \cos 	heta$. Let the velocity of the second part be $v_2$.By conservation of linear momentum:$M u \cos 	heta = \frac{M}{2} (-u \cos 	heta) + \frac{M}{2} v_2$$u \cos 	heta = -\frac{1}{2} u \cos 	heta + \frac{1}{2} v_2$$\frac{3}{2} u \cos 	heta = \frac{1}{2} v_2 \implies 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 $E_2 = \frac{1}{2} (M/2) (3 u \cos 	heta)^2 = \frac{1}{4} M (9 u^2 \cos^2 	heta) = \frac{9}{4} M u^2 \cos^2 	heta$.Comparing the two, we get $E_2 = 9 E_1$.

$A$ cannon shell fired breaks into two equal parts at its highest point. One part retraces the 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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