Question

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

• A. $E_2=15E_1$
• B. $E_2=E_1$
• C. $E_2=4E_1$
• D. $E_2=9E_1$

Answer 1

Answer: (D)

A shell fired at an angle ${}^{\prime }{\theta }^{\prime }$ , velocity of the shell at the highest point is $ucos\theta$
Let its mass be $m$
The speed of one piece is $ucos\theta$ and let the speed of other piece be $v$ .
Hence at the highest point, from conservation of momentum
$mucos\theta =-\frac{m}{2}ucos\theta +\frac{m}{2}v$
$mucos\theta +\frac{m}{2}ucos\theta =\frac{m}{2}v$
$\frac{2mucos\theta +mucos\theta }{2}=\frac{m}{2}v$
$mv=3mucos\theta$
$v=3ucos\theta$ .......(i)
The kinetic energy
$E_1=\frac{1}{2}\times \frac{m}{2}{u}^{2}co{s}^2\theta$
$E_1=\frac{1}{4}m{u}^{2}co{s}^2\theta$ .......(ii)
Similarly,
$E_2=\frac{1}{2}\times \frac{m}{2}\times 9u^{2}co{s}^2\theta$
$E_2=\frac{9}{4}m{u}^{2}co{s}^2\theta$ ......(iii)
The relation between the $E_1$ and $E_2$
$E_2=9\times E_1$ ( from the equation (ii) )
$E_2=9E_1$