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 $'\theta '$ , 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
$mucos \theta = \, -\frac{m}{2}ucos ⁡ \theta +\frac{m}{2}v$
$mucos \theta +\frac{m}{2} \, ucos ⁡ \theta =\frac{m}{2}v$
$\frac{2 m u cos \theta + m u cos ⁡ \theta }{2}=\frac{m}{2}v$
$mv=3mucos \theta $
$v=3ucos \theta $ .......(i)
The kinetic energy
$E_{1}=\frac{1}{2}\times \frac{m}{2}u^{2}cos^{2} \theta $
$E_{1}=\frac{1}{4}mu^{2}cos^{2} \theta $ .......(ii)
Similarly,
$E_{2}=\frac{1}{2}\times \frac{m}{2}\times 9u^{2}cos^{2} \theta \, $
$E_{2}=\frac{9}{4}mu^{2}cos^{2} \theta $ ......(iii)
The relation between the $E_{1}$ and $E_{2}$
$E_{2}=9\times E_{1}$ ( from the equation (ii) )
$E_{2}=9E_{1}$