Question

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 kinetic energy of the second part is $E_2$ , 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)

At highest point
$mu\cos \theta =-\frac{m}{2}u\cos \theta +\frac{m}{2}v$
$mu\cos \theta +\frac{m}{2}u\cos \theta =\frac{m}{2}v$
$\frac{2mu\cos \theta +mu\cos \theta }{2}=\frac{m}{2}v$
$mv=3mu\cos \theta$
$v=3u\cos \theta ...(i)$
The kinetic energy
$E_1=\frac{1}{2}\times \frac{m}{2}{u}^2{\cos }^2\theta$
$E_1=\frac{1}{4}m{u}^2{\cos }^2\theta ...(ii)$
Similarly,
$E_2=\frac{1}{2}\times \frac{m}{2}\times 9u^2{\cos }^2\theta$
$E_2=\frac{9}{4}m{u}^2{\cos }^2\theta ...(iii)$
The relation between the $E_1$ and $E_2$
$E_2=9\times E_1$(from the Eq.(i))
$E_2=9E_1$