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} = 15 \, E_{1} $
• B. $ E_{2} = E_{1} $
• C. $ E_{2} = 4 \, E_{1} $
• D. $ E_{2} = 9 \, E_{1} $

Answer 1

Answer: (D)

At highest point
$m u \,\cos \,\theta=-\frac{m}{2} u \,\cos \,\theta+\frac{m}{2} v $
$m u\, \cos\, \theta+\frac{m}{2} u\, \cos \,\theta=\frac{m}{2} v$
$\frac{2 \,m u \,\cos\, \theta + m u \,\cos\, \theta}{2}=\frac{m}{2} v$
$m v=3 m u \,\cos \,\theta $
$v=3 u \,\cos \,\theta\,\,\,...(i)$
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 9 \,u^{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))
$E_{2}= 9 \,E_{1}$