Question

A cannon ball is fired with a velocity of $200m/s$ at an angle of ${60}^{\circ }$ with the horizontal. At the highest point of its flight, it explodes into three equal fragments, one going vertically upwards with a velocity of $100m/s$, the second one falling vertically downwards with a velocity of $100m/s$. The third fragment will be moving with a velocity of :-

• A. $100m/s$ in the horizontal direction
• B. $300m/s$ in the horizontal direction
• C. $300m/s$ in a direction making an angle of ${60}^{\circ }$ with the horizontal
• D. $200m/s$ in a direction making an angle of ${60}^{\circ }$ with the horizontal

Answer 1

The correct option is B $300m/s$ in the horizontal direction

before explosion velocity at $A$
$=v_3$
$=v_0\cos {60}^o$
$v_2=\frac{v_0}{2}$
$\because$ after $A$ it explodes in $3$ part
$\Rightarrow$ Conserving momentum in $x-$ direction
$m{v}_2=\frac{m}{3}\times v_4\cos \alpha$
$m\frac{v_0}{2}=\frac{v_4}{3}\cos x$ .....(1)
$\because$ Covering momentum in $y$ direction
$0=\frac{m{v}_1}{3}-\frac{m{v}_2}{3}+\frac{m{v}_4}{3}\sin \alpha$
$0=100-100+v_2\sin \alpha$
$v_2\sin \alpha =0$
$a=0$ .....(ii)
(i) and (ii)
$\frac{v_0}{2}=\frac{v_4}{3}\cos \theta$
$v_4=\frac{3v_0}{2}$
$v_4=200\times \frac{3}{2}$

$=300m/s$ in the horizontal direction