Question

A projectile is moving at $60m/s$ at its highest point, where it breaks into two equal parts due to an internal explosion .One part moves vertically up at $50m/s$ with respect to the ground.The other part will move at a speed of :

• A. $110m/s$
• B. $120m/s$
• C. $130m/s$
• D. $10\sqrt{61}m/s$

Answer 1

The correct option is C $130m/s$

From conservation of lenear momentum ,
$m\overrightarrow{v}=\frac{m}{2}\overrightarrow{v_1}+\frac{m}{2}{\overrightarrow{v}}_2$
(Here we take particle and earth as a system so in this case external force is zero and linear momentum is conserved)
Where $\overrightarrow{v}$is velocity of particle before explosion $\overrightarrow{v_1}$ and $\overrightarrow{v_2}$ are velocity of its equal pieces.
Here
$\overrightarrow{v}=60\hat{i}......($ in $x-$directon$)$
$\overrightarrow{v_1}=50\hat{j}......(iny-direction)$
So$\overrightarrow{v_2}=120\hat{i}-50\hat{j}$
or
$|\overrightarrow{v_2}|=130m/s$
[From conservation of linear momentum]
$\tan \theta =\frac{-50}{120}$