Question

A $5kg$ shell kept at rest suddenly splits up into three parts. If two parts of mass $2kg$ each are found flying due north and east with a velocity of $5m/s$ each, what is the velocity of the third part after explosion?

• A. $10m/s$ due north-east
• B. $\frac{10}{\sqrt{2}}m/s$ due south-east
• C. $10\sqrt{2}m/s$ due south-west
• D. $10\sqrt{2}m/s$ due south-east

Answer 1

Answer: (C)

$p_x=\text{ momentum along east }=10kgm/s$
$p_y=\text{ momentum along north }=10kgm/s$
Resultant momentum,
$p=\sqrt{p_x^2+p_y^2}=\sqrt{10^2+10^2}=10\sqrt{2}kgm/s$ due north-east
According to the law of conservation of momentum
$p+p_3=0\text{ or }{p}_3=-p\text{ or }{p}_3=-10\sqrt{2}kgm/s$
or $p_3=10\sqrt{2}kgm/s$ due south-west
$\therefore v_3=\frac{p_3}{m_3}=\frac{10\sqrt{2}}{1}=10\sqrt{2}m/s\text{ due south-west }$