Question

A body of mass 5 kg explodes at rest into three fragments with masses in the ratio 1 : 1 : 3. The fragments with equal masses fly in mutually perpendicular directions with speeds of 21 m/s. The velocity of the heaviest fragment will be

• A. 11.5 m/s
• B. 14.0 m/s
• C. 7.0 m/s
• D. 9.9 m/s

Answer 1

The correct option is D

9.9 m/s

$\overrightarrow{P_1}+\overrightarrow{P_2}+\overrightarrow{P_3}=0\overrightarrow{P_1}=m\times v_x=1\times 21=21kgm/s\overrightarrow{P_2}=m\times v_y=1\times 21=21kgm/s\therefore Resultant=\sqrt{P_x^2+P_y^2}=21\sqrt{2}kgm/s$

The momentum of heavier fragment should be equal to resultant of $\overrightarrow{P_1}and\overrightarrow{P_2}$

$3\times v=\sqrt{P_x^2+P_y^2}=21\sqrt{2}\therefore v=7\sqrt{2}=9.89m/s$