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 heaviest fragment in m/s will be

• A. $7\sqrt{2}$
• B. $5\sqrt{2}$
• C. $3\sqrt{2}$
• D. $\sqrt{2}$

Answer 1

Answer: (A)

Since $5\, kg$ body explodes into three fragments with masses in the ratio $1: 1: 3$ thus, masses of fragments will be $1 \, kg , 1 \, kg$ and $3\, kg$ respectively.
The magnitude of resultant momentum of two fragments each of mass $1\, kg$, moving with velocity $21\, m / s$, in perpendicular directions is
$\sqrt{\left(m_{1} v_{1}\right)^{2}+\left(m_{2} v_{2}\right)^{2}}$
$m^{\prime} v^{\prime}=\sqrt{(21)^{2}+(21)^{2}}=21 \sqrt{2} \, kg\, m / s$
According to law of conservation of linear momentum
$m_{3} v_{3}=m^{\prime} v^{\prime}=21 \sqrt{2}$
or $3 v_{3}=21 \sqrt{2}$
or $v_{3}=7 \sqrt{2}\, m / s$