A shell of mass 'M' initially at rest suddenly explodes in three fragments. Two of
these fragments are of mass 'M \(/ 4\) ' each, which move with velocities \(3 \mathrm{~ms}^{-1}\) and
\(4 \mathrm{~ms}^{-1}\) respectively in mutually perpendicular directions. The magnitude of velocity of the third fragment is

The mass \(M\) is initially at rest, hence its initial momentum is zero. By law or conservation of momentum, the net momentum of the three pieces should be zero. The moments of the two pieces are
\(\mathrm{p}_{1}=\frac{3 \mathrm{M}}{4} \text { and } \mathrm{p}_{2}=\frac{5 \mathrm{M}}{4}\)
Three are at right angles to each other.
Their resultant will be \(p=\sqrt{p_{1}^{2}+p_{2}^{2}}\)
\(\begin{array}{l}
=\sqrt{\left(\frac{3 M}{4}\right)^{2}+\left(\frac{4 M}{4}\right)^{2}} \\
=\frac{M}{4} \sqrt{(3)^{2}+(4)^{2}} \\
=\frac{M}{4} \sqrt{25}=\frac{5 M}{4}
\end{array}\)
The momentum of the third piece will be equal and opposite to this.
The mass of the third piece will be \(\frac{M}{2}\)
If its velocity is \(\mathrm{V}\), then
\(\begin{array}{l}
\frac{M}{2} \cdot V=\frac{5 M}{4} \\
\therefore V=2.5 \mathrm{~ms}^{-1}
\end{array}\)