Question

a bomb at rest explodes into 3 fragments of equal masses . two fragments fly off at right angles to each other with velocities 9m/s and 10m/s respectively . what is the speed of the third fragment ?

Answer 1

Dear Student,
$$\begin{gathered} Theresul\tan tmomentum \\ ofboththeparticlewillbeoppositeofthethirdparticle \\ \frac{m}{3}v=\sqrt{{\left(\frac{m\times 9}{3}\right)}^2+{\left(\frac{m\times 10}{3}\right)}^2} \\ \frac{v}{3}=\frac{\sqrt{81+100}}{3} \\ v=\sqrt{181}m/s \end{gathered}$$
Regards