Location of centre on mass for multiple point masses
Question

# A body of mass(4m) is lying in x–y plane at rest. It suddenly explodes into three pieces. Two pieces, each of mass (m) move perpendicular to each other with equal speeds ($\nu$). The total kinetic energy generated due to explosion is

Difficult
Solution

## Let ${\stackrel{\to }{v}}^{\mathrm{\prime }}$be velocity of third piece of mass 2 m. Initial momentum, ${\stackrel{\to }{p}}_{i}=0$(As the body is at rest)Final momentum, ${\stackrel{\to }{p}}_{f}=mv\stackrel{^}{ı}+mv\stackrel{^}{ȷ}+2m{\stackrel{\to }{v}}^{\mathrm{\prime }}$$\begin{array}{l}{\stackrel{\to }{p}}_{i}={\stackrel{\to }{p}}_{f}\\ 0=mv\stackrel{^}{i}+mv\stackrel{^}{j}+2m{\stackrel{\to }{v}}^{\mathrm{\prime }}\\ {\stackrel{\to }{v}}^{\mathrm{\prime }}=-\frac{v}{2}\stackrel{^}{i}-\frac{v}{2}\stackrel{^}{j}\end{array}$$v\text{'}=\sqrt{{\left(-\frac{v}{2}\right)}^{2}+{\left(-\frac{v}{2}\right)}^{2}}=\frac{v}{\sqrt{2}}$$\begin{array}{l}=\frac{1}{2}m{v}^{2}+\frac{1}{2}m{v}^{2}+\frac{1}{2}\left(2m\right){v}^{\mathrm{\prime }2}\\ =\frac{1}{2}m{v}^{2}+\frac{1}{2}m{v}^{2}+\frac{1}{2}\left(2m\right){\left(\frac{v}{\sqrt{2}}\right)}^{2}\\ =m{v}^{2}+\frac{m{v}^{2}}{2}=\frac{3}{2}m{v}^{2}\end{array}$

Get Instant Solutions
When in doubt download our app. Now available Google Play Store- Doubts App