Question

A stationary body of mass m explodes into the three parts having masses in the ratio 1:3:3. The two fractions with equal masses move at right angles to each other with a velocity of $ 1.5\,m{{s}^{-1}}. $ The velocity of the third part is:

• A. $ 4.5\sqrt{2}\,m{{s}^{-1}} $
• B. $ 5\,m{{s}^{-1}} $
• C. $ 5\sqrt{32}\,m{{s}^{-1}} $
• D. $ 1.5\,m{{s}^{-1}} $

Answer 1

The Correct Option is A

Equate the momenta of the system along two perpendicular axes. Let $ u $ be the velocity and $ \theta $ the direction of the third piece as shown.
Equating the momenta of the system along OA and OB to zero, we get $ 3m\times 1.5-m\times v\cos \theta =0 $ ..(i)
and $ 3m\times 1.5-m\times v\sin \theta =0 $ ..(ii)
These give $ mv=\,\cos \theta =mv\sin \theta $ or $ \cos \theta =\sin \theta $
$ \therefore $ $ \theta ={{45}^{o}} $
Thus, $ \angle AOC=\angle BOC={{180}^{o}}-{{45}^{o}}={{135}^{o}} $
Putting the value of $ \theta $ in E (i), we get $ 4.5\,m=mv\cos {{45}^{o}}=\frac{mv}{\sqrt{2}} $
$ \therefore $ $ v=4.5\sqrt{2}\,m{{s}^{-1}} $
The third piece will go with a velocity of
$ 4.5\sqrt{2}\,m{{s}^{-1}} $
in a direction making an angle of $ {{135}^{o}} $ with either piece.
The square of momentum of third piece is equal to sum of squares of momentum first and second pieces. As from key idea,
$ p_{3}^{2}=p_{1}^{2}+p_{2}^{2} $ or $ {{p}_{3}}=\sqrt{p_{1}^{2}+p_{2}^{2}} $ or $ m{{v}_{3}}=\sqrt{{{(3m\times 1.5)}^{2}}+{{(3m\times 1.5)}^{2}}} $ or $ {{v}_{3}}=4.5\sqrt{2}\,m{{s}^{-1}} $