Question

$1$ kg body explodes into three fragments. The ratio of their masses is $1:1:3$. The fragments of same mass move perpendicular to each other with speeds $30$ m/s, while the heavier part remains in the initial direction. The speed of heavier part is:

• A. $ \frac{10}{\sqrt{2}}\text{m/s} $
• B. $ \text{10}\sqrt{2\,}\text{m/s} $
• C. $ \text{20}\sqrt{2\,}\text{m/s} $
• D. $ \text{30}\sqrt{2\,}\text{m/s} $

Answer 1

Answer: (B)

Key Idea: 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
$ m\times 30-3\,m\times v\cos \theta =0 $ ... (i)
and $ m\times 30-3\,m\times v\sin \theta =0 $... (ii)
These give $ 3\,mv\,\cos \theta =3mv\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 Eq. (i), we get $ 30\,m=3mv\cos {{45}^{o}}=\frac{3mv}{\sqrt{2}} $
$ \therefore $ $ v=10\sqrt{2}\,m/s $
The third piece will move with a velocity of $ 10\sqrt{2}\,m/s $ in a direction making an angle of $ {{135}^{o}} $ with either piece.
Alternative: Key Idea: The square of momentum of third piece is equal to sum of squares of momentum of first and second piece. As from key idea,
$ {{p}_{3}}^{2}={{p}_{1}}^{2}+{{p}_{2}}^{2} $
or $ {{p}_{3}}=\sqrt{{{p}_{1}}^{2}+{{p}_{2}}^{2}} $
or $ 3m{{v}_{3}}=\sqrt{{{(m\times 30)}^{2}}+{{(m\times 30)}^{2}}} $
or $ {{v}_{3}}=\frac{30\sqrt{2}}{3}=10\sqrt{2}\,m/s $