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)

$\text{10}\sqrt{2\,}\text{m/s}$

Answer 2

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 E (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 ${{v}_{3}}=\frac{30\sqrt{2}}{3}=10\sqrt{2}\,m/s$