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

Answer: (A)

$4.5\sqrt{2}\,m{{s}^{-1}}$

Answer 2

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 ${{v}_{3}}=4.5\sqrt{2}\,m{{s}^{-1}}$