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
$m\times 30-3m\times v\cos \theta =0$ ... (i)
and $m\times 30-3m\times v\sin \theta =0$... (ii)
These give $3mv\cos \theta =3mv\sin \theta$
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 $30m=3mv\cos {45}^o=\frac{3mv}{\sqrt{2}}$
$\therefore$ $v=10\sqrt{2}m/s$
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 $v_3=\frac{30\sqrt{2}}{3}=10\sqrt{2}m/s$