A stationary bomb explodes into three pieces. One piece of $2kg$mass moves with a velocity of $8m{s}^{-1}$ at right angles to the other piece of mass $1kg$ moving with a velocity of $12m{s}^{-1}$. If the mass of the third piece is $0.5kg$, then its velocity is

The correct option is D

$40m{s}^{-1}$

Step 1: Given data

Mass of one piece, $m_1=2kg$

The velocity of one piece, $v_1=8m{s}^{-1}$

Mass of another piece, $m_2=1kg$

The velocity of another piece, $v_2=12m{s}^{-1}$

Mass of the third piece, $m_3=0.5kg$

Step 2. Find the velocity of the third piece, ${\mathit{v}}_{\mathbf{3}}$

The momentum of$2kg$is $P_1=m_1{v}_1$

$$\begin{gathered} =2x8 \\ =16kgm{s}^{-1} \end{gathered}$$

The momentum of $1kg$is $P_2=m_2{v}_2$

$$\begin{gathered} =1x12 \\ =12kgm{s}^{-1} \end{gathered}$$

Both are at right angles so the results will be$P=\sqrt{\left({P_1}^2+{P_2}^2\right)}$

$$\begin{gathered} =\sqrt{\left({16}^2+{12}^2\right)} \\ =\sqrt{\left(256+144\right)} \\ =\sqrt{400} \\ =20kgm{s}^{-1} \end{gathered}$$

Initial momentum is zero since the ball is stationary. Therefore, the final momentum should also be zero.

So, the third piece will have momentum in the opposite direction to make the net momentum zero.

Hence, the velocity of the third piece, $v_3=\frac{P}{m_3}$

$$\begin{gathered} =\frac{20}{0.5} \\ =40m{s}^{-1} \end{gathered}$$

Hence, option (D) is correct.