An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are,…
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, $1 \mathrm{~kg}$ first part moving with a velocity of $12 \mathrm{~ms}^{-1}$ and $2 \mathrm{~kg}$ second part moving with a velocity of $8 \mathrm{~ms}^{-1}$. If the third part flies off with a velocity of $4 \mathrm{~ms}^{-1}$, its mass would be
$5 \mathrm{~kg}$
$7 \mathrm{~kg}$
$17 \mathrm{~kg}$
$3 \mathrm{~kg}$
Solution
Key Idea Apply law of conservation of linear momentum.
Momentum of first part $=1 \times 12=12 \mathrm{~kg} \mathrm{~ms}^{-1}$
Momentum of the second part
$=2 \times 8=16 \mathrm{~kg} \mathrm{~ms}^{-1}$
$\therefore$ Resultant momentum
$=\sqrt{(12)^2+(16)^2}=20 \mathrm{~kg} \mathrm{~ms}^{-1}$
The third part should also have the same momentum.
Let the mass of the third part be, then
$\begin{aligned}
4 \times M & =20 \\
M & =5 \mathrm{~kg}
\end{aligned}$
Alternative :
$\begin{aligned}
\mathrm{Mv} \cos \theta & =12 \\
\mathrm{Mv} \sin \theta & =16 \\
\tan \theta & =\frac{16}{12}=\frac{4}{3} \\
M & =\frac{12 \times 5}{4 \times 3}=\frac{60}{12}=5 \mathrm{~kg}
\end{aligned}$
$\begin{aligned}
\mathrm{Mv} \cos \theta & =12 \\
\mathrm{Mv} \sin \theta & =16 \\
\tan \theta & =\frac{16}{12}=\frac{4}{3} \\
M & =\frac{12 \times 5}{4 \times 3}=\frac{60}{12}=5 \mathrm{~kg}
\end{aligned}$