Question

An explosion blows a rock into three parts. Two pieces go off at right angles to each other; 1.0 kg piece with a velocity of 12 m/s and other 2.0 kg piece with a velocity 8 m/s. If the third piece flies off with a velocity 40 m/s, compute the mass of third piece.

• A. 1 kg
• B. 0.5 kg
• C. 2 kg
• D. kg

Answer 1

The correct option is B

0.5 kg

As we know,

${\overrightarrow{x}}_{COM}=\frac{m_1\overrightarrow{v_1}+m_2\overrightarrow{v_2}+m_3\overrightarrow{v_3}}{m_1+m_2+m_3}$

But the Rock was initially at Rest and No Net external force acts on it.
Therefore the COM will remain at rest of the system, after the explosion.
${\overrightarrow{x}}_{COM}=0$
Taking Rightwards as the x and upwards as +ve y axis
$(V_x{)}_{COM}=0=\frac{m_1\left(v_1\right)x+m_2(v_2{)}_x+m_1(v_1{)}_x}{m_1+m_2+m_3}$
$\Rightarrow \frac{1x12+0+m_3(v_3{)}_x}{1+2+m_3}=0$
12 + $m_3(V_3{)}_x$ = 0 ........(i) $(V_y{)}_{COM}=0=\frac{m_1\left(v_1\right)y+m_2(v_2{)}_y+m_1(v_1{)}_y}{m_1+m_2+m_3}$
$\Rightarrow \frac{2x9+0+m_3(v_3{)}_y}{1+2+m_3}=0$
16 + $m_3(V_3{)}_y$ = 0 ........(ii)
Also given, $\sqrt{(v_3{)}^{2}x+(v_3{)}_y^2}=40$
Using (i) & (ii)
${\left(\frac{-12}{m_3}\right)}^2+{\left(\frac{-16}{m_3}\right)}^2=1600$
$\frac{144}{m_3^2}+\frac{256}{m_3^2}=1600$
$\frac{400}{1600}=m_3^2$
$m_3=\sqrt{\frac{400}{1600}}=\frac{20}{40}=0.5kg$