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 $0.2$ kg piece with a velocity pf $8 m / s$ . If the third piece flies off with a velocity of $40 m / s$ . compute mass of the third piece.

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Solution

Let $m_{1}, m_{2}$ and $m_{3}$ be the masses of the three pieces. $m_{1} = 1.0 k g, m_{2} = 2.0 k g$
Let $ν_{1} = 12 m / s, ν_{2} = 8 m / s, ν_{3} = 40 m / s$ . Let $ν_{2}$ be directed along x-axis, respectively and $ν_{3}$ be directed as shown.
By the principle of conservation of momentum, initial momentum is zero. Hence,
along $x - a x i s : 0 = m_{1} ν_{1} - m_{3} ν_{3} C o s θ$
along $y - a x i s : 0 = m_{2} ν_{2} - m_{3} ν_{3} C o s θ$
$⟹ m_{1} ν_{1} = m_{3} ν_{3} C o s θ$ and $m_{2} ν_{2} = m_{3} ν_{3} C o s θ$
By squaring and adding, we get $m_{1}^{2} ν_{1}^{2} + m_{2}^{2} ν_{2}^{2} = m_{3}^{2} ν_{3}^{2}$
$m_{2}^{3} = \frac{1^{2} (12)^{2} + (2)^{2} (8)^{2}}{(40)^{2}} ⟹ m + 3 = 0.5 k g$

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