Question

An explosive blows a rock into three parts. Two pieces go off to right angles to each other. 1.0 kg piece with a velocity 12 m/s and another 0.2 kg piece with a velocity of 8 m/s. If the third piece flies off with a velocity of 40 m/s, compute the mass of the third piece.

Answer 1

Hint: The initial momentum of the rock before the blow is zero. Use the law of conservation of momentum. Assign a direction for the first and second piece of the rock and use the formula for the linear momentum.

Formula used:
The linear momentum of the particle of mass m and velocity v is,
\[P = mv\]

Complete step by step answer:
We assume the first piece of mass 1.0 kg moves in the direction of the x-axis and the second piece moves in the direction of the y-axis as they go off to right angles to each other. We know that, x-axis is represented by the unit vector \[\hat i\] and the y-axis is represented by the unit vector \[\hat j\].

We have the initial momentum of the rock is zero. According to the conservation of momentum, we can write,
\[0 = {\vec P_1} + {\vec P_2} + {\vec P_3}\]
\[ \Rightarrow \left| {{{\vec P}_3}} \right| = \left| {{{\vec P}_1}} \right| + \left| {{{\vec P}_2}} \right|\]

Here, \[{P_1}\] is the momentum of the first piece, \[{P_2}\] is the momentum of the second piece and \[{P_3}\] is the momentum of the third piece.

In the above equation, we have taken the magnitude of the momentum because we don’t know the direction of the third particle after the blow.

We know that the momentum of the body is the product of its mass and velocity. Therefore, we can write the above equation,
\[{m_3}{\vec v_3} = {m_1}{v_1}\hat i + {m_2}{v_2}\hat j\]


We substitute 40 m/s for \[{v_3}\], 1.0 kg for \[{m_1}\], 0.2 kg for \[{m_2}\], 12 m/s for \[{v_1}\], and 8 m/s for \[{v_2}\] in the above equation.
\[{m_3}\left( {40} \right) = \left( 1 \right)\left( {12} \right)\hat i + \left( {0.2} \right)\left( 8 \right)\hat j\]
\[ \Rightarrow 40{m_3} = 12\hat i + 1.6\hat j\]
\[ \Rightarrow 40{m_3} = \sqrt {{{12}^2} + {{1.6}^2}} \]
\[ \Rightarrow {m_3} = 0.30\,kg\]

Therefore, the mass of the third piece is 0.30 kg.

Note:
If the direction of bodies after the collision is not linear, students should never ignore the direction of the bodies. While solving these types of questions, one should assign a proper direction for each particle. In the above solution, we have ignored the direction of the third piece as we want to determine its mass and not the velocity.