Question

a bomb at rest explodes into 3 fragments of equal mass. 2 fragments fly off at right angles to each other with velocities 9m/s and 12 m/s respectively. calculate the speed of the third.

Answer 1

not sure about the answer.....
just an intelligent guess.....

Answer 2

Solution :

Let m be the mass of each fragment.

$\tt \therefore \: \vec{ p_1} = m \times 9 \\ \: \: \: \: \: \tt \vec{p_2} = m \times 12$

As $\tt\vec{p_1}$ and $\tt\vec{p_2}$ are perpendicular to each other, therefore, resultant momentum of the two fragments

$\tt \vec p = \sqrt{ { (\vec{p_1})}^{2} + { (\vec{p_2})}^{2} } \\ \\ \tt \vec p = \sqrt{ {(9)}^{2} + {(12)}^{2} } \\ \\ \tt \vec p = \sqrt{81 + 144} \\ \\ \tt \vec p = \sqrt{225} \\ \\ \tt \vec p = 15$

According to the principle of conservation of linear momentum

$\tt \vec p + \vec{p_3} = 0 \\ \\ \tt\vec{p_3} = - \vec p \\ \\\tt\vec{p_3} = - 15 \: m\\ \\ \tt m \times {v_3} = - 15 \: m\\ \\ \large \boxed{ \tt\blue{ v_3 = - 15 \: {m \: s}^{ - 1}} }$

Negative sign implies that third fragment will fly in a direction opposite to the direction of resultant momentum of two fragments.