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# A bomb at rest explodes into two pieces of masses $20\,kg$ and $30\,kg$ . If the first one moves with a velocity of $30\,m{s^{ - 1}}$ . Find the velocity of the other?A. $- 20\,m{s^{ - 1}}$B. $0.5\,m{s^{ - 1}}$C. $+ 20\,m{s^{ - 1}}$D. $0.3\,m{s^{ - 1}}$

Last updated date: 19th Jun 2024
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Hint: Since, there are two masses here. So, we have to use the law of conservation of linear momentum here.
According to the law of conservation of linear momentum, the overall momentum of the system is often maintained for an object or a system of objects if no external force acts on them.

Mass, ${m_1} = 20\,kg$

Mass, ${m_2} = 30\,kg$

So, total mass, $m = 50\,kg$

First velocity, ${v_1} = 30\,m{s^{ - 1}}$

Let second velocity be ${v_2}$

Since, the bomb was at rest, so, total velocity $v$ will be zero.
We know that,
According to law of conservation of linear momentum,
$mv = {m_1}{v_1} + {m_2}{v_2} \\ 50 \times 0 = 20 \times 30 + 30 \times {v_2} \\ {v_2} = - 20\,m{s^{ - 1}} \\$

Since, the bomb explodes the second piece of the bomb moves in the opposite direction from the first one. So, we take the second velocity as negative.
Hence, option A is correct.