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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.

Complete step by step answer:Given,
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.

Additional information:

The linear momentum can be defined as the product of the mass and the velocity of a particle. A particle’s conservation of momentum is a property shown by any particle where the total quantity of momentum never varies.
Launching rockets is one of the applications of momentum conservation. The exhaust gases are forced backward by the rocket fuel fires and because of this the rocket is forced upward. Motorboats also run on the same principle, moving forward to maintain momentum is response.
The total momentum of a rocket and its fuel is zero prior to launch. The downward momentum of the expanding exhaust gases during launch only matches the upward momentum of the rising rocket in magnitude, so that the system’s overall momentum remains stable at zero value in this case.

Note:Here we have to be careful while writing the answer of the second velocity since the answer may be negative or positive.