
A particle of mass $4m$, at rest, explodes into three fragments. Two of the fragments each of mass m are found to move with a speed $v$ in mutually perpendicular directions. The total energy released in the explosion is:
(A) $2m{v^2}$
(B) $\dfrac{1}{2}m{v^2}$
(C) \[m{v^2}\]
(D) $\dfrac{3}{2}m{v^2}$
Answer
554.1k+ views
Hint: We are looking for the change in momentum when it blasts. It could be simply solved by the momentum formula.
Formula used: To calculate the momentum $p$
$p = mv$
Here, $p$ is the momentum of the body,
$m$ is the mass of the body
$v$is the velocity of the body,
Also, we use the equation of motion.
Complete step by step answer:
We are given that the mass of the big particle is 4m and velocity of two fragments are mutually perpendicular to each other.
If \[{p_1}{\text{ and }}{p_2}\] are the assumed momentum of the fragments then momentum of third fragment is ${p_3} = \sqrt {p_1^2 + p_2^2} = \sqrt 2 mv$
Velocity of third fragment $ = \dfrac{v}{{\sqrt 2 }}$
Total energy released in the explosion $\dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2} + \dfrac{1}{2}2m{\left( {\dfrac{v}{{\sqrt 2 }}} \right)^2}$
On solving we get, $ = \dfrac{3}{2}m{v^2}$
Thus, the total energy released in the explosion is $\dfrac{3}{2}m{v^2}$
So, we need to see from the above options, and select the correct value.
Thus, the correct answer is option D.
Additional Information : Momentum is the property of a moving body by virtue of its mass and motion and that is equal to the product of the body's mass and velocity. In everyday life momentum is used many times. This is because the momentum of vehicles running at high speeds is very high and causes a lot of damage to the vehicles and injuries to passengers during the collision. A bullet, although small in mass, has a large momentum because of an extremely large velocity
Note: The mistake that should be avoided here is while considering the number of fragments.
Some examples of change in momentum from daily life are:
A bowling ball (large mass) moving very slowly (low velocity) can have the same momentum as a baseball (small mass) that is thrown fast (high velocity). A bullet is another example where the momentum is very-very high, due to the extraordinary velocity.
Formula used: To calculate the momentum $p$
$p = mv$
Here, $p$ is the momentum of the body,
$m$ is the mass of the body
$v$is the velocity of the body,
Also, we use the equation of motion.
Complete step by step answer:
We are given that the mass of the big particle is 4m and velocity of two fragments are mutually perpendicular to each other.
If \[{p_1}{\text{ and }}{p_2}\] are the assumed momentum of the fragments then momentum of third fragment is ${p_3} = \sqrt {p_1^2 + p_2^2} = \sqrt 2 mv$
Velocity of third fragment $ = \dfrac{v}{{\sqrt 2 }}$
Total energy released in the explosion $\dfrac{1}{2}m{v^2} + \dfrac{1}{2}m{v^2} + \dfrac{1}{2}2m{\left( {\dfrac{v}{{\sqrt 2 }}} \right)^2}$
On solving we get, $ = \dfrac{3}{2}m{v^2}$
Thus, the total energy released in the explosion is $\dfrac{3}{2}m{v^2}$
So, we need to see from the above options, and select the correct value.
Thus, the correct answer is option D.
Additional Information : Momentum is the property of a moving body by virtue of its mass and motion and that is equal to the product of the body's mass and velocity. In everyday life momentum is used many times. This is because the momentum of vehicles running at high speeds is very high and causes a lot of damage to the vehicles and injuries to passengers during the collision. A bullet, although small in mass, has a large momentum because of an extremely large velocity
Note: The mistake that should be avoided here is while considering the number of fragments.
Some examples of change in momentum from daily life are:
A bowling ball (large mass) moving very slowly (low velocity) can have the same momentum as a baseball (small mass) that is thrown fast (high velocity). A bullet is another example where the momentum is very-very high, due to the extraordinary velocity.
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