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A bag of sand of mass M is suspended by a rope. A bullet of mass m traveling with speed v gets embedded in it the loss of kinetic energy is ???


sumansv1966 13th Aug, 2020
Answer (1)
Prabhav Sharma 13th Aug, 2020

Hello,

Initial Velocity of bullet is v ( Given)

Let mass of the bullet be m

Therefore, Initial momentum = mv

and, Initial Kinetic Energy of the bullet will be: 1/2 * m * v^2

Now after the collision, let final velocity of the system (sand bag + bullet) be u

Mass of the bag is M ( Given)

Therefore, Final momentum of the system = (m + M) * u

and, Final kinetic energy of the system = 1/2 * M * u^2

Now according to the law of conservation of momentum, momentum remains conserved, therefore we have initial momentum = final momentum

Therefore, mv = (m + m) * u

= u = mv / (m + M)

Now, loss in kinetic energy = Initial Kinetic energy - final kinetic energy

Therefore, loss in kinetic energy:

=  1/2 * m * v^2 - 1/2 * M * u^2

Substituting the values of u obtained above, we get:

= 1/2 * m * v^2 - 1/2 * (m + M) * (mv / (m + M))^2

Solving the above equation we get, loss in kinetic energy =

(m * M * v^2 ) / 2 * (m + M) (Answer)



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