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 ???
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)
