# A baseball bat of a mass of 2 kg and a velocity of 15 m/s moves towards a 0.4-kg baseball that is moving at 33 m/s. There is an elastic collision between the bat and the ball. If the final velocity of the ball is 47 m/s, what is the total kinetic energy of the bat after the collision?

The final kinetic energy of the baseball bat after the collision is 1 J.

## Expert Answers When two objects undergo an elastic collision, both momentum and kinetic energy are conserved. There is no net loss of the total kinetic energy in the collision.

A baseball bat of mass of 2 kg moving at 15 m/s has an elastic collision with a baseball of mass 0.4 kg baseball moving at 33 m/s. After the collision, the velocity of the ball is 47 m/s. The final kinetic energy of the bat has to be determined.

The total initial kinetic energy of the baseball-baseball bat system was

`KE_0 = (1/2)*2*(15)^2 + (1/2)*0.4*(33)^2 = 442.8`

After the collision, the velocity of the ball is 47 m/2. The kinetic energy of the ball is `KE_1= (1/2)*0.4*47^2 = 441.8`

The kinetic energy of the bat after the collision is the difference of `KE_0` and `KE_1` . This is equal to 442.8 - 441.8 = 1 J

The final kinetic energy of the baseball bat is 1 J.

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