A 0.32-kg ball is moving horizontally at 30 m/s just before suddenly bouncing off a wall Just after the bounce, it is moving horizontally at 25 m/s but in the opposite direction. DOK 1 (4 marks) (a) What is the magnitude of the change in momentum of the ball during the bounce? (b) What percentage of the ball's original kinetic energy was lost in the collision?

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facundo3141592 facundo3141592 · Answer 1 · 2022-03-31T13:31:07+02:00

We will see that the change in momentum is P' - P = -17.6 kg*m/s, and the percentage of kinetic energy lost is 30.56%.

The momentum is given by the product between the mass and the speed.

So the original momentum is:

P = 0.32kg*30m/s = 9.6 kg*m/s

After the bounce, the new velocity is -25m/s, then the new momentum is:

P' = 0.32kg*-25 m/s = -8 kg*m/s

The change in momentum is:

P' - P = -8 kg*m/s - 9.6 kg*m/s = -17.6 kg*m/s

The kinetic energy is:

KE = (1/2)*M*v^2

The initial kinetic energy is:

KE = (1/2)*0.32kg*(30 m/s)^2 = 144 J

The final kinetic energy is:

KE = (1/2)*0.32kg*(-25 m/s)^2 = 100 J

The percentage lost is:

P = 100%*(144 J - 100J)/(144 J) = 30.56%

If you want to learn more about the momentum, you can read:

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