A tennis player practices against a wall, hitting a 0.1 kg ball towards the
wall with a velocity of +20 m/s. The ball bounces straight back from the
wall with a velocity of -20 m/s. What is the average force the wall exerts on
the ball if the ball is in contact with the wall for 0.05 s?*

80 Newton

Explanation:

Initial velocity of ball = +20 m/s.

Final velocity of ball = -20 m/s

Mass of ball = 0.1kg

Time taken = 0.05 seconds

Average force = (Change in momentum of moving ball / Time taken)

Since, change in momentum = Mass (final velocity - initial velocity)

Change in momentum =0.1 x (-20 - (+20))

= 0.1 x (-20-20)

= 0.1 x (-40)

= -4.0 kgm/s

Then, put -4.0 kgm/s in the equation of force when Average Force = (Change in momentum / Time taken)

= (-4.0kgm/s / 0.05 seconds)

= 80Newton (note that the negative sign does not reflect on the magnitude of force)

Thus, the average force exerted on the ball is 80N

pe = weightxheight = 2000x1000=2,000,000joules