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

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.