2 Answers | Add Yours
The speed of the ball as it approaches the wall is 10m/s. When it bounces off the wall there is a change in velocity of 20m/s. Now the change in velocity takes place over an instantaneous duration of time but we cannot say exactly how long that is, so it is better to consider that as a constant C.
So the change in velocity of the ball is 20 m/s in C sec. This gives an acceleration of 20/C. The change in velocity is due to a force exerted by the wall. As an action has an equal and opposite reaction a force is exerted on the wall and that is equal to mass* acceleration = 20/C* (500/1000) = 20/C *0.5 N = 10/C N.
The force exerted on the wall is 10/C N.
The force extend on the wall by the ball is reacted with an equal but opposite force in accordance with the 3rd law of motion .
So the mass of the ball = 500g. = 1/2 kg.
Velocity of the ball = 10m/sec.
Time or duration the in which thel ball lost its velocity to zero is say t secs. (But here the information of time is not given)
Therefore the acceleration (in this case retrdation ) of the ball = (10m/s)t s = (10/t)m/s^2
Therefore the force = ma = ((1/2)Kg * (10/t)m/s^2 = (5/t) Newton is exerted on the wall by the ball.
We’ve answered 317,858 questions. We can answer yours, too.Ask a question