Question

A ball of mass 0.5kg moving with a speed of 2m/s strikes a rigid wall in a direction perpendicular to the wall and is reflected back after a perfectly eristic collision. If the ball has remained in contact with the wall for 0.5s calculate the avrege force exerted on the ball by the wall . please solved this numerical ​

Answer 1

Answer:

4N

Explanation:

Initial velocity = 2m/s

Final velocity = -2m/s (Since it is in the opposite direction)

mass = 0.5kg

Initial momentum (Pi) = 0.5*2 = 1kgm/s

Final momentum (Pf) = 0.5*(-2) = -1kgm/s

Change in momentum = Pf-Pi = -1 -1 = -2kgm/s

From Newtons second law

Change in momentum/time duration of the change = force applied

Given time = 0.5

F = -2/0.5

= -4N

Note the sign : It is negative because it is the force by the wall on the ball.

According to newtons 3rd law, force applied by ball on wall = force applied by wall on ball. (but in opposite direction )

Therefore, the required answer = 4N

Answer 2

Answer:

Average Force= 4N

Explanation:

Hope it will help you.