Question

A METAL PLATE WEIGHING 200g IS BALANCED IN MID AIR BY THROWING 40 BALLS PER SECOND VERTICALLY UPWARDS FROM BELOW. AFTER COLLISION BALLS REBOUND WITH THE SAME SPEED WHAT IS THE SPEED WITH WHICH THE BALLS STRIKE THE PLATE ? (GIVEN : MASS OF EACH BALL IS 200g)

Answer 1

In this case the plate will be suspended only if weight of the plate = force exerted by 40 balls per second. So the weight of plate, W = M Force exerted by 40 balls per second, F = 40 x ma = 40 x mv/t = 40 x mv [as t=1s] So Mg  = mv 

 

Mass of plate M = 200g  = 0.2kg 

Mass of ball = 200g = 0.2kg 

So 0.2 x 10 = 40 x 0.2 x v 

Or v = 10/40 

So the velocity of each ball will be  

v = 0.25 m/s