Question

a ball of mass 10gm dropped from a height of 5m hits the floor and rebounds to a height of 1.25m. if the ball is in contact with ground for 0.1 s the force exerted by the group on the ball is ( g= 20)​

Answer 1

Answer:

The answer will be 2.11 N

Explanation:

According to the problem the mass of the ball,m = 10 gm = 0.01 kg

The height from which the ball is dropped, h1 = 5 m

and the height the ball gained after touching the ground , h2 = 1.25 m

The collision time, Δ t = 0.1 s

Now we need to find the initial velocity of the ball, v1^2= u^2+2gh1

         As u = 0

Therefore,

v1 = √2gh1 = √2 x 20 x 5 = 10√2 m/s

Now the initial momentum p1 =  mv1 = 0.01 x (-10√2) = - 0.141 N.s

Now we need to find the final velocity of the ball, v2^2 = u^2+2gh2

             Therefore v2 =√2gh2 = √ 2 x 20 x 1.25 = 5√2 m/s

Therefore, the final momentum p2 = mv2 = 0.01 x 5√2 = 0.070 N.s

Therefore the force ,f = Δ p/Δ t = (0.070-(-0.141))/0.1 = 2.11 N