Question

A 1.0 kg ball falling vertically hits a floor with a velocity of 3.0 m/s and bounces vertically up with a velocity of 2.0 m/s . If the ball is in contact with the floor for 0.10 s, the average force on the floor by the ball is:

Answer 1

Answer:

50 N

Explanation:

given,

mass of ball = 1 Kg

initial velocity = 3 m/s

final velocity = 2 m/s

time = 0.10 s

impulse = change in momentum

I = m ( v - u)            

I = 1( 2 -(-3))                

I = 5 kg.m/s          

Force is equal to impulse per unit time

average force = $\dfrac{I}{t}$

                       = $\dfrac{5}{0.1}$

                       = 50 N

Average force on the floor will be equal to 50 N