Physics Mechanics Problem. PLEASE help!!

Question

A ball of mass m is dropped from a height of h₁ and rebouns to a height of h_2. The ball is in contact with the floor for a time interval of t. 

a) Show that the average net force on the ball is given by: (fraction)

F=m√2gh₁√2gh₂⁄2

b) If h1=8 m, h2=6 m, t=0.125s and m=.250 kg, calculate the average force exerted by the ball on the floor.

 

 

Thanks so much for any help!

Answer 1

I am not sure about your answer.    This problem appears to be a standard   Impulse = change of momentum type.

 

A ball dropped from height h₁ will reach the floor with speed  s₁ = sqrt(2gh₁) .  This result can be obtained a number of ways.   For example the kinematic formula  vf² - vi² =  2 g d      

 

By a similar approach, the velocity of the ball leaving the floor on its way to h₂  must leave with s₂ = sqrt(2gh₂)

 

Since the velocities are oppositely directed, the change in momentum resulting from the collision with the floor is

 

     Δp =  m [ sqrt(2gh₁) + sqrt(2gh₂)   ]

 

     The impulse is   J = F t        since impulse =   Δp    we get

 

  m [ sqrt(2gh1) + sqrt(2gh2)]   =   F t   so

 

  F =  m [ sqrt(2gh1) + sqrt(2gh2) ] /t

 

Plugging in the numbers gives   F = 46.7  N

I used g = 9.8 m/s²