Question

A rubber ball of mass 50g falls from a height of 10cm and rebounds to a height of 50cm.Determine the change in linear momentum and average force between the ball and the ground taking time of contact as 0.1s.Hint : mgh and 1/2mv2(square) is usedAns :: 0.226 N/s , 2.26 NPLS. GIVE METHOD

Answer 1

Given: Mass of Ball = 50gm
Height To Fall = 10cm = 0.1 m
Rebounce = 50 cm = 0.5m
Time of Contact = 0.1 sec
Velocity  Just before it hits the Ground ⇒ v² - u² = 2as
⇒v² - 0 = 2 x 9.8 x 0.1
⇒v= √(1.96)
⇒v = 1.4 m/sec
after rebounce final velocity = 0
v² - u² = 2as
0 - u² = 2 x 9.8 x 0.5
⇒u² = 9.8
⇒u = √9.8 = 3.13 m/s
Change In Momentum
⇒ Pf- Pi
⇒m(v-u)
⇒m(v-(-u) ∵Both the velocities are in opposite direction
=0.05(1.4 + 3.13) = 0.2265 kgm/s
 
Average Focce between THe Groun and The Ball
⇒ dP/dT = 0.2265/0.1 = 2.265N