Question

A 290 g ball is dropped from a height of 2.2 m , bounces on a hard floor, and rebounds to a height of 1.7 m .

(Figure 1) shows the impulse received from the floor.

What maximum force does the floor exert on the ball?

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Mass of the ball = m = 290 g = 0.29 kg

Height from which the ball is dropped = h₁ = 2.2 m

Height to which the ball rises after rebounding off the floor = h₂ = 1.7 m

Velocity of the ball before hitting the floor = V₁

Velocity of the ball after rebounding off the floor = V₂

The potential energy of the ball at the beginning is converted into kinetic energy before the ball hits the floor.

mgh₁ = mV₁²/2

/2gh1

V(2) (9.81) (2.2) Vi =

V₁ = 6.57 m/s

The kinetic energy of the ball after rebounding off the floor is converted to potential energy as it rises.

mV₂²/2 = mgh₂

22gh2 V2=

2(2) (9.81) (1.7) V2=

V₂ = 5.77 m/s

Let us take the downward direction to be negative and the upward direction to be positive.

The velocity of the ball before hitting the floor is directed downwards therefore,

V₁ = -6.57 m/s

The velocity of the ball after rebounding off the floor is directed upwards therefore,

V₂ = 5.77 m/s

Time period the ball is in contact with the floor = t = 5 ms = 5 x 10^-3 s

Maximum force exerted on the ball = F_max

The change in momentum of the ball is equal to the area under the force vs time graph

mV₂ - mV₁ = F_maxt/2

m(V₂ - V₁) = F_maxt/2

(0.29)(5.77 - (-6.57)) = F_max(5x10^-3)/2

F_max = 1431.44 N

Maximum force the floor exerts on the ball = 1431.44 N