Question

A golf ball of mass m = 0.19 kg is dropped from a height h. It interacts with the floor for t = 0.13 s, and applies a force of F = 11.5 N to the floor when it elastically collides with it.

Answer 1

a) The equation for ball´s velocity is V₁ = (F*t)/m

b) The ball velocity after collision is 7.87 m/s

c) The heigh from which the ball is dropped is 3.16 m

Step-by-step explanation:

Given data:

m = mass of golf ball = 0.19 kg

t = interaction time = 0.13 s

F = force = 11.5 N

a) Due the collision is eslastic, the change of momentum is:

ΔV*m = F*t

V₁ - V₀ = (F*t)/m

V₁ = ((F*t)/m) + V₀

Where V₀ = 0

V₁ = (F*t)/m

b) Replacing:

`V_(1) =(11.5*0.13)/(0.19) =7.87m/s`

c) According conservation of kinetic energy:

(Ek - Ep)initial = (Ek - Ep)final

0 + mgh = (1/2)mV₁² + 0

Clearing h:

`h=(1)/(2)(V_(1)^(2) )/(g) =(7.87^(2) )/(2*9.8) =3.16m`