Question

Isaac drops a rubber ball drom height of 2.0m and it bounces to a height of 1.5m. a) What fraction of it's initial energy is lost during the bounce? b)What is the rubber ball's speed just before and just after it bounces? c)Where did the energy go?

Answer 1

a) ΔE = 25 %

b) v = 8,85 m/s

c) The energy was used against air resistance

Step-by-step explanation:

In any situation total energy of a body is equal to potential energy plus

kinetic energy, then, just at the moment when Isaac dop the ball the situation is:

Ei = Ep + Ek where Ep = m*g*h and Ek = 1/2*m*v²

As v = 0 (Isaac drops the ball)

Ei = Ep = m*g*h = 2*m*g

At the end (when the ball bounced to 1,5 m

E₂ = Ep₂ + Ek₂ again at that point v =0 and

E₂ = 1,5*m*g*

Ei = E₂ + E(lost)

E(lost) = Ei - E₂

E(lost) = 2*m*g* - 1,5*m*g and the fraction of energy lost is

E(lost)/Ei

ΔE = (2*m*g* - 1,5*m*g )/ 2*m*g

ΔE = 0,5*m*g / 2*m*g

ΔE = 0,5/2

ΔE = 0,25 or ΔE = 25 %

b) The speed of the ball is

Potential energy is converted in kinetic energy just when the ball is touching the ground, then

m*g*h = 1/2*m*v²

2*h*g = 1/2 *v²

v² = 4*g*h

v² = 4*2*9,8

v² = 78,4

v = 8,85 m/s

If the impact is an elastic collision, then Ek before and after the impact is the same.