Question

A 0.70-kg basketball dropped on a hardwood floor rises back up to 65% of its original height. (a) If the basketball is dropped from a height of 1.5 m, how much energy is dissipated in the first bounce? (b) How much energy is dissipated in the fourth bounce? (c) To which type of incoherent energy is the dissipated energy converted?

Answer 1

Part a)

`Loss = 3.6 J`

Part b)

`Loss = 0.99 J`

Part C)

This is loss in terms of thermal energy due to collision with the floor

Step-by-step explanation:

Part a)

Since we know that the ball rises up by 65% of initial height

so after first bounce it will lose 35% of its initial energy

so we will have

`U = mgH`

Energy Loss = 0.35 mgH[/tex]

`Loss = 0.35(0.70)(9.81)(1.5)`

`Loss = 3.6 J`

Part b)

Energy of the ball after first bounce

`U_1 = 0.65 mgH`

energy of ball after 2nd Bounce

`U_2 = 0.65(0.65 mgH)`

energy of the ball after 3rd bounce

`U_3 = (0.65)(0.65^2)mgH`

`U_3 = 0.65^3(0.70)(9.81)(1.5)`

`U_3 = 2.83 J`

Now we will have energy loss in fourth bounce given as

`Loss = 0.35 U_3`

`Loss = 0.35(2.83)`

`Loss = 0.99 J`

Part C)

This is loss in terms of thermal energy due to collision with the floor