Physics Force Question

Question

A baseball player throws a ball at a nearby boulder. The ball hits the boulder with a speed of 39 m/s and then bounces straight up to a height of 15 m above the boulder.

Assuming that the rotational motion of the ball is negligible, what fraction 𝑓 of the ball's initial mechanical energy dissipates in the collision with the boulder?

Answer 1

Hey Zandria,

Mechanical energy is the total of potential and kinetic energy.

From Conservation of energy Law we know that energy only transforms or transfers.

Applying this law to the problem we know that the initial energy is the same as the final energy + dissipated energy.

We can write this in an equation as follows:

ME(1) = ME(2) + ME(dissipated) (ME stands for mechanical energy and its equal to Potential + Kinetic)

KE(1) + PE(1) = KE(2) + PE (2) + ME(dissipated)

rearrange like this : KE(dissipated) = KE(1) + PE(1) - KE(2) - PE (2)

Since the Question is asking for the ratio of the initial ME, divide both sides of the equation by ME(1)

ME(dissipated) / ME(1) = [ KE(1) + PE(1) - KE(2) - PE (2) ] / ME(1)

Now notice how PE(1) and KE(2) are zero and rewrite

ME(dissipated) / ME(1) = [ KE(1) - PE (2) ] / KE(1)

Use formula for KE=1/2 * m *v^2 and PE= m*g*h where m is mass, v is velocity, g gravitational acceleration and h is height.

ME(dissipated) / ME(1) = f ( the fraction)

f= KE(1) / KE(1) - PE(2) / KE(1) = 1- [ m*g*h / (1/2 * m *v ^2)

Notice that you can simplify m on the right hand side of the equation.

Rewrite Equation as:

f= 1- [ (2*g*h) / (v^2) ]

plug in the values for h =15 and v = 39.

I got f= 0.807 using g= 9.8 m/s^2

Let me know if you have any questions

Answer 2

The energy before the collision is E_i = 1/2 m v². The kinetic energy after the collision gets converted to potential energy and is E_f = mgh. The energy lost relationship would be E_i = E_f + E_L. The fraction of energy dissipated would be

E_L/E_i = (E_i - E_f)/E_i = (1/2 m v² - mgh)/(1/2 m v²) = (v² - 2gh)/v² ≈ 0.81 .