Calculating Average Force in a Ball Rebound Momentum Problem

[lmc489]

A .3 kg ball is dropped from a height of 2 m. It strikes the group and rebounds to a height of 1.5 m. The collision with the ground lasts for .01 seconds. The average force exerted by the ground on the ball is:

a)280
b) 540
c) 430
d) 350

So, i know that we have to use some type of formula involving mgh which equals velocity?

I did:

.3 [ rad 2mgh +rad 2 mgh] / .01 = 350

is this correct?

 

[einstein_a_go_go]

yes - from whence came your equation?

 

[baba89]

I would say that your calculation is on the right track, but it is missing a crucial component. In order to calculate the average force, we need to use the formula F=ma, where F is force, m is mass, and a is acceleration. In this case, the acceleration is due to the change in velocity during the collision with the ground.

To calculate the change in velocity, we can use the formula vf^2 = vi^2 + 2ad, where vf is the final velocity, vi is the initial velocity (which in this case is 0 since the ball is dropped from rest), a is the acceleration, and d is the distance traveled (in this case, the distance the ball drops and rebounds, which is 0.5m).

So, using this formula, we can calculate the final velocity of the ball as it rebounds:

vf^2 = 0^2 + 2(-9.8)(0.5)
vf = 1.4 m/s

Now, we can use the formula F=ma to calculate the average force:

F = 0.3(1.4)/0.01 = 42 N

Therefore, the average force exerted by the ground on the ball is 42 N.

Looking at the answer choices provided, none of them match with our calculated value. It is possible that there was a mistake in the given information or the answer choices. As scientists, it is important to always double check our calculations and question any discrepancies in the data.