Solve Impulse Force Problem: 75g Ball Dropped from 2.2m

[master_333]

Homework Statement

A 75-g ball is dropped from rest from a height of 2.2 m. It bounces off the floor and rebounds to a maximum height of 1.7 m. If the ball is in contact with the floor for 0.024 s, what is the magnitude and direction of the average force exerted on the ball by the floor during the collision?

Homework Equations

p = Ft

The Attempt at a Solution

I found the speed of the ball right before the drop using vfinal ^2 = vinitial^2 +2ad
The answer I got was 6.56 m/s. Now I don't know what to do after that.

 

[gneill]

Show the details of your calculation. It's not clear which velocity you were calculating --- If the ball was dropped, it should have zero velocity "right before the drop".

What is your plan for calculating the average force? How is average force defined in terms of momentum?

 

[master_333]

Okay tell me if what I did is correct

Velocity right before the ball hit the ground
vfinal^2 = vinitial^2 + 2ad
vfinal^2 = (0)+2(9.8)(2.2m)
vfinal = 6.57m/s

If this was the velocity that the ball hit the ground with then the force is
F = p/t
F = (0.075kg)(6.57m/s)/0.024s
F = 20.53N

If 20.53N was the force, then the ball would be sent up to 2.2m

Now find the force required to send it to 1.7m

(20.53)(1.7)/2.2
F = 15.86 N

 

[gneill]

Okay tell me if what I did is correct

Velocity right before the ball hit the ground
vfinal^2 = vinitial^2 + 2ad
vfinal^2 = (0)+2(9.8)(2.2m)
vfinal = 6.57m/s

That's fine so far.

If this was the velocity that the ball hit the ground with then the force is
F = p/t
F = (0.075kg)(6.57m/s)/0.024s
F = 20.53N

If 20.53N was the force, then the ball would be sent up to 2.2m

Now find the force required to send it to 1.7m

(20.53)(1.7)/2.2
F = 15.86 N

Nope. You need to go to the definition of average force based on change in momentum.

What's the formula for the average force based on change in momentum?

Hint: You need to find the momentum of the ball when it first contacts the floor and when it just loses contact with the floor. Remember that momentum is a vector quantity.

 

[master_333]

Okay, the momentum of the ball when it first contacts the floor is (0.075kg)(6.57m/s) = 0.49275 kg.m/s

I don't know how to find the momentum of the ball when it just loses contact with the floor

Please, can you show me how to do it. Rather than giving me hints that I don't understand

 

[gneill]

What speed must it have when it just leaves the floor? It's a projectile at that point, and reaches a certain maximum height...

 

[master_333]

Okay thanks for trying to help me. But the way your trying to help me, I clearly do not understand anything. Whatever, I guess I will just skip this question.

 

[gneill]

Please, can you show me how to do it. Rather than giving me hints that I don't understand

Okay thanks for trying to help me. But the way your trying to help me, I clearly do not understand anything. Whatever, I guess I will just skip this question.

Sorry master_333, but the forum rules are clear that helpers cannot simply provide answers or do your homework for you. This includes telling you step by step how to solve a problem.

We can only offer hints or point out errors or suggest things to investigate so that you can gain the knowledge to solve the problem yourself.

I have suggested that you look up the definition of average force in terms of change of momentum. Have you looked it up in your notes, text, or on the web?

 

[rude man]

Homework Statement

A 75-g ball is dropped from rest from a height of 2.2 m. It bounces off the floor and rebounds to a maximum height of 1.7 m. If the ball is in contact with the floor for 0.024 s, what is the magnitude and direction of the average force exerted on the ball by the floor during the collision?

Homework Equations

p = Ft

The Attempt at a Solution

I found the speed of the ball right before the drop using vfinal ^2 = vinitial^2 +2ad
The answer I got was 6.56 m/s. Now I don't know what to do after that.

OK so far.
Now relate the change in kinetic energy between 2.2m and 1.7m to the work done by the force over the distance of floor contact.
Since you don't know the contact distance s you might consider the chain rule: ds = ds/dt * dt.