# Find the momentum of the ball

1. Jun 27, 2010

### mandy9008

1. The problem statement, all variables and given/known data
A 0.25 kg ball is thrown straight up into the air with an initial speed of 12 m/s. Find the momentum of the ball at the following locations.
(a) at its maximum height
(b) halfway to its maximum height

2. Relevant equations
p=mv

3. The attempt at a solution
a. p = (0.25 kg)(12 m/s)
p = 3 kg m/s

b. p = 1.5 kg m/s

2. Jun 27, 2010

### Staff: Mentor

Re: momentum

What's the speed at the highest point? (Not the starting point.) At the halfway point?

3. Jun 27, 2010

### mandy9008

Re: momentum

how am I supposed to know the highest point?

4. Jun 27, 2010

### Staff: Mentor

Re: momentum

This is just an example of projectile motion. If you wanted, you could figure out the height it reaches before falling back down. But you won't need that. Hint for (a): No calculation is required.

5. Jun 27, 2010

### mandy9008

Re: momentum

okay i got you, the momentum when it reaches the maximum height is zero. but this is not the case for it when it is halfway there, correct?

6. Jun 27, 2010

### Staff: Mentor

Re: momentum

Right.
Correct. Here you'll have to figure out the new speed. Hint: What happens to the kinetic energy?

7. Jun 27, 2010

### mandy9008

Re: momentum

Is it the potential energy or the kinetic energy that stays the same throughout?

8. Jun 27, 2010

### Staff: Mentor

Re: momentum

Neither. (But their sum remains constant.)

9. Jun 27, 2010

### mandy9008

Re: momentum

so KE=mgh
v^2 = 2gh
12 m/s ^2 = 2 (9.8 m/s^s)h
h=7.35m

KE= (0.25kg) (9.8 m/s^2)(7.35m)
KE= 18.0 J

10. Jun 27, 2010

### Staff: Mentor

Re: momentum

You found the height reached and the original KE.

What's the KE at the half-way point?

11. Jun 27, 2010

### mandy9008

Re: momentum

KE= (0.25kg) (9.8 m/s^2)(3.675m)
KE= 9.00375 J

12. Jun 27, 2010

### mandy9008

Re: momentum

KE= p^2 / 2m
9.00375J = p^2 / 2(0.25kg)
p=2.1218 kg m/s

is this correct?

13. Jun 27, 2010

### Staff: Mentor

Re: momentum

Good. Now use that to solve for the speed at that point. Then the momentum.

14. Jun 27, 2010

### mandy9008

Re: momentum

KE= p^2 / 2m
9.00375J = p^2 / 2(0.25kg)
p=2.1218 kg m/s

is this correct?

15. Jun 27, 2010

### Staff: Mentor

Re: momentum

Sure, that works.

16. Jun 27, 2010

### mandy9008

Re: momentum

okay, thank you! :)