What is the momentum of the ball at different locations?

[mandy9008]

Homework Statement

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

Homework Equations

p=mv

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

 

[Doc Al]

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

 

[mandy9008]

how am I supposed to know the highest point?

 

[Doc Al]

how am I supposed to know the highest point?

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.

 

[mandy9008]

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?

 

[Doc Al]

okay i got you, the momentum when it reaches the maximum height is zero.

Right.

but this is not the case for it when it is halfway there, correct?

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

 

[mandy9008]

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

 

[Doc Al]

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

Neither. (But their sum remains constant.)

 

[mandy9008]

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

 

[Doc Al]

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

You found the height reached and the original KE.

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

 

[mandy9008]

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

 

[mandy9008]

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

is this correct?

 

[Doc Al]

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

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

 

[mandy9008]

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

is this correct?

 

[Doc Al]

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

is this correct?

Sure, that works.

 

[mandy9008]

okay, thank you! :)