Finding Horizontal Momentum of Cannonball

[elmosworld403]

A 2.00 kg cannonball is fired out of a cannon at an angle of 30.0 to the horizontal. When the cannonball reaches the top of its path, its momentum has a magnitude of 800 kg-m/s. What is the horizontal component of the cannonball's momentum when it left the cannon?




Change in Momentum
p=mΔv
p=fnetΔT
Impulse=Δp


3. Just learning momentum, but I don't know how to use the 30 degree angle with the momentum values to find the horizontal momentum? It does have to do with projectile motion, but I don't know how to use the data.

 

[TSny]

Hi elmosworld403. Welcome to PF!

Do you remember how the horizontal component of velocity of a projectile changes during flight?

 

[elmosworld403]

Horizontal velocity doesn't change at all. Vertical velocity does from gravity.

 

[Saitama]

Horizontal velocity doesn't change at all. Vertical velocity does from gravity.

Right. What are the values of both the components at the highest point?

 

[elmosworld403]

I checked the answer it was 800 kg- m/s

But the picture is confusing me isn't that the momentum of the cannonball at a 30 degree angle taking both the vertical and horizontal into account?

 

[Saitama]

But the picture is confusing me isn't that the momentum of the cannonball at a 30 degree angle taking both the vertical and horizontal into account?

It is but you are asked only the horizontal component of the momentum. As I previously asked, what are the values of both the components of velocity at the highest point?

 

[elmosworld403]

Right. What are the values of both the components at the highest point?

Idk

Horizontal
Distance=?
Velocity=?
Time=?

Vertical
Acceleration= 9.81 m/s squared.
Vinitial=?
VFinal=?
Distance=?
Time=?

The question only gives me the momentum and weight and the angle.
Momentum= 800 kg m/s
weight= 2 kg
angle= 30 degrees

 

[Saitama]

Idk

Horizontal
Distance=?
Velocity=?
Time=?

Vertical
Acceleration= 9.81 m/s squared.
Vinitial=?
VFinal=?
Distance=?
Time=?

The question only gives me the momentum and weight and the angle.
Momentum= 800 kg m/s
weight= 2 kg
angle= 30 degrees

The cannonball goes up due to the vertical component of velocity. When the vertical component of velocity reduces to zero, it starts getting back to the ground. So effectively, only the horizontal component of velocity contributes to the momentum of the ball at the highest point.

 

[elmosworld403]

The cannonball goes up due to the vertical component of velocity. When the vertical component of velocity reduces to zero, it starts getting back to the ground. So effectively, only the horizontal component of velocity contributes to the momentum of the ball at the highest point.

Holy Thanks Dude, my mind is blown right now. Didn't think that the highest point would not take into account the vertical since its velocity is zero. So the at the highest point only the momentum of the horizontal is left.

Until gravity takes over? then the vertical has a velocity again which goes towards the momentum?