Change in Impulse and Momentum

1. Nov 1, 2007

Jtappan

1. The problem statement, all variables and given/known data

An object of mass 3.0 kg is projected into the air at a 45° angle. It hits the ground 3.7 s later. What is its change in momentum while it is in the air? Ignore air resistance.
_____ kg·m/s downward

2. Relevant equations

F$$\Delta$$T = $$\Delta$$P = m$$\vec{v}$$$$_{f}$$-m$$\vec{v}$$$$_{i}$$

3. The attempt at a solution

This is a really basic question i am sure but i am getting thrown off by the 45 degree angle what do I do?

2. Nov 1, 2007

hotcommodity

The object moves in projectile motion, so you need to use the equations for constant acceleration, and the two projectile equations relating each component of the initial velocity to the initial velocity vector. This will allow you to find the initial and final velocities of the mass. Does this make sense?

3. Nov 4, 2007

Jtappan

No this doesnt make any sense to me at all. I know this should be a very simple problem but i dont know wat to do

4. Nov 4, 2007

Antineutron

Is this question worded exactly as you say? I don't know if we have enough info to solve the problem.

5. Nov 4, 2007

Jtappan

it is worded exactly how it was posted. i copied and pasted it

6. Nov 4, 2007

Staff: Mentor

You have all the information needed to figure out the initial and final velocity of the mass. (It's just a change in momentum question, not really an impulse question.)

7. Nov 4, 2007

Jtappan

I still do not understand how to figure it out. I knwo it is a change in momentum question but i dont know how to do it.

8. Nov 4, 2007

Jtappan

can anybody help on this one?

9. Nov 4, 2007

hotcommodity

Start by considering the object's movement in the vertical direction. It moves with constant acceleration, so we can use a kinematic equation. We'd have:

$$y = y_0 + v_{0y}t - .5gt^2$$

And we know that $$v_{0y} = v_0sin\theta$$

Now you can solve for the initial velocity, right?

10. Nov 4, 2007

rl.bhat

At the point of projection the vertical component of the velocity is vsin(theta). By using angle of projection and time of flight you can find this velocity. Just before the object hits the ground this velocity is the same but the direction is changed. So take one as positive and other as negative and find the change in velocity and hence momentum.

11. Nov 4, 2007

Jtappan

if i took one as postive and one as negative wouldnt they just cancel each other out if they were the same? or is it a negative minus a negative?

12. Nov 4, 2007

rl.bhat

Yes. It is a negative minus a negative = positive.

13. Nov 4, 2007

Jtappan

negative minus a negative = a bigger negative

14. Nov 4, 2007

hotcommodity

You can work it out on paper and see if it's true on not.

15. Nov 4, 2007

Jtappan

I have been trying to do that. I just dont know how to do it thats why im having problems with this problem. It is an extremely easy problem, for some reason I cannot figure out how to incorporate the angle into the F(deltaT) = mvf-mvi equation.

16. Nov 4, 2007

hotcommodity

You have the mass, all you need are the initial and final velocities. Lets start by using the equation:

$$y = y_0 + v_{0y}t - .5gt^2$$

where $$v_{0y} = v_0sin\theta$$.

You assume that y is zero, because the object moves up some distance, and comes back down covering that same distance, right?

17. Nov 4, 2007

rl.bhat

F(deltaT) = mvf-(-mvi ) = ?

18. Nov 4, 2007

Jtappan

ok then the Vi = 25.6393 roughly .. then what?

19. Nov 4, 2007

hotcommodity

Yes, correct. Convert that into the initial velocities in the x and y directions using :

$$v_{0x} = v_0cos\theta, v_{0y} = v_0sin\theta$$

and find the final velocity components using $$v_f = v_0 + at$$

So far so good?

20. Nov 4, 2007

Jtappan

ok whats next?