# Homework Help: 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?