# Homework Help: Momentum and angle

1. Mar 20, 2015

### goonking

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

2. Relevant equations

p = mv

3. The attempt at a solution
if an object's velocity at the highest point of a trajectory is 0, how can it still have momentum (since p = mv)? is this question technically correct?

2. Mar 20, 2015

### Suraj M

I think you mean to say, vertical component of velocity, there is no force acting in the horizontal direction, hence it would remain a constant. try doing the rest.
Yes

3. Mar 20, 2015

### goonking

so the object loses horizontal velocity over time?

4. Mar 20, 2015

### Suraj M

There is no force in the horizontal direction, do you think it looses it's horizontal velocity?

5. Mar 20, 2015

### goonking

it went from 6.4 to 4.6 , and mass stays the same so I thought velocity decreased

6. Mar 20, 2015

### Suraj M

yes it did, but only the vertical component of the velocity changed, not the horizontal, this means that the momentum at the top considers only the horizontal velocity.

7. Mar 20, 2015

### goonking

but we don't know the mass to find horizontal velocity

8. Mar 20, 2015

### Suraj M

it's okay don't worry. start off with the problem. Find the horizontal component of the initial momentum. in terms of $\theta$

9. Mar 20, 2015

### goonking

cos theta x initial velocity or in this case Vinitial = p/m = 6.4 / m

10. Mar 20, 2015

### Suraj M

the momentum is 6.4. Find the initial horizontal component of it.
This is not the horizontal component, it's just the initial velocity.

11. Mar 20, 2015

### goonking

nvm, answer is 44 degrees, i figured it out. thank you!

Last edited: Mar 20, 2015
12. Mar 20, 2015

### Suraj M

Good, now what happens to this value as the particle reaches its highest position?

13. Mar 20, 2015

### goonking

it stays the same

14. Mar 20, 2015

### Suraj M

Isn't the final momentum given? so find the final velocity and equate to the horizontal velocity you got.

15. Mar 20, 2015

### goonking

how do you find the final velocity if you don't know mass?

16. Mar 20, 2015

### Suraj M

Again don't worry keep it as a variable, and find the final velocity and equate, m WILL get cancelled.

17. Mar 20, 2015

### goonking

6.4 / m times cos theta = 1/2 m Vf^2?

18. Mar 20, 2015

### Suraj M

NO.
i told you to equate the final VELOCITY and $\frac{6.4}{m}\cos\theta$

19. Mar 20, 2015

### goonking

20. Mar 20, 2015

### Suraj M

Yes
I have noticed you are a frequent visitor(&poster) to this forum, you write many mathematical equations too, why don't you take a look at the first part of this

21. Mar 20, 2015

### goonking

Ok, I'm back.
Ill be honest, I'm still 100% sure on this problem. I 'kind of' understand it but what would did your "equate the final VELOCITY and 6.4/m cosθ" look like?

and about the latex, I will learn that soon. thank you for the link

22. Mar 20, 2015

### Suraj M

Summary:
Initial horizontal momentum = final horizontal momentum.....eq1
notice i'm using momentum,
initial momentum= 6.4 units
initial horizontal momentum = $6.4 \cos\theta$........eq2
final horizontal momentum(highest position) = final momentum =4.6units......eq3
because vertical momentum = 0
hence from eq 1,2, and 3
we have $$6.4\cos\theta = 4.6$$
hence get $cos\theta$
ok?