Can momentum exist when velocity is zero at a trajectory's highest point?

[goonking]

Homework Statement

Homework Equations

p = mv

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?

 

[Suraj M]

if an object's velocity at the highest point of a trajectory is 0,

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.

? is this question technically correct?

Yes

 

[goonking]

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

so the object loses horizontal velocity over time?

 

[Suraj M]

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

 

[goonking]

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

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

 

[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.

 

[goonking]

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.

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

 

[Suraj M]

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

 

[goonking]

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

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

 

[Suraj M]

the momentum is 6.4. Find the initial horizontal component of it.

Vinitial = p/m = 6.4 / m

This is not the horizontal component, it's just the initial velocity.

 

[goonking]

cos theta x initial velocity

the momentum is 6.4. Find the initial horizontal component of it.

This is not the horizontal component, it's just the initial velocity.

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

 

[Suraj M]

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

 

[goonking]

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

it stays the same

 

[Suraj M]

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

 

[goonking]

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

 

[Suraj M]

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

 

[goonking]

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

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

 

[Suraj M]

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

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

 

[goonking]

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

answer is 44 degrees?

 

[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
It would help you in the future.

 

[goonking]

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
It would help you in the future.

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

 

[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?