What Is the Change in Momentum of a 3.0 kg Object Projected at a 45° Angle?

[Jtappan]

Homework Statement

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

Homework Equations

F\DeltaT = \DeltaP = m\vec{v}_{f}-m\vec{v}_{i}

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?

 

[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?

 

[Jtappan]

No this doesn't make any sense to me at all. I know this should be a very simple problem but i don't know wat to do

 

[Antineutron]

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

 

[Jtappan]

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

 

[Doc Al]

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.)

 

[Jtappan]

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

 

[Jtappan]

can anybody help on this one?

 

[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?

 

[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.

 

[Jtappan]

if i took one as postive and one as negative wouldn't they just cancel each other out if they were the same? or is it a negative minus a negative?

 

[rl.bhat]

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

 

[Jtappan]

negative minus a negative = a bigger negative

 

[hotcommodity]

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

 

[Jtappan]

I have been trying to do that. I just don't know how to do it that's why I am 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.

 

[hotcommodity]

You have the mass, all you need are the initial and final velocities. Let's 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?

 

[rl.bhat]

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

 

[Jtappan]

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

 

[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?

 

[Jtappan]

ok what's next?

 

[hotcommodity]

You should get 18.13 m/s for the final velocity's x component, and -18.13 for the final velocity's y component. Now we can add them vectorially to get the final velocity:

v_f = \sqrt{(18.13)^{2} + (-18.13)^{2}} = 25.63

Now you can plug that into your equation to find that the overall change in momentum is zero. Does that make sense?

 

[Jtappan]

yea that makes complete sense but Zero is not the correct answer. So i don't know what is wrong..

 

[hotcommodity]

What's the answer given?

 

[Jtappan]

it doesn't give me an answer. that's the problem it just tells me whether or not i put in the right answer. It is internet homework.

 

[hotcommodity]

Hmmm, I suppose if we think of the final velocity as negative we can get a negative change in momentum. Maybe something like -2 X 10^2 kg*m/s, but anything other than zero just seems counter intuitive...

 

[Jtappan]

i know... i don't understand it. It is just like a problem where a ball bounces off the wall and you are suppose to find the momentum change. Any other thoughts? Thanks for spending this much time helping me

 

[hotcommodity]

I'm happy to spend the time, but I'm sorry we didn't arrive at the correct answer. I'll keep thinking it over, and I'll post if I come up with something.

 

[Jtappan]

awesome thanks a bunch!

 

[learningphysics]

careful about signs...

final momentum in y-direction = -18.13*mass

initial momentum in y-direction = 18.13*mass

what is final momentum - initial momentum ?

the answer wants the momentum downwards... so leave off the minus sign in the final answer.

 

[rl.bhat]

In the projectile motion y-component of the intial velocity changes, where as x- component remains constant.Therefore change in the momentum of the body = change in the y-component of velocity x mass. If you take final velocity (which is downwards) as positive and intial velocity ( which is upwards) as negative, then change in momentum = [18.13 -(-18.13)]*3Ns