Collision Physics 1 problem with distances and angles

[jamesbiomed]

Homework Statement

A 3000-kg Cessna airplane flying north at v1 = 70 m/s at an altitude of 1750 m over the jungles of Brazil collided with a 7000-kg cargo plane flying at an angle of θ = 34° north of west with speed v2 = 111 m/s. As measured from a point on the ground directly below the collision, the Cessna wreckage was found 1000 m away at an angle of 25° south of west, as shown in the figure. The cargo plane broke into two pieces. Rescuers located a 4000-kg piece 1800 m away from the same point at an angle of 22° east of north. Where should they look for the other piece of the cargo plane? Give a distance and a direction from the point directly below the collision. (Let to the east be the +x-direction and to the north be the +y-direction.)

http://www.webassign.net/bauerphys1/7-p-064a-alt.gif
http://www.webassign.net/bauerphys1/7-p-064b.gif

Homework Equations

P0=P.
Px=7000kg(111 m/s) cos34
Py=3000kg(70m/s)+7000kg(111 m/s)sin34

P=mv

Perhaps kinematics to find distance.

The Attempt at a Solution

px=7000(111)cos34=644,162.2 kg m/s
py=7000(111)sin34+(70*3000)=644,492.9 kg m/s
P=sqrt (px^2+py^2)=911216.8 kg m/s

Now here's where I'm grasping for straws:

Py/Px=1.00. arctan (1.00)==45

Final angles: 22+25+90=137.

Tan(137)=-.933.
1.00x=.933=>x=.933

.933*911216.8=849718.8.
849718.8/ (mass of third piece=3000kg)=283.24 m/s

Tried kinematics to get a final distance traveled, but to no avail.
For the angle: 1.0-.93=.07. arctan.07=4 degrees. which was incorrect.

Thanks!

 

[TSny]

Hi jamesbiomed. Welcome.

Px=7000kg(111 m/s) cos34
Py=3000kg(70m/s)+7000kg(111 m/s)sin34

[EDIT: The problem wants you to take west as the negative x direcion.]

Now here's where I'm grasping for straws:

Py/Px=1.00. arctan (1.00)==45

Final angles: 22+25+90=137.

Tan(137)=-.933.
1.00x=.933=>x=.933

.933*911216.8=849718.8.
849718.8/ (mass of third piece=3000kg)=283.24 m/s

Don't think you grabbed even one straw there .

Tried kinematics to get a final distance traveled, but to no avail.

You might explain how you set up your kinematics.

So far you've avoided the most important principle that should be invoked in a collision problem.

 

[jamesbiomed]

Hi jamesbiomed. Welcome.

Thx

[EDIT: The problem wants you to take west as the negative x direcion.]

True, forgot about that.

Don't think you grabbed even one straw there .

k.

You might explain how you set up your kinematics.

So far you've avoided the most important principle that should be invoked in a collision problem.

Conservation of momentum (p0=p). I totalled P0. But with distances and no times, there are how can you total final momentum?

Kinematics I tried were vy^2=(0)-g(0-alt). This could give me the final velocity in the y-direction--That in turn could give me the time through vy=vy0-gt.

If I knew the velocity in the x-direction, than the distance would be doable. If I initial velocity (v (terminal) in this case) then I could separate vy, get vx (which would be constant) and then get the distance that way.

So the gap in my efforts is in going from the theory that momentum is conserved into appropriating the velocities, and finding these angles. Do you understand what I'm saying?

 

[TSny]

Conservation of momentum (p0=p). I totalled P0. But with distances and no times, there are how can you total final momentum?

Kinematics I tried were vy^2=(0)-g(0-alt). This could give me the final velocity in the y-direction--That in turn could give me the time through vy=vy0-gt.

OK, good. Conservation of momentum is central and you can get the time of free fall.

If I knew the velocity in the x-direction, than the distance would be doable. If I initial velocity (v (terminal) in this case) then I could separate vy, get vx (which would be constant) and then get the distance that way.

You are given the horizontal distance traveled by two of the pieces. Can you use this info along with the time to get the horizontal speed of these pieces right after the collision?

 

[jamesbiomed]

yes, actually I believe I can. Let me try that :)

 

[jamesbiomed]

So I'm doing that, and getting all the initial velocities, which look reasonable. Then I'm multiplying those initial x velocities (vy0=0) by the mass and using the conservation of momentum to solve for the momentum for the final object.

P0=Mcessna*vxcessna+ Mpiece*vxpiece+Ppeice2

Then I can divide by the remaining mass to get its initial velocity, then use that with x=x0 + vx0*t to get the distance.

Unfortunately, that distance value is coming out incorrect. One a practice version, I'm off by a factor of a few thousand. (I've double checked the math)

I guess my question is, is my method correct so far? Do I need to take direction into account when multiplying their initial velocities?

Thanks a lot for the help so far !

 

[TSny]

Well, let's suppose you want to find the speed of the Cessna immediately after the collision (still 1750 m high). Use the horizontal distance the Cessna travels while falling and the time of fall. What do you get for the speed of the Cessna immediately after the collision?

[Sorry, I meant to add that your method looks correct.]

 

[TSny]

Work with the x and y components separately.

The x-component of the total momentum must be conserved in the collision.

Likewise for the y-component.

Thus, you an find the x and y components of the momentum of the second piece after the collision.

 

[jamesbiomed]

No problem. Sorry for the late reply also, I see you're answering almost immediately. I keep checking but it seems delayed to show up...anyway--

For the Cessna:

first I need time:

v^2=v0^2-(2)9.81(0-1750)=> Vf=185.30
185.30=0-9.81t=> (ignoring negative) t=18.89 s

Cessna lands 1000 m away: 1000=0+vx0t. =1000/18.89=vx=> 52.94 m/s

again, sorry for the delay. hope you're still on. I'll keep the page up this time and be more timely if so

 

[TSny]

Good. That's what I got. Can you now get the x and y components of the momentum of the Cessna after the collision?

 

[jamesbiomed]

Well, total momentum should be (52.94*3000kg)=158826.1 kg m/s

Do you mean x and y as in the horizontal plane the planes were originally flying on? Because in that case, no, I don't know how to do that.

The problem I'd have would be how to divide the initial momentum into x and y.

 

[TSny]

Yes, the horizontal plane: x-axis points east, y-axis points north.

You can get the x and y components of the momentum using the angle that specifies the direction of the momentum in the x-y plane.

 

[jamesbiomed]

On the vertical plane, then momentum in the x-direction would be the total momentum, and in the y-direction would be zero.

But I think you mean horizontal, and since the problem for the angles, it seems there must be a way.

Since I know the original x and y momentum's can that help me?

 

[jamesbiomed]

sorry, didn't see that. Let me read that first:)

 

[jamesbiomed]

The direction of the total momentum? Or x and y's seperately?

(sorry, this is not intuitive yet)

I know 34 degrees is the direction of part of the inital momentum and then 90 degrees to the horizontal is the direction of the other part. Could that help?

 

[TSny]

Yes. But I thought we were working with the momentum of the Cessna immediately after the collision. You found the magnitude of this momentum. Now you need the x and y components of this momentum.

[You'll need to go to page 2 to see further comments.]

 

[jamesbiomed]

would arctan of the momentum give me the angle?

 

[jamesbiomed]

Would the arctan of the momentum give me the angle?

 

[TSny]

would arctan of the momentum give me the angle?

Yikes!

Can you see the direction of the momentum from
http://www.webassign.net/bauerphys1/7-p-064b.gif

Remember, momentum has the same direction as velocity.

 

[jamesbiomed]

about 45 degrees north West?

 

[TSny]

about 45 degrees north West?

We're still talking about the Cessna, right?

 

[jamesbiomed]

Ouch. True. so 25 degrees South West

 

[jamesbiomed]

And in that case, per your original question, the x and y components of the cessna:

x: -158826 (cos25)
y: -158826 sin 25

 

[jamesbiomed]

Sorry,

X:-158820cos25
Y:-158820sin25

 

[TSny]

Yes, so since you know the magnitude of the momentum of the Cessna after the collision and you know the angle, you can find the x and y components of this momentum. (Careful with signs.)

{Edit: I see you already did this. Good.}

You can similarly work out the x and y components of the final momentum of the first piece of the cargo plane.

Use conservation of the x-component of momentum to find the x-component of final momentum of the second piece of the cargo plane. That is, set up an equation stating that if you add the x-components of the momentum of the Cessna and cargo plane before the collision, that should equal the sum of the x-components of the momenta of the Cessna and two pieces of the cargo plane after the collision.

Similarly for the y-component.

You can then put the x and y components of the momentum of the second piece together to find the magnitude of the momentum of the second piece. From this you can get the speed. And you know how to get the distance from the speed and time. Finally you can get the direction to the landing of the second piece using the x and y components of the momentum and the tangent function.

 

[jamesbiomed]

Thank you, that makes a lot of sense. I'm going to try all that and let you know how it goes

 

[TSny]

OK, I'm sure you can get it. Careful with the signs of the various components and note the 22^o is given relative to the y-axis rather than the x-axis.

 

[jamesbiomed]

Victory! Sir/Ma'am, this meant a lot, thank you very much!

I like how you don't give it away, because I really feel like I learned it. Thanks!

 

[jamesbiomed]

I wondered about the 22 degrees, but as long as I remembered the sin and cos would be different, it didn't affect the answer

 

[TSny]

Great! Happy problem solving

 

[jamesbiomed]

Thanks!