Momentum and Collision Question

[mohabitar]

A blue car with mass mc = 438 kg is moving east with a speed of vc = 23 m/s and collides with a purple truck with mass mt = 1237 kg that is moving south with a speed of vt = 10 m/s . The two collide and lock together after the collision.

What is the angle that the car-truck combination travel after the collision? (give your answer as an angle South of East)

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I really have no idea how to approach this one. I've tried many combinations based off just random guesses, but none were right. Any ideas how I can begin this problem?

 

[JaredJames]

I think this is the best explanation for this problem, it fits perfectly for your question:

http://www.physicsclassroom.com/mmedia/momentum/2di.cfm

Calculate the momentum east for the car and then the momentum south for the truck. Then using pythagoras (exactly like adding forces / velocities to get the resultant), combine them to get the final momentum and direction.

If you would like your answer checked, post it and I'll have a look.

Jared

 

[mohabitar]

Yes very relevant, but do not see anything about angles-this is the tricky part that I am stuck on..

 

[JaredJames]

M = mv, where M = momentum, m = mass and v = velocity.

Do this for both vehicles and you get the momentum eastwards for the car and southwards for the truck.

Now you can combine them (a^2 + b^2 = c^2) to get the resultant momentum (where a = car momentum, b = truck momentum and c = resultant momentum). Once you have that, you can calculate the final angle using Soh Cah Toa.

Jared

 

[mohabitar]

I already have total momentum, its 15953, but how do I use this find an angle?

 

[JaredJames]

Once you have that, you can calculate the final angle using Soh Cah Toa.

Have you done trigonometry? You need to apply the Soh Cah Toa rules to resolve the angle.

Jared

 

[mohabitar]

Ok I know trig, but I am not understanding how to apply it. I would need the arc of something (arctan for example) but to which sides do I use it on?

 

[mohabitar]

And the final momentum is the hypotenuse, why does that page say to use tan somehow, which only involves opposite and adjacent?

 

[JaredJames]

Sin(A) = Opposite / Hypotenuse

Cos(A) = Adjacent / Hypotenuse

Tan(A) = Opposite / Adjacent

Where A = the angle in question, Adjacent is the side touching the angle you require, Opposite is the side opposite the angle you require and Hypotenuse is the hypotenuse of the right angle triangle.

So in your case, you know Hypotenuse = 15953 and you know the horizontal momentum = 10074, plug those in and then using 2nd function button Cos on your calculator you get the angel from east.

So: Cos(A) = 10074 / 15953, rearrange to get Cos^-1(10074 / 15953) = A

To use Tan, you would use:

Tan(A) = 12370 / 10074, rearrange to get Tan^-1(12370 / 10074) = A

Jared

 

[JaredJames]

And the final momentum is the hypotenuse, why does that page say to use tan somehow, which only involves opposite and adjacent?

Based on this statement, I recommend you look over trigonometry rules for Soh Cah Toa to improve your understanding. Tan = Toa which means you can use your two momentums (without the need for hypotenuse) to get the resultant angle.

Arctan doesn't come into this at all. I believe you are over thinking things.

Jared

 

[mohabitar]

That wasnt the issue, I just hadnt figured out visually what or where the angle that I was supposed to find looked like. I guess now that I know the answer it makes sense that south of east is the angle right under the east axis...thanks.

 

[JaredJames]

Do you draw free body diagrams? If not, I'd definitely recommend you do so as it will help you visualise roughly what resultant you should expect.

Jared