2D completely inelastic collision question

  • Thread starter Hoodoo
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  • #1
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Homework Statement


Andy (mass 92 kg) and Sam (mass 75 kg) are both on toboggans (each toboggan has a mass of 4.6 kg) travelling on a crash course towards each other. Just as they hit, they lock hands so they move together as one. What is their new speed? In what direction do they travel?

mA = 96.6 kg
vA = 8.8 m/s at 8.7 degrees above horizontal (headed +x direction)
mS = 79.6 kg
vS = 6.7 m/s at 7.3 degrees below horizontal (headed +x direction)
mA + mS = 176.2 kg
vf = ?


Homework Equations


mAvA + mSvS = mAvA' + mSvS'




The Attempt at a Solution


Tried finding x and y components and unsure what to do next.. since Andy is headed southeast, do I make that y-component negative? Am I combining too much into these equations? I am so confused with these questions where 2 things are moving at the same time or at non-right angles to each other. Can someone break these problem types down for me so that I have a step by step approach to them? I end up with equations and have no idea how to follow through. Here, I have 2 equations and 2 unknowns in each equation. I don't know what to do next or if this is even correct.

(96.6)(8.8 cos 8.7) + (79.6)(6.7 cos 7.3) = (176.2)(vfcosσ)
7.77 = (vfcosσ)

(96.6)(8.8 sin 8.7) + (79.6)(6.7 sin 7.3) = (176.2)(vfsinσ)
1.11 = (vfsinσ)
 
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Answers and Replies

  • #2
ehild
Homework Helper
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mA = 96.6 kg
vA = 8.8 m/s at 8.7 degrees above horizontal (headed +x direction)
mS = 79.6 kg
vS = 6.7 m/s at 7.3 degrees below horizontal (headed +x direction)
mA + mS = 176.2 kg
vf = ?


Homework Equations


mAvA + mSvS = mAvA' + mSvS'




The Attempt at a Solution


Tried finding x and y components and unsure what to do next.. since Andy is headed southeast, do I make that y-component negative?
Andy is headed North-East, so the y component of its velocity is positive. But Sam travels to South-East, use negative sign with the y component of his velocity.

(96.6)(8.8 cos 8.7) + (79.6)(6.7 cos 7.3) = (176.2)(vfcosσ)
7.77 = (vfcosσ)

(96.6)(8.8 sin 8.7) - (79.6)(6.7 sin 7.3) = (176.2)(vfsinσ)
[STRIKE]1.11 = (vfsinσ)[/STRIKE]

Use the negative sign in the second equation and then you get a numerical value for (vfcosσ) and (vfsinσ).

If you square these equations and add them, the result is vf2.
If you divide them you can get the the tangent of angle σ.

ehild
 
  • #3
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Andy is headed south-east and Sam is heading north-east.
 
  • #4
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Alright, so I got:

-0.3452/7.77 = vfsinσ/vfcosσ
-0.0444 = tanσ

-2.5 degrees for σ

Does that imply it's below the horizontal?

Also, followed the instructions on adding the equations together and then squaring. It wasn't until I got to the end that I remembered that sin squared + cos squared = 1 so I wasn't realizing that they would eventually go away.

So vf = 7.8 m/s at 2.5 degrees below the horizontal

My problem with this question was not remembering my trig! Thanks for opening up my brain.
 
  • #5
ehild
Homework Helper
15,543
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Alright, so I got:

-0.3452/7.77 = vfsinσ

vfsinσ=0.3452/7.77, positive. So is σ.

ehild
 
  • #6
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Answer from the text agrees with what I got. You are picturing them heading in the wrong directions. They are both heading towards each other. Andy is headed south-east (-y) and Sam is headed north-east (+y). The -2.5 degrees tells me that it's below the horizontal which is what my text tells me.

Is that wrong?
 
  • #7
ehild
Homework Helper
15,543
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Well, I think "above" and "below" horizontal refers to the direction of velocity vectors, and not to arrows representing the velocities. So "8.7° above horizontal" means a positive angle enclosed with the +x axis. "7.3° below horizontal" is a negative angle with respect to the positive x axis. I might be wrong.

ehild
 

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