2D completely inelastic collision question

[Hoodoo]

Homework Statement

Andy (mass 92 kg) and Sam (mass 75 kg) are both on toboggans (each toboggan has a mass of 4.6 kg) traveling 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?

m_A = 96.6 kg
v_A = 8.8 m/s at 8.7 degrees above horizontal (headed +x direction)
m_S = 79.6 kg
v_S = 6.7 m/s at 7.3 degrees below horizontal (headed +x direction)
m_A + m_S = 176.2 kg
v_f = ?

Homework Equations

m_Av_A + m_Sv_S = m_Av_A' + m_Sv_S'

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)(v_fcosσ)
7.77 = (v_fcosσ)

(96.6)(8.8 sin 8.7) + (79.6)(6.7 sin 7.3) = (176.2)(v_fsinσ)
1.11 = (v_fsinσ)

 

[ehild]

m_A = 96.6 kg
v_A = 8.8 m/s at 8.7 degrees above horizontal (headed +x direction)
m_S = 79.6 kg
v_S = 6.7 m/s at 7.3 degrees below horizontal (headed +x direction)
m_A + m_S = 176.2 kg
v_f = ?

Homework Equations

m_Av_A + m_Sv_S = m_Av_A' + m_Sv_S'

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)(v_fcosσ)
7.77 = (v_fcosσ)

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

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

If you square these equations and add them, the result is v_f².
If you divide them you can get the the tangent of angle σ.

ehild

 

[Hoodoo]

Andy is headed south-east and Sam is heading north-east.

 

[Hoodoo]

Alright, so I got:

-0.3452/7.77 = v_fsinσ/v_fcosσ
-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.

 

[ehild]

Alright, so I got:

-0.3452/7.77 = v_fsinσ

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

ehild

 

[Hoodoo]

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?

 

[ehild]

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