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    dks2114's Avatar
    dks2114 Posts: 32, Reputation: 2
    Junior Member
     
    #1

    Oct 14, 2007, 07:49 PM
    Inelastic collisions
    AT THE CENTER OF A 50 M DIAMETER CIRCULAR ICE RINK, A 75 KG SKATER TRAVELING NORTH AT 2.5 M/S COLLIDES WITH AND HOLDS ON TO A 60 KG SKATER WHO HAS BEEN HEADING WEST AT 3.5M/S. HOW LONG WILL IT TAKE THEM TO GLIDE TO THE EDGE OF THE RINK? WHERE WILL THEY REACH IT- GIVE THE ANSWER AS A ANGLE NORTH OF WEST.

    Okay so can anyone help me with this problem. I found that the northward velocity is 1.4 m/s and the westward velocity is 1.6 m/s and that the total final velocity which is 2.13 m/s NE. And I am thinking that to find the time it might have something to do with impulse is equal to change in momentum. But it doesn't work because how would you find "Force" (Ft= mVf-mVi) or which mass do you put into the equation and which initial velocity do you put inti the equation. My other thought was to use the vf^2 = Vi^2 + 2at formula but I encountered the same problem with Vi and I didn't know how to find acceleration. So what should I do am I even on the right path?
    str82hell's Avatar
    str82hell Posts: 30, Reputation: 5
    Junior Member
     
    #2

    Oct 14, 2007, 09:03 PM
    Vector addition. The x components of skater 1 plus the skater 2, likewise the y components. Then plug them into an arctangent to get the angle. The angle will be from standard position so you will need to take the reference angle from the y axis.

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