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Inelastic Collision - 2D, find mass, theta, x-component

  1. Mar 13, 2014 #1
    Football Player 1 (mass m1 = 72 kg) has just caught a pass at the 50 yard line when he is tackled by Football Player 2 (of mass m2). Each player was initially moving at an angle of theta with respect to the 50 yard line at 8.0m/s. the tackle is a perfectly inelastic collision and the players move off with a combined velocity V. The y-component of V is 4.0m/s. Friction with the ground (uk=0.5) stops the players after a distance of 1.69m. Find m2, theta, and the x-component of V.

    there is a picture with this problem - or diagram really. Player two is moving towards (0,0) at an angle theta from the x-axis, in the 3rd quadrant. Player one is moving towards (0,0) at the same angle theta in the 4th quadrant. The final angle is in the first quadrant. So it's like an upside down V that meets at (0,0) on a regular x-y graph.

    momentum is conserved, so delta p = 0, and pf=pi. After the collision you can use work-energy stuff to find the work done by friction: -uk*(total mass)*98*1.69. I broke the momentum equations up into their components, and used the y-component equation to solve for theta - I got 30 degrees, which sounds probable.

    I tried to solve for the initial velocity after the collision using delta K = Work done by friction. I'm not entirely sure that's right. I ended up with -1/2(m1+m2)Vi^2 = the above expression for work done by friction. I got Vi = 4.1m/s. I was kind of dismissed that because I didn't think it was reasonable and wasn't sure about setting up delta K to equal work done by friction - thoughts?

    If 4.1m/s IS right, does that mean that the velocity in the x-direction after collision is 0.1m/s, because velocity in y-direction is 4.0m/s? And if I am on the wrong track there, what is a better relationship to use?

    Is there some way to solve for m2 that I'm not seeing? I keep getting that they equal each other, so I think solving for the x-component of V is probably the next step.
  2. jcsd
  3. Mar 15, 2014 #2


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    4.1 m/s is correct, BUT it's a rounded-off result. I recommend going another 2 decimal places (4 sig figs) until you have calculated a final answer, then round off at the end as appropriate.
    That approach doesn't work. Remember that a vector and its components can be represented by a right triangle: the x and y components are represented by the legs, and the hypotenuse represents the vector's magnitude. How do you find the unknown side length in a right triangle?

    I haven't gotten that far with the problem myself, but probably solving conservation of momentum in the x-direction (the one thing that hasn't been done yet) would help.
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