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Collisions Deriving some equations for specific situations in the most general forms.

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Presentation on theme: "Collisions Deriving some equations for specific situations in the most general forms."— Presentation transcript:

1 Collisions Deriving some equations for specific situations in the most general forms

2 Momentum KE

3 Solving for the simplest case of equal masses and a stationary target, lets try to find the final velocity of the target as a function of the original speed v1 of the car which strikes it and the masses. If all masses are the same m1 = m2 = m mv1 = mvf1 + mvf2Factor and cancel the mass m v1 = vf1 + vf2 So in this special case, the mass doesn’t matter! Now I’d like to vf2 as a function of v1 only. So I want to eliminate vf1, by using conservation of kinetic energy (elastic case). KE initial = KE final ½ mv1 2 = ½ mvf1 2 + ½ mvf2 2 Factor out and divide away the ½ and m v1 2 = vf1 2 + vf2 2

4 v1 = vf1 + vf2 Solve simultaneously to eliminate Vf2 We solve the left equation and plug into the right or vice versa. Which do you think will be less work? V1= vf1+ √v1 2 - vf1 2 v1 2 = { vf1+ √v1 2 - vf1 2 v1 2 - vf1 2 = vf2 2 √v1 2 - vf1 2 = vf2 Take square root and substitute To group like terms, square both sides } x {vf1+ √ v1 2 - vf1 2 } Now FOIL v1 2 = vf1 2 + 2vf1 √ v1 2 - vf1 2 + (v1 2 –vf1 2 ) 0 = 2vf1 √(v1 2 - vf1 2 ) v1 2 = vf1 2 + 2vf1 √v1 2 - vf1 2 + (v1 2 –vf1 2 ) Which can only happen if Vf1 = 0 v1 = vf1 + vf2 v1 = vf2 This means that the incoming car stops on collision and the target car goes off with all the speed the incoming car had.

5 What the CoM is doing? It is moving rightward at constant speed.

6 What if the masses aren’t equal?

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9 The above equations will work for any elastic collision where v2 starts at rest as a stationary target, no matter what the initial masses.

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