A ball of mass 0.440 kg moving east with a speed of 3.70 m/s collides head-on with a 0.220-kg mass at rest. If the collision is perfectly elastic, what is the speed and direction of each ball after the collision?(adsbygoogle = window.adsbygoogle || []).push({});

momentum conservation:

m1v1 + m2v2 = m1v1¢ + m2v2¢;

(0.440 kg)(3.70 m/s) + (0.220 kg)(0) = (0.440 kg)v1¢ + (0.220 kg)v2¢.

1.628 kg m/s = (0.440 kg)v1¢ + (0.220 kg)v2¢

elastic, kinetic energy is constant:

½ m1 v21 + ½m2 v22 = ½ m1v1’2 + ½ m2v2’2,

½ (0.440 kg)(3.70 m/s)2 + 0 = ½ (0.440 kg)v1’2 + ½ (0.220 kg)v2’2

6.02 kg m2/s2 = (0.440 kg)v1’2 + (0.220 kg)v2’2

Whats the quickest way to solve with 2 variables? This one gets a little messy because of the velocity squares.

My teacher was talking about using a derterminant. Could I use one for this question? If not, can you please given an example of a problem in which I can use one?

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# Homework Help: Very Quick momentum equation question

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