Question

Starting with equations $m_1v_1 = m_1v_1'\cos(\theta_1) + m_2v_2'\cos(\theta_2)$ and $ 0 = m_1v_1'\sin(\theta_1) + m_2v_2'\sin(\theta_2)$ for conservation of momentum in the - and -directions and assuming that one object is originally stationary, prove that for an elastic collision of two objects of equal masses, $\dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_1'^2 + \dfrac{1}{2}mv_2'^2 + mv_1'v_2'\cos(\theta_1 - \theta_2)$ as discussed in the text.

Final Answer

see video for derivation.

Solution Video

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This is College Physics Answers with Shaun Dychko.
In this solution we're going to show that this and this combined together can make this expression here. So these first two lines, equations one and two come from the conservation of momentum in the x direction and conservation of momentum in the y direction for a scenario where one thing is initially at rest and then some other thing comes to hit it and then deflects off at some angle and this thing deflects off at some angle as well. So, in this expression here in green, well, I mean, the writing is in green and my highlighting is in blue, we have this trigonometric thing here and it has the difference of two angles. That should make you think about trigonometric identities. The trig identity that has cos of an angle minus a different angle is cos

*theta one*times cos*theta two*plus sine*theta one*times sine*theta two*. It's useful to keep this in mind because it helps us figure out what should we do to these equations one and two in order to show that this third part is true. So we need to create these products here of cosine of one angle times the other then sine of one angle times the other. We can do that by squaring each of these equations because if we square -- this is a binomial and if we square it we end up with -- by the way, squaring a binomial falls this pattern. You have one thing plus another and you square that, you get the first thing squared plus two times the first times the second, plus the second thing squared. So when you square the right hand side of equation one, you end up with this m one squared*v one prime*squared, times cos squared*theta one*plus two times this times that. Then that is what is going to get you the cos*theta one*multiplied by cos theta two part. Then continuing the squaring, we get*m two*squared*v two prime*squared times cos squared*theta two*. All those terms are shown here. Then you square the second equation and of course the left side is zero and the right side is following this pattern of squaring a binomial and you get these terms here. On the left side we get*m one*squared*v one*squared and then we're adding these two equations together so that we accomplish this addition that's needed in order to make this trigonometric identity. A couple of handy things happen here, not only do we get to use this trig identity, but we also get to use the Pythagorean identity to make some angles disappear completely. So we have cos squared*theta one*here and then we have a sine squared*theta one*here and they both have the same factor in front. So we can factor out that common factor between these two like terms and we get*m*squared*v one prime*squared times sine squared*theta one*plus cos squared*theta one*. The handy thing with that is that sine squared*theta one*plus cos squared*theta*equals the number one. So this just becomes*m*squared*v one prime*squared which is what I've written here. Then combining this cos squared*theta two*term with this sine squared*theta two*term, they also have a common factor in front and so combining those together gives us*m*squared*v two prime*squared. Then lastly, we have these terms containing the cos*theta one*times cos*theta two*and sine*theta one*times sine*theta two*and they have common factors here so we can factor that out. Two*m*squared*v one prime*times*v two prime*times cos*theta one*cos*theta two*plus sine*theta one*sine*theta two*and there is the magic. We have this bracket here which is this trigonometric identity that we're looking for. Then we can replace all that with cos*theta one**minus**theta two*. We're also going to divide every term by two*m*and then we get this last line. One half*m*to the power of one now,*v one*squared equals one half*m v one prime*squared plus one half*m v two prime*squared plus*m v one prime v two prime*times cos*theta one*minus*theta two*.