TomHart said:
You're actually not that far away. First of all, when you write those equations, eliminate all of the units (kg, m/s, etc.).
Then multiply the whole equation by the 120 on the bottom. Then multiply out the squares inside of the parentheses on the left side. If you notice, when you multiply those out, 2 of the terms will be (25)(190)cos2θ1 and (25)(190)sin2θ1. Those reduce to (25)(190) because of the sin2 + cos2 = 1 trig identity. So you should be left with an equation of the form acosx + bsinx = c.
It seems that these completely elastic collision problems are more complicated mathematically. If there is an easier way to solve this problem, I don't know what it is.
By the way, I still don't even think this problem is possible in the real world. How can a lighter object with a speed delta of +5 m/s (relative to the heavier object) increase the velocity of the heavier object by 5 m/s?
By far the easiest way to deal with such collision problems is to first transform them into the so-called
center of momentum (CM) frame, look at the final velocities in that frame, and then transform those back into the original (lab) frame. I will go through most of the steps; I don't think this would volate PF policy, since you have already solved the problem.
Denote the initial lab-frame velocities of particles 1 and 2 by ##\vec{v}_{i1}## and ##\vec{v}_{i2}## and the initial lab-frame momenta by ##\vec{p}_{i1}= m_1 \vec{v}_{i1}## and ##\vec{p}_{i2} = m_2 \vec{v}_{i2}##. The total initial lab-frame momentum is ##\vec{p} = \vec{p}_{i1} + \vec{p}_{i2}##. Of course, this is also the total final lab-frame momentum, because of momentum conservation.
The CM frame is a new coordinate system moving at a constant velocity ##\vec{V}##, so that in the CM frame the initial velocities are ##\vec{u}_{i1}= \vec{v}_{i1} - \vec{V}## and ##\vec{u}_{i2} = \vec{v}_{i2} - \vec{V}##. (This just uses the standard kinematics of transforming velocities between frames in constant relative motion.) The CM frame is the one in which
total momentum = 0. That means that ##\vec{p} - (m_1+m_2) \vec{V} = \vec{0}##, or
$$\vec{V} = \frac{m_1 \vec{v}_{i1} + m_2 \vec{v}_{i2} }{m_1+m_2}. $$
Now comes the clincher: in the CM frame, the individual particle speeds remain the same before and after a perfectly elastic collision. (This follows from conservation of total kinetic energy.) So, in the CM frame, the final speed of particle 2 is the same as it was initially, before the collision. Also, in the CM frame the two particles are heading in exactly opposite directions. After the collision, the deflection angle can be anything from ##0^o## (for a glancing collision in which the particles just barely touch) to ##180^o## (for a head-on collision, in which each particle bounces back). If we let ##\omega## be the initial angle and ##\theta## be the deflection between ##\vec{u}_{i2}## and ##\vec{u}_{f2}## we have ##\vec{u}_{f2} = |\vec{u}_{i2}| (-\cos(\omega + \theta), \sin(\omega +\theta))##. (Here we choose the direction vector to agree with the initial velocity when the deflection angle is zero, so the final angle is the same as the initial angle.)
By the way, the deflection angle is limited in the lab frame even though it is unlimited in the CM frame.
Now we can transform that back into the lab frame to get the final lab-frame velocity of particle as ##\vec{v}_{f2} = \vec{u}_{f2} + \vec{V}##, and we can get the final lab-frame speed as ##|\vec{v}_{f2}|##.
For your data we have ##\vec{v}_{i1} = (20/\sqrt{2},-20/\sqrt{2})##, ##\vec{v}_{i2} = (15/\sqrt{2},15/\sqrt{2})##. The total momentum is ##\vec{p} = (95 \sqrt{2}, -5 \sqrt{2})##, and ##\vec{V} = \vec{p}/(m_1+m_2) = (95\sqrt{2}/11, -5\sqrt{2}/11)##. We have
##\vec{u}_{i1} = \vec{v}_{i1}- \vec{V} = (15 \sqrt{2}/11,-105 \sqrt{2}/11) ## and ##\vec{u}_{i2} = \vec{v}_{i2} - \vec{V} = (-25 \sqrt{2}/11,175 \sqrt{2}/11).##
The initial speed of particle 2 in the CM frame is ##S_{CM,2} = |\vec{u}_{i2}| = 125/11##, so ##\vec{u}_{f2} = (125/11)(-\cos(\omega), \sin(\omega))##. The final lab-frame velocity of particle 2 is ##\vec{v}_{f2} = \vec{u}_{f2} + \vec{V}##.
We can determine ##\omega## from the direction of ##\vec{u}_{i2}##: ##\omega = \arctan(7)##. Substituting that into the above and expanding ##\cos(\omega + \theta)## and ##\sin(\omega + \theta)## we get the final lab-speed of particle 2 as
$$S_{LAB,2} = (5/11)\sqrt{1349 - 260 \cos(\theta) + 1320 \sin(\theta)}.$$
This has a maximum value of about 23.59 m/s, when the deflection angle in the CM frame is about ##\theta = 101.143^o##.
So, indeed, in the lab frame the heavier particle can be going faster after the collision; on the other hand, the lighter particle is going much slower after the collision. You can work the details for particle 1.
So: believe it or not, this way of working things out is about as simple as it gets! With a bit of practice, it becomes almost second-nature, and it removes a lot of the uncertainty about how to deal with the effects of deflection angles and the like.
