Solving 2D Elastic Collision Problem in Physics

[mgrantbaker]

Came across this one in my freshman physics book:

A particle of mass 1m is traveling along the x-axis at velocity V1. It collides elastically with a second particle of mass 3m traveling at velocity V2. The first particle (1m) moves off at 0.92m/s at 48º to the x-axic. The second particle (3m) moves off at 1.2m/s at 17º to the x-axis. Find two sets of possible values for the initial velocities for both particles and the direction of the second particle (3m).



I've tried brute forcing it with trigonometry/dot-producting the momentum equation and using that with conservation of energy, but I ended up with an equation for the initial direction that was way too complex to solve. Also tried reversing the collision, shifting the angles, and then translating into a rest frame for one of the particles. That approach yielded a direction for the 3m particle that was close to one of the two possible answers, but still not right. Plus, my math (which probably has some mistakes) didn't leave the potential for two results.

I haven't attempted a full vector-based analysis (breaking the collision into normal and trangent vectors) since this is from a freshman level book, and that method is beyond the scope of the chapter. So what am I missing here?

 

[zcd]

If the collision is elastic, then you can use conservation of kinetic energy. Also, since velocities are constant before and after the collision no external forces are acting on the projectiles and momentum is conserved, so you can separate the momentums into equal x and y components.

 

[tomcue]

I would suggest approaching this problem using the principles of conservation of momentum and conservation of energy. These principles state that the total momentum and energy of a system before and after a collision remains the same.

To solve this problem, we can first write the equations for conservation of momentum and energy:

Conservation of momentum:
m1V1 + m2V2 = m1V1' + m2V2'

Conservation of energy:
1/2m1V1^2 + 1/2m2V2^2 = 1/2m1V1'^2 + 1/2m2V2'^2

Where m1 and m2 are the masses of the two particles, V1 and V2 are the initial velocities, and V1' and V2' are the final velocities.

We can then use the given information to solve for the final velocities of the particles. From the given data, we can write the following equations:

V1' = V1cos48º + V2cos17º
V2' = V1sin48º + V2sin17º

Substituting these equations into the conservation of momentum and energy equations, we can solve for V1 and V2. This will give us two sets of possible initial velocities for the two particles.

To find the direction of the second particle (3m), we can use the equation for conservation of momentum and solve for the angle between the initial and final velocities. This will give us two possible angles, one for each set of initial velocities.

If your calculations have resulted in a complex equation for the initial direction, it is possible that there is a mistake in your calculations. I would suggest double checking your work to ensure accuracy.

Additionally, while a vector-based analysis may be beyond the scope of the chapter, it can still be a useful approach to solving this problem. It may be worth exploring this method further to see if it yields the correct results.