I don't exactly understand what you mean by "photogate" time. Those times tells you relative velocity of the carts after/before collision (twice the time, half the velocity). With that, you should be able to solve it. It really depends on what Photogate time means.
This is a long problem for an exam.
So I had an exam today and one of the questions is bugging me. Our professor is out of town for a few days, so we had a proctor and I still can't figure it out using my notes. It was a question about two carts and deals with momentum. Cart A is at rest between two photogates, while cart B is off to the right and it's pushed. An elastic collision occurs. Mass of cart A is 100 g and the first photogate time is 0.2 seconds and second time is 0.5 seconds going back through. Find mass of cart B and the initial and final energy of the system.
I didn't think there was enough info supplied, so I was really lost. Maybe I missed something in the question, but this is what I believe it was. Help Please.
4 answers
Unless someone on here knows exactly what your experiment was, I do not see how we can help. I do not know where these photogates are for example.
If you go to google and type "conservation of momentum photogate lab penn state" this is just like the experiment we performed. It should be the top result "Physics 250 Laboratory: Conservation of Momentum. Hope this helps.
You need the velocities of both masses before and after the collision.
Since the initial velocity of the big mass is zero, you need to measure 3 velocities with the two photogates
You need the length of the two masses that pass through the photogates (your Penn state link calls them "flag lengths)
Then you can get for your initially moving mass (B) its initial velocity = +L1/(t forward)
and its final velocity on rebound = -L1/(t back)
However you need that third velocity, the velocity forward of A after it is hit. +L2/t mass A after)
Then mB * VinitialB = mB * VfinalB + 100 * VfinalA
solve that for mB
Note that Vfianal B must be negative or the photogate for B will not trigger the second time.
That means the stationary mass must be much bigger than the moving mass, or the moving one would not rebound.
Now since the collision was elastic the initial and final energies should be the same. However the world is not perfect
masses should now be in kilograms and lengths in meters so Joules result.
.5mB *VBi^2 + 0 and compare to .5mB * vBf^2 + .5mA * vAf^2
Since the initial velocity of the big mass is zero, you need to measure 3 velocities with the two photogates
You need the length of the two masses that pass through the photogates (your Penn state link calls them "flag lengths)
Then you can get for your initially moving mass (B) its initial velocity = +L1/(t forward)
and its final velocity on rebound = -L1/(t back)
However you need that third velocity, the velocity forward of A after it is hit. +L2/t mass A after)
Then mB * VinitialB = mB * VfinalB + 100 * VfinalA
solve that for mB
Note that Vfianal B must be negative or the photogate for B will not trigger the second time.
That means the stationary mass must be much bigger than the moving mass, or the moving one would not rebound.
Now since the collision was elastic the initial and final energies should be the same. However the world is not perfect
masses should now be in kilograms and lengths in meters so Joules result.
.5mB *VBi^2 + 0 and compare to .5mB * vBf^2 + .5mA * vAf^2
1 answer