Solving Two Object Collision Problem

[gaimon]

I've been having some trouble with this as it isn't as simple as the two objects stick together after collision or anything...
So two objects: m1=1.96 kg m2=4.95 kg v1 initially=9.91 m/s v2 initially=15.74
the question is first what is the magnitude of velocity of the first object after collision and then the second regards the magnitude of velocity of the second after collision.
I first thought to try the m1aV1a+m2aV=m1bV1b+m2bV2b and solve for either of the final velocities using that equation to plug back into the original equation but I'd get something that would look like 97=1.96x+(97-1.96x)...obviously not going to work as they just go ahead and cancel each other out.
So I'm at a loss currently now. Any help would be appreciated of course.

 

[OlderDan]

I've been having some trouble with this as it isn't as simple as the two objects stick together after collision or anything...
So two objects: m1=1.96 kg m2=4.95 kg v1 initially=9.91 m/s v2 initially=15.74
the question is first what is the magnitude of velocity of the first object after collision and then the second regards the magnitude of velocity of the second after collision.
I first thought to try the m1aV1a+m2aV=m1bV1b+m2bV2b and solve for either of the final velocities using that equation to plug back into the original equation but I'd get something that would look like 97=1.96x+(97-1.96x)...obviously not going to work as they just go ahead and cancel each other out.
So I'm at a loss currently now. Any help would be appreciated of course.

If the objects are not sticking together, but the collision is not elastic, you don't have enough information. Did you leave something out?

 

[gaimon]

All I was given was that they were on a ramp released from 5.11 m off of the ground, that m1 and m2 were respectively 1.96 kg and 4.95 kg. I figured out the velocity each was traveling before the collision by v=sqrt(mgh) and so that gave v1=9.91m/s and v2=15.74m/s.

That's all that I have.

 

[OlderDan]

All I was given was that they were on a ramp released from 5.11 m off of the ground, that m1 and m2 were respectively 1.96 kg and 4.95 kg. I figured out the velocity each was traveling before the collision by v=sqrt(mgh) and so that gave v1=9.91m/s and v2=15.74m/s.

That's all that I have.

It is not true that v=sqrt(mgh). If they are released from the same ramp they will have the same velocity all the time. Something is very wrong here. State the full problem please.

 

[gaimon]

Two blocks of masses m1 = 1.96 kg and m2 = 4.95 kg are each released from rest, at opposite sides of a ramp, at a height of h = 5.11 m on a frictionless track and undergo an elastic head-on collision.
a)Determine the velocity of the m1 = 1.96 kg block just before the collision.
this I used v=sqrt(mgh) finding v=9.91m/s which the computer told me was correct. I just went ahead and found v2=15.74 for the heck of it.
b.)Determine the magnitude of the velocity of the m2 = 4.95 kg block just after the collision.

 

[OlderDan]

Two blocks of masses m1 = 1.96 kg and m2 = 4.95 kg are each released from rest, at opposite sides of a ramp, at a height of h = 5.11 m on a frictionless track and undergo an elastic head-on collision.
a)Determine the velocity of the m1 = 1.96 kg block just before the collision.
this I used v=sqrt(mgh) finding v=9.91m/s which the computer told me was correct. I just went ahead and found v2=15.74 for the heck of it.
b.)Determine the magnitude of the velocity of the m2 = 4.95 kg block just after the collision.

First, you titled the thread Inelastic Collision, but the problem says the collision is elastic. Second, there is no such equation as v = sqrt(mgh). You will need to correct this to find the correct velocities before the collision. Once you have the correct velocities before the collision, you can find the velocities after the collision using conservation of momentum and conservation of total kinetic energy.