I should know this, but i dont.

[MasterKung]

collisions

I have another question but i didnt want to create two consecutive threads so I am going to edit this one.

A bumper car at an amusement park ride traveling at .8m/s colides with an identical car at rest. This second car moves away with a speed of .5m/s What is the velocity of the first car after the collision.

so do i just use the law of conservation of momentum and disregard mass since they are all the same. if i do it that way i get .3m/s is that right?


Done with this-->{a .1 kg ball is thrown straight up into the air with an initial speed of 15m/s. Find the momentum of the ball (a) at its maximum height, and (b) halfway up to its maximum height.

The question I am wondering is, at its maximum height is the velocity zero, since its going straight up, it eventually stops then goes down right, so would the final velocity be zero, or do i need to find the height then find the velocity. Its asking for the momentum at the maximum height, but at its maximum height it will have stopped making the momentum zero right? i think my question is a bit redundant, gets the point across though.
Ive missed a few days at school and this momentum thing is new to me and I am lost.
thanks}--Done with this, hope i did it right

 

[Doc Al]

You are correct about the momentum at the top of its path: since speed is zero, momentum is zero as well.

Now find the height of the ball at that point.

Or, if you'd like to use energy methods: ask yourself what happens to the KE when the ball reaches the half-way point?

 

[MasterKung]

i used the conservation of energy to find the height to be 11.5 then using one of the big four (final velocity equals the square root of the inital velocity squared plus two times acceleration times distance, distance being the height, acceleration was gravity) i said the final velocity was .63 meters per second, so the momentum would be .06 it would have been easier to just call it zero, but i had extra time.
to find the second part i used KEi=PEf+KEf (there was no initial potential energy) in other words the conservation of energy, but i used the height 5.75 (half of the maximum) and came out with a final velocity of 10.6m/s that multiplied by the mass (.1) gave me a momentum of 1.06(kg)(m/s)

sorry for the ugly way i wrote the equations, i thought it would be a time saver for now to type them out instead of learning the script. but those are my answers, hope theyre right.

oh, and thanks for your help.