Calculating Post-Collision Momentum and Speed of Two Balls

[ShaddollDa9u]

Homework Statement

Homework Equations

I have thought about using the momentum formulas here.

The Attempt at a Solution

As the momentum is a conservative quantity, I have thought that p (before collision) = p(after collision).
Since p (before collision) = 12, I wanted to use that result to find p (after collision) and then find the speed of the ball B.
So p(after collision) = ma x v + mb x v = 12
So 12 = v ( 2 + mb)
However I have two unknown variables here, so I can't find the speed and obviously, the mass of the ball B.

 

[TSny]

Suppose you take the positive direction to be toward the right. Do both balls have the same sign of momentum after the collision? Did you take this into account?

Try to use all the information in the problem. What does "perfectly elastic collision" mean to you?

 

[RedDelicious]

Do you know what it means for the collision to be perfectly elastic?

 

[ShaddollDa9u]

Wikipedia says "An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter."

So according to that, we will also have :
(1/2)ma*va² + (1/2) mb*vb² = (1/2)ma*va'² + (1/2) mb*vb'² (where va' and vb' are the speed after the collision)
So we have 36 = v² * (1 + m/2 )

Then I planned to use the first equation found (12 = v ( 2 + mb)) and solve the system to find the speed and the mass, is that correct ??

 

[TSny]

Wikipedia says "An elastic collision is an encounter between two bodies in which the total kinetic energy of the two bodies after the encounter is equal to their total kinetic energy before the encounter."

So according to that, we will also have :
(1/2)ma*va² + (1/2) mb*vb² = (1/2)ma*va'² + (1/2) mb*vb'² (where va' and vb' are the speed after the collision)
So we have 36 = v² * (1 + m/2 )

Good.

Then I planned to use the first equation found (12 = v ( 2 + mb)) and solve the system to find the speed and the mass, is that correct ??

This equation is written incorrectly. Please see the first part of post #2.

 

[ShaddollDa9u]

I couldn't get it, does the sign of the momentum of A is inversed after the collision ?

So ma * va + mb * vb = - (ma * va') + mb * vb' ??

 

[TSny]

I couldn't get it, does the sign of the momentum of A is inversed after the collision ?

So ma * va + mb * vb = - (ma * va') + mb * vb' ??

This is correct if va' and vb' represent the final speeds of the particles.

If you are uncertain about this, we can discuss it in more detail.

 

[I have a Dream]

m1v1/(m2-m1)=V
This mass thing is getting to my head. Are you still trying to figure it out?