Collision question?

Question

Two pool balls of equal mass move toward each other. One travels at 1.5 m/s while the other travels at .5 m/s. If the collision is elastic, what are the velocities of each pool ball after the collision?
For the ball traveling at 1.5 m/s I got a final velocity of 0 m/s
The other ball I have traveling at 1 m/s
Is that right? thanks

ok what about -1/2 m/s and 3/2 m/s?
nikki...im using .5 m/s and not 5 m/s

Answer 1

No math required because of special condition of equal masses and head-on collision.

From conservation of momentum ...
the one originally travelling at 1.5 goes in the reverse direction at .5

the one going .5 reverses direction going 1.5 m/s

Answer 2

There is actually a secret equation that applies to elastic collisions. I have seen a simple derivation for the equation for a very specialized case, and I have heard that the equation has been empiracally verified in general. The equation basically says that in an elastic collision the relative velocity of the objects before the collision is equal to minus the relative velocity after the collision or:

V1Before - V2Before = V2after - V1after

(This equation is much easier to use as a second equation than the conservation of Kinetic energy, which the previous answerer invoked, and was correct about.)

So I would use momentum conservation:

m(0.5) + m(-1.5) = mV1A+mV2A

and the "secret" elastic question:

0.5 - (-1.5) = V2A + V1A

Now you have two equations with two unknowns. Solve for the two velocities and you are in business.

Answer 3

your answer is wrong.
according to the consevation of momentum, 1.5M-5M=Mv1+MV2
according to the conservation of energy, 1/2 M(1.5)square plus 1/2 M(5) square = 1/2 M (V1)square +1/2 M(V2) square.
calculate those two equations, and V1 , V2.