2 pool balls of equal mass experience a perfectly elastic head on collision. 1 ball is traveling at 2m/s and the other at 3m/s in the opposite direction.What is thier speed just after impact? I know the balls swap speed and change direction but can't find the correct formula to prove this. Please help!
There is a good derivation here. Be sure to set m1 = m2 to prove that thay switch velocities and directions.
http://blog.weber.k12.ut.us/jrhoades/files/2010/11/Ch-7-AP-Notes-Elastic-Collision-Formula.pdf
To solve this problem, you can use the principle of conservation of momentum and the formula for the velocity of two objects after an elastic collision.
The principle of conservation of momentum states that the total momentum of a system remains constant before and after a collision, provided no external forces are acting on it. This means that the initial momentum of the two balls should equal their final momentum after the collision.
Let's denote the mass of the two balls as m1 and m2, and their initial velocities as u1 and u2, respectively.
According to the problem, m1 = m2 and u1 = -2 m/s (negative sign indicating opposite direction) and u2 = 3 m/s.
The formula for the final velocity of two objects after an elastic collision is:
v1 = ((m1 - m2)/(m1 + m2)) * u1 + ((2 * m2)/(m1 + m2)) * u2
v2 = ((2 * m1)/(m1 + m2)) * u1 + ((m2 - m1)/(m1 + m2)) * u2
Plugging in the given values:
v1 = ((m1 - m2)/(m1 + m2)) * -2 + ((2 * m2)/(m1 + m2)) * 3
v2 = ((2 * m1)/(m1 + m2)) * -2 + ((m2 - m1)/(m1 + m2)) * 3
Since m1 = m2, we can simplify:
v1 = ((1 - 1)/(1 + 1)) * -2 + ((2 * 1)/(1 + 1)) * 3
v2 = ((2 * 1)/(1 + 1)) * -2 + ((1 - 1)/(1 + 1)) * 3
Simplifying further:
v1 = 0 * -2 + (2/2) * 3
v2 = (2/2) * -2 + 0 * 3
v1 = 3
v2 = -2
Therefore, after the collision, the first ball will be traveling at 3 m/s in the same direction as its initial velocity, and the second ball will be traveling at 2 m/s in the opposite direction of its initial velocity.