Two billiard balls of equal mass undergo a perfectly elastic head-on collision.
ID: 2259657 • Letter: T
Question
Two billiard balls of equal mass undergo a perfectly elastic head-on collision.If one ball's initial speed was 2.00m/s , and the other's was 4.10m/s in the opposite direction, what will be their speeds after the collision? Enter your answers numerically separated by a comma. Two billiard balls of equal mass undergo a perfectly elastic head-on collision. Two billiard balls of equal mass undergo a perfectly elastic head-on collision. Two billiard balls of equal mass undergo a perfectly elastic head-on collision.
If one ball's initial speed was 2.00m/s , and the other's was 4.10m/s in the opposite direction, what will be their speeds after the collision? Enter your answers numerically separated by a comma.
If one ball's initial speed was 2.00m/s , and the other's was 4.10m/s in the opposite direction, what will be their speeds after the collision? Enter your answers numerically separated by a comma.
If one ball's initial speed was 2.00m/s , and the other's was 4.10m/s in the opposite direction, what will be their speeds after the collision? Enter your answers numerically separated by a comma.
Explanation / Answer
Using the Law of Conservation of Momentum,
M1V1 + M2V2 = M1V3 + M2V4
where
M1 = M2 = mass of the billiard ball
V1 = initial speed of ball 1 = 2 m/sec.
V2 = initial speed of ball 2 = - 4.1 m/sec (going opposite of ball 1)
V3 = final speed of ball 1
V4 = final speed of ball 2
Since M1 = M2, then the above reduces to
V1 - V2 = V3 + V4
and substituting appropriate values,
2 - 4.1 = V3 + V4
V3 + V4 = -2.1 --- call this Equation 1
Using the Law of Conservation of Energy (since this is an elastic collision),
Kinetic energy before collision = Kinetic Energy after collision
(1/2)(M1)(V1)^2 + (1/2)(M2)(V2)^2 = (1/2)(M1)(V3)^2 + (1/2)(M2)(V4)^2
Simplifying the above,
(V1)^2 + (V2)^2 = (V3)^2 + (V4)^2
Substituting values,
(2)^2 + (-4.1)^2 = (V3)^2 + (V4)^2
4 + 16.81 = (V3)^2 + (V4)^2
(V3)^2 + (V4)^2 = 20.81 --- call this Equation 2
At this point, you have two equations with two unknowns. I will stop my actual solution here. At this point, the solution requires simple algebra and I trust that you can do the required procedure from hereon in.