Two billiard balls, one heading north at 15.0 m/s and a secondheading south at 1

Question

Two billiard balls, one heading north at 15.0 m/s and a secondheading south at 10.0 m/s, collide head-on. Take the collision tobe perfectly elastic and choose the positive direction north. What is the post-impact speed of the first ball? Answer: m/s
What is the post-impact speed of the second ball? Answer: m/s
I don't even know where to start, since they don't give themass how do i do it? Two billiard balls, one heading north at 15.0 m/s and a secondheading south at 10.0 m/s, collide head-on. Take the collision tobe perfectly elastic and choose the positive direction north. What is the post-impact speed of the first ball? Answer: m/s
What is the post-impact speed of the second ball? Answer: m/s
I don't even know where to start, since they don't give themass how do i do it?

Explanation / Answer

Let the mass of the two billiard balls be m.
The initial speed of the first billiard ball is u1 =15.0 m/s The initial speed of the second billiard ball is u2= -10.0 m/s The speed of the two billiard balls after the elasticcollision is v1 = [u1 * (m1 -m2) + 2m2u2/(m1 +m2)] and v2 = [u2 * (m2 -m1) + 2m1u1/(m1 +m2)] Substituting the values of u1 and u2 inthe above two equations we get the speed of the two billiard ballsafter the elastic collision.

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