Answer to Question #1045 in Mechanics | Relativity for Eric
I was wondering if i could get help with this problem:
Two billiard balls of equal mass undergo a head-on collision. The speed of ball 1 was initially 6.00 m/s to the right, and that of ball 2 was 8.00 m/s in the opposite direction (to the left). After they bounce off of each other the speed of ball 1 is 8.00 m/s.
What is the speed of ball 2 after the collision?
According to the Momentum Concervation Law for a head-on collision: m1v1 + m2v2 = m1v1' + m2v2'; as m1 = m2, v1 +v2 = v1' + v2';We chose the direction to the right as positive. 6 - 8 = -8 + V; V = 6 - the second ball would have the speed on 6 m/s to the right.