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# Answer on Mechanics | Kinematics | Dynamics Question for Eric

Question #1045
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&#039; + m2v2&#039;;
as m1 = m2,
v1 +v2 = v1&#039; + v2&#039;;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.