# Answer to Question #1045 in Mechanics | Relativity 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?

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?

Expert's answer

According to the Momentum Concervation Law for a head-on collision:

m

as m

v

6 - 8 = -8 + V;

V = 6 - the second ball would have the speed on 6 m/s to the right.

m

_{1}v_{1}+ m_{2}v_{2}= m_{1}v_{1}' + m_{2}v_{2}';as m

_{1}= m_{2},v

_{1}+v_{2}= v_{1}' + v_{2}';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.

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