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?

Expert's answer

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

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.

m

as m

v

6 - 8 = -8 + V;

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

## Comments

## Leave a comment