Question

In a game of pool the cue ball is rolling at 2.00 m / s in a direction 30.0° north of east when it collides with the eight ball (initially at rest). The mass of the cue ball is 170 g, but the mass of the eight ball is only 156 g. After the collision the cue ball heads off at 10.0° north of east, and the eight ball moves off due north. What are the final speeds of each ball after the collision?

Answer 1

`v_(f1)=1.758m/s`

`v_(f1)=1.758m/s`

Step-by-step explanation:

The collision is elastic so we can use the conservation of momentum

`P_i=P_f`

`m_1*v_1+m_2*v_2=m_1*v_(f1)+m_2*v_(f2)`

Describe the motion in axis x'

`170g*2.0m/s*cos(30)+156g*0m/s=170*cos(10)*v_(f1)+156g*0m/s`

`294.44 g*m/s=167.41g*v_(f1)`

`v_(f1)=1.758m/s`

Describe the motion in axis y'

`170g*2.0m/s*sin(30)=170g*1.75m/s*sin(10)+156*v_(f2)`

`v_(f2)=1.55m/s`