Question

One billiard ball is shot east at 1.8 m/s . A second, identical billiard ball is shot west at 0.80 m/s . The balls have a glancing collision, not a head-on collision, deflecting the second ball by 90∘ and sending it north at 1.37 m/s .

Answer 1

Velocity of ball1 after the collision is 0.35 m/s at 53.87° due south of east.

Step-by-step explanation:

By conservation of the linear momentum:

`m*V_(1o)+m*V_(2o)=m*V_(1f)+m*V_(2f)` Since both masses are the same, and expressing the equation for each axis x,y:

`V_(1ox)+V_(2ox)=V_(1fx)` eqX

`0=V_(1fx)+V_(2fx)` eqY

From eqX:
`V_(1fx)=1m/s`

From eqY:
`V_(1fy)=-1.37m/s`

The module is:

`V_(1f)=√(1^2+(-1.37)^2)=0.35m/s`

The angle is:

`\theta=atan(-1.37/1)=-53.87\°` This is 53.87° due south of east