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

Answer:

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

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:

$V_{1f}=\sqrt{1^2+(-1.37)^2}=0.35m/s$

The angle is:

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