Question

One billiard ball is shot east at 2.2m/s. A second, identical billiard ball is shot west at 1.2m/s. The balls have a glancing collision. not a head on collision, deflecting the second ball by 90 degrees and sending it north at 1.6m/s. What are the speed and direction of the first ball after the collision?
_____________m/s

________degrees south of east

Answer 1

Answer:

$v_{1}=1.886 \frac{m}{s}$

β= 57.99 south of east

Explanation:

$v_{1}=2.2 \frac{m}{s} \\v_{2}=1.2 \frac{m}{s} \\m_{1}=m_{2}=m\\v_{fx}=1.6 \frac{m}{s} \\v_{fy}=$?

Velocity in axis x the two balls come one from east and west

$m_{1}*v_{1x}+m_{2}*v_{2x}=m_{1}*v_{fx1}+m_{2}*v_{fx2}\\m*(v_{1x}+v_{2x})=m*(v_{fx1}+v_{fx2})\\v_{fx2}=0\\v_{1x}+v_{2}=v_{f1}+0\\v_{fx1}=2.2 \frac{m}{s}+(1.2\frac{m}{s})\\ v_{fx1}=1 \frac{m}{s}$

Velocity in axis y initial is zero so:

$v_{y1}+v_{y2}=v_{y1f}+v_{y2f}\\v_{y1}=0\\v_{y2}=0\\v_{y1f}+v_{y2f}=0\\v_{y1f}=-v_{y2f}\\v_{y2f}=1.6\frac{m}{s}$

$v=\sqrt{v_{1fx}^{2}+v_{1fy}^{2}}\\ v=\sqrt{1^{2}+1.6^{2}}\\v=1.886 \frac{m}{s}$

Angle is find using:

tan(β)=$\frac{v_{fy}}{v_{fx}}$

$\beta =tan^{-1}*\frac{1.6}{1}=57.99$