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

`v_(1)=1.886 (m)/(s)`

β= 57.99 south of east

Step-by-step explanation:

`v_(1)=2.2 (m)/(s) \\v_(2)=1.2 (m)/(s) \\m_(1)=m_(2)=m\\v_(fx)=1.6 (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 (m)/(s)+(1.2(m)/(s))\\  v_(fx1)=1 (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(m)/(s)`

`v=\sqrt{v_(1fx)^(2)+v_(1fy)^(2)}\\ v=\sqrt{1^(2)+1.6^(2)}\\v=1.886 (m)/(s)`

Angle is find using:

tan(β)=
`(v_(fy))/(v_(fx))`

`\beta =tan^(-1)*(1.6)/(1)=57.99`