Respuesta :
Answer:
The speed of the cue ball is 0.2170 m/s
Explanation:
We can resolve this problem using conservation of linear momentum.
Calling [tex]\hat{i}[/tex] the unit vector pointing in the initial direction of the cue ball, we get that the initial momentum is
[tex]\vec{p}_{i} = m_{cue} \ \vec{v}_{cue_i} + m_{8 ball} \ \vec{v}_{8ball_i}[/tex]
As the 8 ball starts at rest, this means that
[tex]\vec{v}_{8ball} = 0[/tex]
and, calling m the masses of both balls
[tex]\vec{p}_{initial} = m_{cue} \ \vec{v}_{cue} = m \ 0.80 \frac{m}{s} \hat{i} [/tex]
Now, as there are no external forces in the horizontal direction, we know that the linear momentum is conserved. Calling [tex]\hat{j}[/tex] the unit vector at 90° from [tex]\hat{i}[/tex] and pointing at 52° from the 8-ball movement, we get:
[tex]\vec{v}_{8ball_f} = 0.27 \frac{m}{s} ( cos (38 \°) , sin(38\°) )[/tex]
The velocity of the cue ball will be
[tex]\vec{v}_{cue_f} = x ( cos(\theta) , sin(\theta) )[/tex]
the momentum is
[tex]\vec{p}_{f} = m_{cue} \ \vec{v}_{cue_f} + m_{8 ball} \ \vec{v}_{8ball_f}[/tex]
[tex]\vec{p}_{f} = m \ x ( cos(\theta) , sin(\theta) ) + m \ 0.27 \frac{m}{s} ( cos (38 \°) , sin(38\°) )[/tex]
and this must be equal to the initial momentum
[tex]m \ 0.80 \frac{m}{s} \hat{i} = m \ x ( cos(\theta) , sin(\theta) ) + m \ 0.27 \frac{m}{s} ( cos (38 \°) , sin(38\°) )[/tex]
We get the equations
[tex]m \ 0.80 \frac{m}{s} = m \ x \ cos(\theta) + m \ 0.27 \frac{m}{s} \ cos (38 \°)[/tex]
and
[tex] 0 = m \ x \ sin(\theta) + m \ 0.27 \frac{m}{s} \ sin (38 \°)[/tex]
From the second equation, we get
[tex] \ x \ sin(\theta) = \ 0.27 \frac{m}{s} \ sin (38 \°)[/tex]
[tex] \ x = \ 0.27 \frac{m}{s} \frac{ \ sin (38 \°) } { sin(\theta)} [/tex]
putting this in the first equation
[tex]\ 0.80 \frac{m}{s} = \ 0.27 \frac{m}{s} \frac{ \ sin (38 \°) } { sin(\theta)} cos(\theta) + \ 0.27 \frac{m}{s} \ cos (38 \°)[/tex]
[tex]\ 0.80 \frac{m}{s} = \ 0.27 \frac{m}{s} (\ sin (38 \°) \frac{ cos(\theta) } { sin(\theta)} + \ cos (38 \°))[/tex]
[tex]\frac{ 0.80 \frac{m}{s}}{ 0.27 \frac{m}{s} } = (\ sin (38 \°) \frac{ cos(\theta) } { sin(\theta)} + \ cos (38 \°))[/tex]
[tex]2.9629 - \ cos (38 \°) = sin (38 \°) \frac{ cos(\theta) } { sin(\theta)} [/tex]
[tex]2.9629 - \ cos (38 \°) = sin (38 \°) \frac{1 } { tan(\theta)} [/tex]
[tex]tan(\theta)= \frac{ sin (38 \°)}{2.9629 - \ cos (38 \°) }[/tex]
[tex]\theta= arctan ( \frac{ sin (38 \°)}{2.9629 - \ cos (38 \°) } )[/tex]
[tex]\theta= 50 \°[/tex]
Now, using the equation
[tex] \ x = \ 0.27 \frac{m}{s} \frac{ \ sin (38 \°) } { sin(\theta)} [/tex]
[tex] \ x = \ 0.27 \frac{m}{s} \frac{ \ sin (38 \°) } { sin(50 \°)} [/tex]
[tex] \ x = \ 0.2170 \frac{m}{s} [/tex]