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A cue ball traveling at 0.80 m/s hits the stationary 8-ball, which moves off with a speed of 0.27 m/s at an angle of 38° relative to the cue ball's initial direction. Assuming that the balls have equal masses and the collision is inelastic, what will be the speed of the cue ball?

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]

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