Question

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?

Answer 1

Answer:

The speed of the cue ball is 0.2170 m/s

Explanation:

We can resolve this problem using conservation of linear momentum.

Calling $\hat{i}$ the unit vector pointing in the initial direction of the cue ball, we get that the initial momentum is

$\vec{p}_{i} = m_{cue} \ \vec{v}_{cue_i} + m_{8 ball} \ \vec{v}_{8ball_i}$

As the 8 ball starts at rest, this means that

$\vec{v}_{8ball} = 0$

and, calling m the masses of both balls

$\vec{p}_{initial} = m_{cue} \ \vec{v}_{cue} = m  \ 0.80 \frac{m}{s} \hat{i}$

Now, as there are no external forces in the horizontal direction, we know that the linear momentum is conserved. Calling $\hat{j}$ the unit vector at 90° from $\hat{i}$ and pointing at 52° from the 8-ball movement, we get:

$\vec{v}_{8ball_f} = 0.27 \frac{m}{s} ( cos (38 \°) , sin(38\°) )$

The velocity of the cue ball will be

$\vec{v}_{cue_f} = x ( cos(\theta) , sin(\theta) )$

the momentum is

$\vec{p}_{f} = m_{cue} \ \vec{v}_{cue_f} + m_{8 ball} \ \vec{v}_{8ball_f}$

$\vec{p}_{f} = m \ x ( cos(\theta) , sin(\theta) ) + m \ 0.27 \frac{m}{s} ( cos (38 \°) , sin(38\°) )$

and this must be equal to the initial momentum

$m  \ 0.80 \frac{m}{s} \hat{i} = m \ x ( cos(\theta) , sin(\theta) ) + m \ 0.27 \frac{m}{s} ( cos (38 \°) , sin(38\°) )$

We get the equations

$m  \ 0.80 \frac{m}{s}  = m \ x  \ cos(\theta)  + m \ 0.27 \frac{m}{s}  \ cos (38 \°)$

and

$0  = m \ x  \ sin(\theta)  + m \ 0.27 \frac{m}{s}  \ sin (38 \°)$

From the second equation, we get

$ \ x  \ sin(\theta)  = \ 0.27 \frac{m}{s}  \ sin (38 \°)$

$ \ x    = \ 0.27 \frac{m}{s} \frac{  \ sin (38 \°) } {  sin(\theta)}$

putting this in the first equation

$\ 0.80 \frac{m}{s}  = \ 0.27 \frac{m}{s}  \frac{ \ sin (38 \°) } {  sin(\theta)} cos(\theta)  +  \ 0.27 \frac{m}{s}  \ cos (38 \°)$

$\ 0.80 \frac{m}{s}  = \ 0.27 \frac{m}{s}   (\ sin (38 \°) \frac{ cos(\theta) } {  sin(\theta)} +  \ cos (38 \°))$

$\frac{ 0.80 \frac{m}{s}}{  0.27 \frac{m}{s} } =  (\ sin (38 \°) \frac{ cos(\theta) } {  sin(\theta)} +  \ cos (38 \°))$

$2.9629 -  \ cos (38 \°) =   sin (38 \°) \frac{ cos(\theta) } {  sin(\theta)}$

$2.9629 -  \ cos (38 \°) =   sin (38 \°) \frac{1 } {  tan(\theta)}$

$tan(\theta)=  \frac{ sin (38 \°)}{2.9629 -  \ cos (38 \°) }$

$\theta= arctan ( \frac{ sin (38 \°)}{2.9629 -  \ cos (38 \°) } )$

$\theta= 50 \°$

Now, using the equation

$ \ x    = \ 0.27 \frac{m}{s} \frac{  \ sin (38 \°) } {  sin(\theta)}$

$ \ x    = \ 0.27 \frac{m}{s} \frac{  \ sin (38 \°) } {  sin(50 \°)}$

$ \ x    = \ 0.2170 \frac{m}{s}$