Question

The cue ball is deflected so that it makes an angle of 30.0° with its original direction of travel.
(a) Find the angle between the velocity vectors of the two balls after the collision.
(b) Find the speed of each ball after the collision. cue ball m/s target ball m/s.

Answer 1

Answer:

(a) θ= 43.89°

(b) $v_{1} = 1.88\sqrt{3} \frac{m}{s}$

    $v_{2} = 6.79 \frac{m}{s}$

Explanation:

Ball 1:

$u_{1} = 7.52\frac{m}{s}$

Ball 2:

$u_{2} = 0\frac{m}{s}$

As the collision is elastic, it means that kinetic energy and momentum are conserved. Following that, we apply the law of conservation of energy and momentum:

$\frac{1}{2} m_{1}u_{1}^{2}   = \frac{1}{2} m_{1}v_{1}^{2} + \frac{1}{2} m_{2}v_{2}^{2}$

and

$m_{1}u_{1} =   m_{1}v_{1} + m_{2}v_{2}$

Where u is the velocity before the collision, and v are the velocities after the collision. Both previous equations can be simplified as:

$u_{1}^{2}   = v_{1}^{2} + v_{2}^{2}$

and

$u_{1} =   v_{1} + v_{2}$

This because the two balls have the same mass. We know that the cue ball is deflected and makes an angle of 30°. From conservation in x-direction, we get::

$56.55 = v_{1} ^{2} + v_{2} ^{2}$

and

$u_{1} = v_{1}cos(30) +  v_{2}cos(\theta)$

Solving we get:

$(\frac{2u_{1}-\sqrt{3}v_{1}  }{2cos(\theta)})^{2}  = 56.55-v_{1} ^{2}$

From conservation in y-direction, we get:

$0 = v_{1}sin(30) -  v_{2}sin(\theta)$

From this, we solve this equation system and get the answers. Remember to add 30° to the angle obtained.