Question

A 105 g cue ball is traveling to the right at 12.5 m/s. Meanwhile, a 104 g 8-ball is moving at 13.4 m/s to the left. After the collision, the cue ball is moving off to the left at a velocity of 8.9 m/s. What is the magnitude and direction of the final velocity of the 8-ball?

Answer 1

Answer:

The velocity of the 8-ball after collision is  $v_2 = 8.12 \ m/s$ and the direction is towards that right

Explanation:

From the question we are told that

 The mass of the cue ball is  $m_1  =  105 \  g$

   The speed of the cue ball is  $u_1 =  12.5 \  m/s$

   The mass of the 8-ball is  $m_2  =  104 \  g$

   The velocity of the 8 ball is  $u_2 =  - 13.4 \  m/s$

The negative sign is because it is moving left  

    The  velocity of the cue ball after collision is $v_1 =  -8.9 \  m/s$

Generally from the law of momentum conservation

     $m_1 *  u_1  +  m_2 *  u_2  =  m_1 * v_1  +  m_2 *  v_2$

      $105 *  12.5  +  104 *  - 13.4 =   104* -8.9 +  104*  v_2$

=>     $-81.1 = -925.6 + 104v_2$

=>    $104v_2 = 844.5$

=>    $v_2 = 8.12 \ m/s$