Question

A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is the velocity of the ball of mass 2m after the collision?

Answer 1

Answer:

The velocity of mass 2m is  $v_B = 0.67 m/s$

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = $v_1$ =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = $v_A$

Let The finial speed  of the billiard ball  B = $v_B$

      According to the law of conservation of Energy

                 $\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

              Substituting values  

                $\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2$

Multiplying through by $\frac{1}{2}m$

                $1 =v_A^2 + 2 v_B ^2 ---(1)$

    According to the law of conservation of Momentum

            $mv_1 + 2m(0) = mv_A + 2m v_B$

    Substituting values

            $m(1) = mv_A + 2mv_B$

Multiplying through by $m$

           $1 = v_A + 2v_B ---(2)$

making $v_A$ subject of the equation 2

            $v_A = 1 - 2v_B$

Substituting this into equation 1

         $(1 -2v_B)^2 + 2v_B^2 = 1$

         $1 - 4v_B + 4v_B^2 + 2v_B^2 =1$

          $6v_B^2 -4v_B +1 =1$

          $6v_B^2 -4v_B =0$

Multiplying through by $\frac{1}{v_B}$

          $6v_B -4 = 0$

            $v_B = \frac{4}{6}$

            $v_B = 0.67 m/s$