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 $\frac{1}{v_B}$

          $6v_B -4 = 0$

            $v_B = \frac{4}{6}$

            $v_B = 0.67 m/s$

Answer 2

The velocity of the ball of mass 2m after the collision is 0.67 m/s assuming that the collision is an elastic collision.

Elastic collision:

The masses of the two billiard balls are m and 2m respectively.

The initial velocity of the ball of mass m is v₁ = 1 m/s.

The initial velocity of the ball of mass 2m is 0

let the velocity of the second ball after the collision be V₂

Let us assume that the collision is elastic, that is the balls collide with each other and move separately and there is no loss in kinetic energy of the system.

For an elastic collision of two masses m₁ and m₂, such that initially, the mass m₁ is moving with a velocity u and mass m₂ is at rest, the final velocity of mass m₂ after the collision is given by:

V = 2m₁u / (m₁ + m₂)

according to the question:

V = 2m×1 / (m + 2m)

V = 0.67 m/s

Learn more about elastic collision:

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