A ball with a mass of # 6 kg# is rolling at #25 m/s# and elastically collides with a resting ball with a mass of # 2 kg#. What are the post-collision velocities of the balls?

Let "Sigma P_b" represent the total momentum prior to the collision.

To find the post-collision velocities of the balls, we can use the principle of conservation of momentum and the principle of conservation of kinetic energy for an elastic collision.

Let ( v_1 ) be the velocity of the 6 kg ball after the collision and ( v_2 ) be the velocity of the 2 kg ball after the collision.

According to the principle of conservation of momentum:

( m_1 \times v_{1i} + m_2 \times v_{2i} = m_1 \times v_1 + m_2 \times v_2 )

Substituting the given values:

( 6 \times 25 + 2 \times 0 = 6 \times v_1 + 2 \times v_2 )

( 150 = 6v_1 + 2v_2 )

And according to the principle of conservation of kinetic energy:

( \frac{1}{2}m_1 \times v_{1i}^2 + \frac{1}{2}m_2 \times v_{2i}^2 = \frac{1}{2}m_1 \times v_1^2 + \frac{1}{2}m_2 \times v_2^2 )

Substituting the given values:

( \frac{1}{2} \times 6 \times 25^2 + \frac{1}{2} \times 2 \times 0^2 = \frac{1}{2} \times 6 \times v_1^2 + \frac{1}{2} \times 2 \times v_2^2 )

( 562.5 = 3v_1^2 + v_2^2 )

We now have a system of two equations:

1. ( 150 = 6v_1 + 2v_2 )
2. ( 562.5 = 3v_1^2 + v_2^2 )

Solving this system of equations will give us the values of ( v_1 ) and ( v_2 ).