A ball with a mass of # 6 kg# is rolling at #18 m/s# and elastically collides with a resting ball with a mass of #3 kg#. What are the post-collision velocities of the balls?

In a perfectly elastic collision, momentum and mechanical energy are conserved.

As two objects collide, all of the kinetic energy they possess is transferred into elastic potential energy in their molecular bonds, and then all of that energy which is stored in the bonds is transformed back into the post-collision kinetic energy of the objects. Although energy is transformed into potential energy during the collision, the mechanical energy before and after the collision is purely kinetic energy.

You can determine the final velocities in terms of the given quantities using these equations(1):

Because the second ball is initially at rest, these simplify to:

Therefore, the post-collision velocity of the first ball is:

and the post collision velocity of the second ball is:

To find the post-collision velocities of the balls, we can use the principle of conservation of momentum and kinetic energy. Let ( v_1 ) and ( v_2 ) be the velocities of the 6 kg and 3 kg balls respectively after the collision.

Using conservation of momentum:

[ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_1 + m_2 \cdot v_2 ]

[ 6 \cdot 18 + 0 = 6 \cdot v_1 + 3 \cdot v_2 ]

[ 108 = 6v_1 + 3v_2 ]

Also, since the collision is elastic, we can use conservation of kinetic energy:

[ \frac{1}{2} m_1 v_{1i}^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 ]

[ \frac{1}{2} \cdot 6 \cdot 18^2 = \frac{1}{2} \cdot 6 \cdot v_1^2 + \frac{1}{2} \cdot 3 \cdot v_2^2 ]

[ 324 = 3v_1^2 + \frac{3}{2} v_2^2 ]

Now, we have a system of two equations:

[ 108 = 6v_1 + 3v_2 ] [ 324 = 3v_1^2 + \frac{3}{2} v_2^2 ]

Solving this system of equations will give us the post-collision velocities ( v_1 ) and ( v_2 ).