What is the final direction of the billiard ball after collision?

Analysis of the Billiard Ball Collision

When two billiard balls collide, their momenta are conserved. In this case, the momentum in the x-direction before the collision is equal to the momentum in the x-direction after the collision. The same principle applies to the momentum in the y-direction.

Before the collision:

Momentum of ball 1 in x-direction: \(m_1 \times v_{x1}\) = \(m_1 \times 33.4\) = \(p_1\)

Momentum of ball 2 in y-direction: \(m_2 \times v_{y2}\) = \(m_2 \times 38.9\) = \(p_2\)

After the collision:

Final Momentum of ball 1 in x-direction: \(m_1 \times v_{x1f}\) = \(m_1 \times 50.6\) = \(p_{1f}\)

Final Momentum of ball 2 in y-direction: 0, as the second ball stops

The total momentum before the collision is equal to the total momentum after the collision. Therefore, we can write the equation:

\(p_1 + p_2 = p_{1f} + p_{2f}\)

By solving the equation using the given velocities, we find that the final velocity of the first ball in the x-direction is 33.2 cm/s. This means that the first ball moves at an angle of approximately 53 degrees with the x-axis.

So, the final direction of the first billiard ball after the collision is at an angle of approximately 53 degrees with the x-axis.