Question

A 5 kg bowling ball traveling at 2.6 m/s collides with a stationary 0.5 kg beach ball in an elastic collision. The bowling ball leaves the collision with a velocity of 1.1 m/s traveling in the same direction as the beach ball. Calculate the speed of the beach ball.

Answer 1

The speed of the beach ball after the elastic collision with the bowling ball is 15 m/s. This was calculated using the conservation of momentum, which is a concept in physics concerned with the properties of moving objects.

Step-by-step explanation:

To calculate the speed of the beach ball after the collision, we use the conservation of momentum and the given information that the collision is elastic. The conservation of momentum for two objects in a one-dimensional collision is given by:

m1v1i + m2v2i = m1v1f + m2v2f

where:

• m1 and m2 are the masses of the bowling ball and beach ball, respectively,
• v1i and v2i are the initial velocities of the bowling ball and beach ball, respectively,
• v1f and v2f are the final velocities of the bowling ball and beach ball, respectively.

Since the beach ball was initially stationary, its initial velocity v2i is 0. Substituting the given values:

(5 kg)(2.6 m/s) + (0.5 kg)(0) = (5 kg)(1.1 m/s) + (0.5 kg)v2f

Solving for v2f the beach ball's final velocity:

v2f = ((5 kg)(2.6 m/s) - (5 kg)(1.1 m/s)) / (0.5 kg) = 15 m/s

Therefore, the final speed of the beach ball is 15 m/s.