Question

A steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 5.0 m to one side of the point directly under the ball. When a button is pressed, the ball drops and, simultaneously, the cannon fires a plastic ball. The two balls collide 1.4 m above the floor.What was the launch speed of the plastic ball?

Answer 1

The launch speed of the plastic ball is approximately 8.85 m/s.

Step-by-step explanation:

To find the launch speed of the plastic ball, we can use the principle of conservation of energy. The potential energy of the steel ball before it is dropped is converted into kinetic energy when it reaches its lowest point. This kinetic energy is then transferred to the plastic ball during the collision. We can use the equation:

1/2 mv^2 = mgh

where m is the mass of the steel ball, v is the launch speed of the plastic ball, g is the acceleration due to gravity, and h is the height the steel ball falls. Rearranging the equation:

v = √(2gh)

Substituting the given values:

v = √(2 * 9.8 m/s^2 * 4 m)

= √78.4 m/s

≈ 8.85 m/s

Answer 2

Answer: 7.92 m/s

Step-by-step explanation: see attachment