Question

A 0.226 kg baseball is thrown with a speed of 19.5 m/s. It is hit straight back at the pitcher with a final speed of 21.5 m/s. What is the magnitude of the impulse delivered to the ball?

Answer 1

The magnitude of the impulse delivered to a 0.226 kg baseball, which was thrown at 19.5 m/s and hit back at 21.5 m/s, is 9.266 kg·m/s.

Step-by-step explanation:

The question involves calculating the magnitude of the impulse delivered to a baseball. Impulse is defined as the change in momentum of an object when it is subject to a force. The formula for impulse (J) is given by J = Δp = m(v_f - v_i), where m is the mass of the object, v_f is the final velocity, and v_i is the initial velocity.

In this scenario, a 0.226 kg baseball is initially thrown at a speed of 19.5 m/s and then hit back at a final speed of 21.5 m/s. Since the ball is hit straight back, the change in direction indicates a change in the sign of the velocity. Hence, the impulse can be calculated as:

J = 0.226 kg * (21.5 m/s - (-19.5 m/s)) = 0.226 kg * (21.5 + 19.5) m/s = 0.226 kg * 41 m/s.

So the magnitude of the impulse delivered to the ball is:

J = 0.226 kg * 41 m/s = 9.266 kg·m/s.