To determine the impulse applied to the baseball, we use the formula for impulse:

$\text{Impulse} = \Delta p = m \cdot \Delta v$

Where:

• $m$ is the mass of the baseball (in kg),
• $\Delta v$ is the change in velocity of the baseball.

Given:

• $m = 0.5 \, \text{kg}$,
• Initial velocity ($v_i$) = $9 \, \text{m/s}$ (pitched towards Bob),
• Final velocity ($v_f$) = $-13 \, \text{m/s}$ (opposite direction after being hit).

Step 1: Calculate the change in velocity ($\Delta v$):

$\Delta v = v_f - v_i = -13 - 9 = -22 \, \text{m/s}$

Step 2: Calculate the impulse:

$\text{Impulse} = m \cdot \Delta v = 0.5 \cdot (-22) = -11 \, \text{kg·m/s}$

Step 3: Take the absolute value (magnitude of impulse):

Impulse is typically expressed as the magnitude of the change: $\text{Impulse} = 11.0 \, \text{kg·m/s}$

Final Answer:

The impulse applied to the baseball is 11.0 kg·m/s.

Would you like details or have any questions?

Related Questions:

1. How is impulse related to force and time during the interaction?
2. What is the significance of the negative sign in the velocity change?
3. If the contact time was 0.2 seconds, what was the average force applied to the ball?
4. How does impulse differ in elastic and inelastic collisions?
5. Can impulse be conserved in a system involving external forces?

Tip:

Always account for direction when calculating changes in velocity; it's crucial in momentum problems.