This problem involves calculating the total impulse imparted to the ball when it strikes the floor. Impulse is related to the change in momentum of the object. Let's go through the solution step-by-step.

Given:

• Mass of the ball, $m = 12 \, \text{g} = 0.012 \, \text{kg}$
• Initial drop height, $h_1 = 1.5 \, \text{m}$
• Rebound height, $h_2 = 0.75 \, \text{m}$

Step 1: Find the velocities before and after impact.

The velocities before and after impact can be found using the kinematic energy equation, derived from the conservation of mechanical energy (ignoring air resistance):

Velocity just before hitting the floor (using $h_1$):

$v_1 = \sqrt{2gh_1}$ Where $g = 9.81 \, \text{m/s}^2$ (acceleration due to gravity).

Velocity just after bouncing back (using $h_2$):

$v_2 = \sqrt{2gh_2}$

Step 2: Calculate the impulse.

Impulse $J$ is the change in momentum. Since the ball changes direction, the velocities will have opposite signs:

$J = m(v_2 - (-v_1)) = m(v_1 + v_2)$

Now, let's compute the values:It seems like I can't do more advanced data analysis right now. Please try again later. However, if you still have questions about the problem or need a different approach explained, feel free to ask!

Would you like a detailed explanation of the process, or help with other related questions?

Here are 5 related questions that might interest you:

1. How is momentum calculated in other types of collisions?
2. What role does air resistance play if we don't ignore it?
3. How does impulse relate to force over time in physics?
4. What are other examples of conservation of energy in real-world situations?
5. How would the impulse change if the ball had rebounded to its original height?

Tip: When working with impulse, always remember that it is equivalent to the change in momentum, not just the force applied.