A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.30 m/s and rebounds with a speed of 1.70 m/s, determine the following.
(a)
magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.)
kg · m/s
(b)
change in the magnitude of the ball's momentum in kg · m/s (Let negative values indicate a decrease in magnitude.)
kg · m/s
(c)
Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?

the magnitude of the change in the ball's momentum
They both are equally related to the net force acting on the ball.
Neither are related to the net force acting on the ball.
the change in the magnitude of the ball's momentum

We are asked to analyze the change in momentum of a ball that bounces off the floor, with the following known information:

- Mass of the ball: \( m = 275 \, \text{g} = 0.275 \, \text{kg} \)
- Speed before hitting the floor: \( v_{\text{initial}} = 2.30 \, \text{m/s} \)
- Speed after rebounding: \( v_{\text{final}} = 1.70 \, \text{m/s} \)
- The upward direction is defined as positive.

We will compute:

**(a) The magnitude of the change in the ball's momentum (Δp):**

Momentum is given by \( p = m \times v \). The change in momentum \( \Delta p \) is the difference between the final and initial momentum, taking into account direction (since the velocities are in opposite directions):

\[
\Delta p = m \times v_{\text{final}} - m \times v_{\text{initial}}
\]

Given that the final velocity is upward (positive) and the initial velocity is downward (negative, since it's the opposite direction):

\[
\Delta p = 0.275 \times (1.70) - 0.275 \times (-2.30)
\]
\[
\Delta p = 0.275 \times (1.70 + 2.30)
\]
\[
\Delta p = 0.275 \times 4.00 = 1.10 \, \text{kg·m/s}
\]

So, the **magnitude of the change in momentum** is:

\[
\boxed{1.10 \, \text{kg·m/s}}
\]

**(b) The change in the magnitude of the ball's momentum:**

Here, we only consider the magnitudes of the velocities (ignoring direction). The change in magnitude is the absolute difference between the final and initial momentum magnitudes:

\[
\Delta p_{\text{magnitude}} = m \times \left| v_{\text{final}} - v_{\text{initial}} \right|
\]

\[
\Delta p_{\text{magnitude}} = 0.275 \times \left| 1.70 - 2.30 \right|
\]
\[
\Delta p_{\text{magnitude}} = 0.275 \times 0.60 = 0.165 \, \text{kg·m/s}
\]

So, the **change in the magnitude of the ball's momentum** is:

\[
\boxed{0.165 \, \text{kg·m/s}}
\]

**(c) Which of the two quantities is more directly related to the net force acting on the ball?**

The net force acting on the ball during the collision is related to the **magnitude of the change in the ball's momentum** because force is defined by the impulse-momentum theorem:

\[
F \Delta t = \Delta p
\]

Thus, the force is directly proportional to the total change in momentum (not just the magnitude of the change in momentum).

Therefore, the correct answer is:

\[
\boxed{\text{the magnitude of the change in the ball's momentum}}
\]

### Do you need any clarifications or further details? Here are some related questions:

1. How does the direction of velocity affect momentum changes?
2. What is the impulse-momentum theorem and how does it apply here?
3. How would the result change if the ball was inelastic and didn't bounce back?
4. What is the significance of the sign of momentum in physics problems?
5. How does this problem relate to Newton's third law?

**Tip:** Always pay attention to the sign of velocities when calculating momentum changes, as direction plays a critical role in momentum conservation problems.

A ball with a mass of 275 g is dropped from rest, hits the floor and rebounds upward. If the ball hits the floor with a speed of 2.30 m/s and rebounds with a speed of 1.70 m/s, determine the following. (a) magnitude of the change in the ball's momentum in kg · m/s (Let up be in the positive direction.) kg · m/s (b) change in the magnitude of the ball's momentum in kg · m/s (Let negative values indicate a decrease in magnitude.) kg · m/s (c) Which of the two quantities calculated in parts (a) and (b) is more directly related to the net force acting on the ball during its collision with the floor?

Learn how to calculate the change in momentum of a 275 g ball that hits the floor with a speed of 2.30 m/s and rebounds with a speed of 1.70 m/s. This problem covers momentum, the impulse-momentum theorem, and Newton's laws. Suitable for high school physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation