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. given the magnitude of the change in the ball's momentum in kg · m/s is 1.10 kg m/s (Let up be in the positive direction.)

(b)
change in the magnitude of the ball's momentum in kg · m/s (Let negative values indicate a decrease in magnitude.)

To solve for the change in the magnitude of the ball's momentum, we need to recall that momentum is the product of mass and velocity.

The ball's momentum before and after the collision with the floor can be calculated using the formula:

\[
p = mv
\]

Where:
- \( p \) is momentum,
- \( m \) is mass, and
- \( v \) is velocity.

We are given:
- Mass of the ball: \( 275 \, \text{g} = 0.275 \, \text{kg} \),
- Speed just before hitting the floor: \( 2.30 \, \text{m/s} \) (downward, which is negative since up is positive),
- Speed just after rebounding: \( 1.70 \, \text{m/s} \) (upward, which is positive),
- The magnitude of the change in momentum: \( 1.10 \, \text{kg} \cdot \text{m/s} \).

### Step 1: Calculate the momentum before and after the impact
1. **Momentum before hitting the floor (downward):**
   \[
   p_{\text{before}} = m \cdot v_{\text{before}} = 0.275 \, \text{kg} \cdot (-2.30 \, \text{m/s}) = -0.6325 \, \text{kg} \cdot \text{m/s}
   \]

2. **Momentum after hitting the floor (upward):**
   \[
   p_{\text{after}} = m \cdot v_{\text{after}} = 0.275 \, \text{kg} \cdot 1.70 \, \text{m/s} = 0.4675 \, \text{kg} \cdot \text{m/s}
   \]

### Step 2: Calculate the change in momentum
To find the total change in momentum (\( \Delta p \)), we subtract the initial momentum from the final momentum:

\[
\Delta p = p_{\text{after}} - p_{\text{before}} = 0.4675 \, \text{kg} \cdot \text{m/s} - (-0.6325 \, \text{kg} \cdot \text{m/s}) = 0.4675 + 0.6325 = 1.10 \, \text{kg} \cdot \text{m/s}
\]

This is the given change in momentum's **magnitude**, which confirms that the information provided is consistent.

### Step 3: Answer for the magnitude change

For part (b), the **magnitude change** refers to how much the ball’s momentum has decreased. Since the ball's momentum goes from a higher negative value to a smaller positive value, we see a decrease overall in momentum.

Thus, the change in magnitude of the ball's momentum would simply be the negative of the given magnitude:

\[
\boxed{-1.10 \, \text{kg} \cdot \text{m/s}}
\]

This negative sign indicates a decrease in the magnitude of the momentum.

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**Relative Questions**:
1. How does the concept of momentum apply to collisions in general?
2. What is the relationship between impulse and change in momentum?
3. How would the result change if the mass of the ball was different?
4. Can the speed of the ball after the rebound ever exceed its speed before the impact?
5. What external forces influence the change in momentum during the ball's collision with the floor?

**Tip**: Always consider the direction when dealing with momentum since it is a vector quantity! Positive and negative signs are crucial in understanding the behavior of the system.

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. given the magnitude of the change in the ball's momentum in kg · m/s is 1.10 kg m/s (Let up be in the positive direction.) (b) change in the magnitude of the ball's momentum in kg · m/s (Let negative values indicate a decrease in magnitude.)

Determine the change in the magnitude of a ball's momentum when it rebounds off the floor after being dropped from rest. The problem involves calculating momentum using the mass and velocity of the ball before and after impact, applying the momentum formula and understanding vector directions. Suitable for high school physics students.

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