A 0.1 kg steel ball is dropped straight down onto a hard, horizontal floor and bounces straight up. The ball's speed just before and just after impact with the floor is 10 m/s. Determine the magnitude of the impulse delivered to the floor by the steel ball.

Options

A. zero N.s
B. 1 N.s
C. 2 N.s
D. 10 N.s

Answer

To solve the problem of determining the magnitude of the impulse delivered to the floor by the steel ball, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Mass of the Ball:**
The mass \( m \) of the steel ball is given as \( 0.1 \, \text{kg} \).

2. **Determine the Velocity Before Impact:**
The speed of the ball just before impact with the floor is \( v_i = -10 \, \text{m/s} \) (negative because it is moving downward).

3. **Determine the Velocity After Impact:**
The speed of the ball just after impact with the floor is \( v_f = 10 \, \text{m/s} \) (positive because it is moving upward).

4. **Calculate the Initial Momentum:**
The initial momentum \( p_i \) of the ball just before impact can be calculated using the formula:
\[ p_i = m \cdot v_i = 0.1 \, \text{kg} \cdot (-10 \, \text{m/s}) = -1 \, \text{kg m/s} \]

5. **Calculate the Final Momentum:**
The final momentum \( p_f \) of the ball just after impact can be calculated as:
\[ p_f = m \cdot v_f = 0.1 \, \text{kg} \cdot 10 \, \text{m/s} = 1 \, \text{kg m/s} \]

6. **Calculate the Change in Momentum:**
The change in momentum \( \Delta p \) is given by:
\[ \Delta p = p_f - p_i = 1 \, \text{kg m/s} - (-1 \, \text{kg m/s}) = 1 \, \text{kg m/s} + 1 \, \text{kg m/s} = 2 \, \text{kg m/s} \]

7. **Determine the Impulse:**
The impulse \( I \) delivered to the floor by the steel ball is equal to the change in momentum:
\[ I = \Delta p = 2 \, \text{kg m/s} \]

### Final Answer:
The magnitude of the impulse delivered to the floor by the steel ball is \( 2 \, \text{N s} \) (Newton-seconds).

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