A 0.70 kg ball moving horizontally at 6.0 m/s strikes a vertical wall and rebounds with speed 3.5 m/s. What is the magnitude of the change in its linear momentum?

Answer

To find the magnitude of the change in linear momentum of the ball, we will follow these steps:

### Step 1: Calculate the Initial Momentum
The initial momentum (\(P_i\)) of the ball can be calculated using the formula:
\[ P_i = m \cdot v_i \]
where:
- \(m = 0.70 \, \text{kg}\) (mass of the ball)
- \(v_i = 6.0 \, \text{m/s}\) (initial velocity)

Substituting the values:
\[ P_i = 0.70 \, \text{kg} \cdot 6.0 \, \text{m/s} = 4.2 \, \text{kg m/s} \]

### Step 2: Calculate the Final Momentum
The final momentum (\(P_f\)) after the ball rebounds can be calculated similarly:
\[ P_f = m \cdot v_f \]
where:
- \(v_f = -3.5 \, \text{m/s}\) (final velocity; negative because it is in the opposite direction)

Substituting the values:
\[ P_f = 0.70 \, \text{kg} \cdot (-3.5 \, \text{m/s}) = -2.45 \, \text{kg m/s} \]

### Step 3: Calculate the Change in Momentum
The change in momentum (\(\Delta P\)) is given by:
\[ \Delta P = P_f - P_i \]

Substituting the values:
\[ \Delta P = -2.45 \, \text{kg m/s} - 4.2 \, \text{kg m/s} = -6.65 \, \text{kg m/s} \]

### Step 4: Calculate the Magnitude of the Change in Momentum
The magnitude of the change in momentum is the absolute value of \(\Delta P\):
\[ |\Delta P| = |-6.65 \, \text{kg m/s}| = 6.65 \, \text{kg m/s} \]

### Final Answer
The magnitude of the change in linear momentum is:
\[ \boxed{6.65 \, \text{kg m/s}} \]

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