A golf ball of mass m is hit by a golf club so that the ball leaves the tee with speed v. The club is in contact with the ball for time T. The magnitude of the average force on the club on the ball during the time T is :

Options

A. `mv T`
B. `mv//T`
C. `(1//2) mv^2 T`
D. `mv^2 (2T)`

Explanation

To find the magnitude of the average force exerted by the golf club on the golf ball during the time \( T \), we can use the impulse-momentum theorem. The average force can be calculated using the change in momentum of the ball over the time interval \( T \).

### Step-by-Step Solution:

1. **Identify the given values**:
- Mass of the golf ball: \( m \)
- Final velocity of the ball after being hit: \( v \)
- Initial velocity of the ball before being hit: \( u = 0 \) (the ball is at rest)
- Time of contact between the club and the ball: \( T \)

2. **Calculate the change in momentum (\( \Delta p \))**:
- The momentum before the impact:
\[ p_i = m \cdot u = m \cdot 0 = 0 \]
- The momentum after the impact:
\[ p_f = m \cdot v \]
- Therefore, the change in momentum is:
\[ \Delta p = p_f - p_i = m \cdot v - 0 = m \cdot v \]

3. **Use the impulse-momentum theorem**:
- The average force (\( F_{avg} \)) can be calculated using the formula:
\[ F_{avg} = \frac{\Delta p}{\Delta t} \]
- Substituting the values we found:
\[ F_{avg} = \frac{m \cdot v}{T} \]

4. **Final expression for the average force**:
- Thus, the magnitude of the average force exerted by the club on the ball during the time \( T \) is:
\[ F_{avg} = \frac{m \cdot v}{T} \]

### Final Answer:
\[ F_{avg} = \frac{m \cdot v}{T} \]