A ball of mass 1.00 kg is attached to a loose string fixed to a ceiling. The ball is released from rest and falls 2.00 m, where the string suddenly stops it. Find the impulse on it from the string.

Answer

To solve the problem of finding the impulse on a ball of mass 1.00 kg that falls 2.00 m before being stopped by a string, we can follow these steps:

### Step 1: Understand the situation
The ball is released from rest and falls under the influence of gravity. When it falls 2.00 m, the string becomes taut and stops the ball.

### Step 2: Determine the initial and final velocities
- **Initial velocity (u)**: Since the ball is released from rest, \( u = 0 \, \text{m/s} \).
- **Final velocity (v)**: When the string stops the ball, the final velocity is \( v = 0 \, \text{m/s} \) at the moment of stopping.

### Step 3: Calculate the velocity just before the string stops the ball
To find the velocity just before the string stops the ball, we can use the principle of conservation of energy. The potential energy lost by the ball is converted into kinetic energy.

1. **Potential energy (PE) lost**:
\[ \Delta PE = mgh \]
where:
- \( m = 1.00 \, \text{kg} \) (mass of the ball)
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( h = 2.00 \, \text{m} \) (height fallen)

Thus,
\[ \Delta PE = 1.00 \times 10 \times 2.00 = 20 \, \text{J} \]

2. **Kinetic energy (KE) just before the string stops the ball**:
\[ KE = \frac{1}{2} mv^2 \]
Setting the change in potential energy equal to the kinetic energy:
\[ 20 = \frac{1}{2} \times 1.00 \times v^2 \]
Solving for \( v^2 \):
\[ 20 = 0.5 \times v^2 \implies v^2 = 40 \implies v = \sqrt{40} \approx 6.32 \, \text{m/s} \]

### Step 4: Calculate the impulse
Impulse (\( I \)) is defined as the change in momentum, which can also be expressed as:
\[ I = m \Delta v \]
where \( \Delta v = v - u \).

Since the ball comes to a stop, we have:
- \( u = \sqrt{40} \, \text{m/s} \)
- \( v = 0 \, \text{m/s} \)

Thus,
\[ \Delta v = 0 - \sqrt{40} = -\sqrt{40} \]
Now substituting into the impulse formula:
\[ I = m \Delta v = 1.00 \times (-\sqrt{40}) = -\sqrt{40} \, \text{kg m/s} \]

The negative sign indicates that the impulse is in the opposite direction of the ball's initial motion.

### Step 5: Calculate the numerical value of the impulse
\[ I \approx -6.32 \, \text{kg m/s} \]

### Final Answer
The impulse on the ball from the string is approximately \( 6.32 \, \text{kg m/s} \) in the upward direction.

---