A man of mass `m` climbs to a rope ladder suspended below a balloon of mass `M`. The balloon is stationary with respect to the ground.
(a) If the man begins to climb the ladder at speed `v` (with respect to the ladder), in what direction and with what speed (with respect to the ground) will the balloon move?
(b) What is the state of the motion after the man stops climbing?

Explanation

To solve the problem step by step, we will analyze the situation using the principles of conservation of momentum.

### Step-by-Step Solution

**Step 1: Understand the System**
- We have a man of mass \( m \) climbing a rope ladder attached to a balloon of mass \( M \). Initially, both the man and the balloon are stationary with respect to the ground.

**Step 2: Define the Variables**
- Let \( v \) be the speed of the man climbing the ladder with respect to the ladder.
- Let \( V_B \) be the speed of the balloon with respect to the ground.
- When the man climbs up, the balloon will move down due to the conservation of momentum.

**Step 3: Apply Conservation of Momentum**
- Since there are no external forces acting on the system (the only force acting is gravity, which does not affect horizontal motion), we can apply the conservation of momentum.
- Initially, the total momentum of the system is zero because both the man and the balloon are at rest.

**Step 4: Set Up the Momentum Equation**
- When the man climbs up with speed \( v \) (upward), the balloon moves downward with speed \( V_B \).
- The momentum of the man moving upward is \( m \cdot v \) (upward).
- The momentum of the balloon moving downward is \( M \cdot (-V_B) \) (downward).
- According to conservation of momentum:
\[ m \cdot v + M \cdot (-V_B) = 0 \]
Rearranging gives:
\[ m \cdot v = M \cdot V_B \]

**Step 5: Solve for the Speed of the Balloon**
- From the equation \( m \cdot v = M \cdot V_B \), we can solve for \( V_B \):
\[ V_B = \frac{m \cdot v}{M} \]
- This indicates that the balloon moves downward with speed \( V_B \) while the man climbs upward with speed \( v \).

**Step 6: Analyze the State of Motion After Climbing**
- When the man stops climbing, he will no longer exert any force on the ladder.
- The system's total momentum must remain conserved. Since the man stops (his velocity becomes zero), the balloon will also come to rest.
- Thus, after the man stops climbing, the balloon will also stop moving.

### Final Answers
(a) The balloon moves downward with speed \( V_B = \frac{m \cdot v}{M} \) with respect to the ground.

(b) After the man stops climbing, the balloon comes to rest.