Two men of masses M and `M+m` start simultaneously from the ground and climb with uniform accelerations up from the free ends of a massless inextensible rope whick passes over a smooth pulley at a height h from the ground.
(a) Which man reaches the pulley first?
(b) If the mam who reaches first takes time t to reach the pulley, then find the distance of the second man from the pulley at this instant.

Explanation

To solve the problem step by step, we will break it down into parts (a) and (b) as specified in the question.

### Part (a): Which man reaches the pulley first?

1. **Identify the forces acting on each man**:
- For the man with mass \( M \):
- Weight: \( W_1 = Mg \)
- Tension in the rope: \( T \)
- For the man with mass \( M + m \):
- Weight: \( W_2 = (M + m)g \)
- Tension in the rope: \( T \)

2. **Write the equations of motion**:
- For the first man (mass \( M \)):
\[ T - Mg = Ma_1 \quad \text{(1)} \]
- For the second man (mass \( M + m \)):
\[ T - (M + m)g = (M + m)a_2 \quad \text{(2)} \]

3. **Rearranging the equations to find accelerations**:
- From equation (1):
\[ a_1 = \frac{T}{M} - g \quad \text{(3)} \]
- From equation (2):
\[ a_2 = \frac{T}{M + m} - g \quad \text{(4)} \]

4. **Compare the accelerations**:
- To determine which man reaches the pulley first, we need to compare \( a_1 \) and \( a_2 \):
\[ a_1 > a_2 \quad \text{if} \quad \frac{T}{M} > \frac{T}{M + m} \]
- Since \( M < M + m \), it follows that:
\[ \frac{1}{M} > \frac{1}{M + m} \implies a_1 > a_2 \]

5. **Conclusion for part (a)**:
- The man with mass \( M \) reaches the pulley first.

### Part (b): Distance of the second man from the pulley when the first man reaches the pulley

1. **Time taken by the first man to reach the pulley**:
- Let the time taken by the first man to reach the pulley be \( t \). The distance traveled by the first man is:
\[ h = \frac{1}{2} a_1 t^2 \quad \text{(5)} \]

2. **Substituting \( a_1 \) from equation (3)**:
- From equation (3):
\[ h = \frac{1}{2} \left(\frac{T}{M} - g\right) t^2 \]

3. **Distance traveled by the second man**:
- The distance traveled by the second man in time \( t \) is:
\[ s_2 = \frac{1}{2} a_2 t^2 \quad \text{(6)} \]

4. **Substituting \( a_2 \) from equation (4)**:
- From equation (4):
\[ s_2 = \frac{1}{2} \left(\frac{T}{M + m} - g\right) t^2 \]

5. **Distance from the pulley**:
- The distance of the second man from the pulley when the first man reaches it is:
\[ d = h - s_2 \quad \text{(7)} \]

6. **Substituting \( h \) and \( s_2 \)**:
- From equations (5) and (6):
\[ d = \left(\frac{1}{2} \left(\frac{T}{M} - g\right) t^2\right) - \left(\frac{1}{2} \left(\frac{T}{M + m} - g\right) t^2\right) \]

7. **Simplifying the expression**:
- Factor out \( \frac{1}{2} t^2 \):
\[ d = \frac{1}{2} t^2 \left(\left(\frac{T}{M} - g\right) - \left(\frac{T}{M + m} - g\right)\right) \]
- This simplifies to:
\[ d = \frac{1}{2} t^2 \left(\frac{T}{M} - \frac{T}{M + m}\right) \]

8. **Final expression for distance**:
- The distance of the second man from the pulley is:
\[ d = \frac{1}{2} t^2 \left(\frac{T(M + m) - TM}{M(M + m)}\right) = \frac{1}{2} t^2 \left(\frac{Tm}{M(M + m)}\right) \]