After falling from rest through a height `h`, a body of mass `m` begins to raise a body of mass `M (M gt m)` connected to it through a pulley. Determnethe time it will take for the body of mass `M` to return to its original position

Options

A. `(2m)/(M_m)sqrt((2h)/g)`
B. `(2m)/(M-m)sqrt((2h)/g)`
C. `(2m)/(M-m)sqrt(h/g)`
D. `m/(M-m)sqrt((2h)/g)`

Explanation

To solve the problem, we will break it down into steps:

### Step 1: Determine the speed of mass `m` just before the string becomes taut.
When the mass `m` falls from rest through a height `h`, we can use the equation of motion to find its speed just before the string becomes taut. The equation is given by:

\[ v = \sqrt{2gh} \]

### Step 2: Apply conservation of momentum at the moment the string becomes taut.
At the moment the string becomes taut, the momentum of the system must be conserved. Let `V'` be the common speed of both masses `m` and `M` after the string is jerked. The conservation of momentum gives us:

\[ mv = (M + m)V' \]

Substituting for `v` from Step 1, we have:

\[ m\sqrt{2gh} = (M + m)V' \]

From this, we can solve for `V'`:

\[ V' = \frac{m\sqrt{2gh}}{M + m} \]

### Step 3: Determine the acceleration of the system.
The net force acting on the system when the mass `M` is rising and mass `m` is falling is given by:

\[ F = Mg - mg \]

Using Newton's second law:

\[ F = (M + m)a \]

Setting these equal gives:

\[ Mg - mg = (M + m)a \]

From this, we can solve for `a`:

\[ a = \frac{Mg - mg}{M + m} = \frac{(M - m)g}{M + m} \]

### Step 4: Use the kinematic equation to find the time `t` taken for mass `M` to return to its original position.
We use the kinematic equation:

\[ S = ut + \frac{1}{2}at^2 \]

Since mass `M` returns to its original position, the displacement `S` is equal to `0`, and the initial velocity `u` is equal to `V'`. Thus, we have:

\[ 0 = V't - \frac{1}{2}at^2 \]

Rearranging gives:

\[ \frac{1}{2}at^2 = V't \]

Dividing both sides by `t` (assuming `t ≠ 0`):

\[ \frac{1}{2}at = V' \]

From this, we can solve for `t`:

\[ t = \frac{2V'}{a} \]

### Step 5: Substitute for `V'` and `a` to find `t`.
Substituting the expressions for `V'` and `a` from Steps 2 and 3:

\[ t = \frac{2 \left(\frac{m\sqrt{2gh}}{M + m}\right)}{\frac{(M - m)g}{M + m}} \]

This simplifies to:

\[ t = \frac{2m\sqrt{2h}}{(M - m)\sqrt{g}} \]

### Final Result
Thus, the time it will take for the body of mass `M` to return to its original position is:

\[ t = \frac{2m\sqrt{2h}}{(M - m)\sqrt{g}} \]

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