Two bodies of same mass tied with an inelastic string of length `l` lie together. One of them is projected vertically upwards with velocity `sqrt(6gl)`. Find the maximum height up to which the centre of mass system of the two masses rises.

Options

A. `(3l)/(4)`
B. `(l)/(2)`
C. `(3l)/(2)`
D. `l`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the Initial Conditions
We have two bodies of the same mass tied together with an inelastic string of length \( l \). One of the bodies is projected vertically upwards with an initial velocity of \( v_0 = \sqrt{6gl} \).

### Step 2: Determine the Velocity of the First Body When the String Becomes Taut
As the first body rises, it will continue to move upward until the string becomes taut. At this point, the second body is still at rest. We can use the kinematic equation to find the velocity of the first body when the string becomes taut.

Using the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( v \) = final velocity (when the string becomes taut)
- \( u = \sqrt{6gl} \) = initial velocity
- \( a = -g \) (acceleration due to gravity, acting downwards)
- \( s = l \) (the distance the first body rises until the string becomes taut)

Substituting the values:
\[ v^2 = (\sqrt{6gl})^2 - 2g(l) \]
\[ v^2 = 6gl - 2gl = 4gl \]
\[ v = \sqrt{4gl} = 2\sqrt{gl} \]

### Step 3: Apply Conservation of Momentum
When the string becomes taut, both bodies will start moving together with a common velocity \( v_2 \). By conservation of momentum:
\[ m v_1 = (m + m) v_2 \]
where \( v_1 = 2\sqrt{gl} \) (velocity of the first body just before the string becomes taut).

Substituting the values:
\[ m (2\sqrt{gl}) = 2m v_2 \]
\[ 2\sqrt{gl} = 2v_2 \]
\[ v_2 = \sqrt{gl} \]

### Step 4: Find the Maximum Height of the Center of Mass
Now that we have the common velocity \( v_2 = \sqrt{gl} \), we can find the maximum height the center of mass rises. The center of mass of the two bodies is initially at a height of \( \frac{l}{2} \) (since both bodies are at the same height).

Using the kinematic equation again to find the height \( h \) the center of mass rises from its initial position:
\[ v^2 = u^2 + 2as \]
where:
- \( v = 0 \) (final velocity at the maximum height)
- \( u = \sqrt{gl} \) (initial velocity of the center of mass)
- \( a = -g \) (acceleration due to gravity)
- \( s = h \) (height the center of mass rises)

Substituting the values:
\[ 0 = (\sqrt{gl})^2 - 2g h \]
\[ 0 = gl - 2gh \]
\[ 2gh = gl \]
\[ h = \frac{l}{2} \]

### Step 5: Calculate the Total Height of the Center of Mass
The total height \( H_{max} \) of the center of mass from the ground is:
\[ H_{max} = \frac{l}{2} + \frac{l}{2} = l \]

### Final Answer
The maximum height up to which the center of mass system of the two masses rises is \( \boxed{l} \).

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