Two particles A and B of equal masses lie close together on a horizontal table and are connected by a light inextensible string of length `l`. A is projected vertically upwards with a velocity `sqrt(10gl)`. Find the velocity with which it reaches the table again.

Explanation

To solve the problem, let's break it down step by step:

### Step 1: Understand the System
We have two particles A and B of equal mass connected by a light inextensible string of length \( l \). Particle A is projected vertically upwards with an initial velocity of \( \sqrt{10gl} \).

### Step 2: Determine the Height at Which the String Becomes Tight
As particle A moves upward, it will rise until the string connecting it to particle B becomes taut. This occurs when particle A has moved a distance \( l \) upwards.

### Step 3: Calculate the Velocity of A When the String Becomes Tight
Using the kinematic equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = \sqrt{10gl} \) (initial velocity)
- \( a = -g \) (acceleration due to gravity, negative because it is acting downwards)
- \( s = l \) (distance moved upwards)

Substituting these values into the equation:
\[ v_1^2 = (\sqrt{10gl})^2 - 2g(l) \]
\[ v_1^2 = 10gl - 2gl = 8gl \]
Thus, the velocity \( v_1 \) of particle A when the string becomes taut is:
\[ v_1 = \sqrt{8gl} \]

### Step 4: Apply Conservation of Momentum
When the string becomes taut, both particles A and B will move together. Let \( v_2 \) be the common velocity of both particles just after the string becomes tight. By the conservation of momentum:
\[ m_A \cdot v_1 = (m_A + m_B) \cdot v_2 \]
Since the masses are equal, we can simplify:
\[ m \cdot \sqrt{8gl} = 2m \cdot v_2 \]
\[ v_2 = \frac{\sqrt{8gl}}{2} = \frac{\sqrt{8}}{2} \sqrt{gl} = \sqrt{2gl} \]

### Step 5: Determine the Final Velocity When A Reaches the Table
As both particles A and B fall back to the table, they will have the same final speed when they reach the table. We can use the kinematic equation again to find this final speed.

Using the equation:
\[ v^2 = u^2 + 2as \]
where:
- \( u = v_2 = \sqrt{2gl} \) (initial velocity just after the string becomes taut)
- \( a = g \) (acceleration due to gravity)
- \( s = l \) (distance fallen)

Substituting these values:
\[ v^2 = (\sqrt{2gl})^2 + 2g(l) \]
\[ v^2 = 2gl + 2gl = 4gl \]
Thus, the final velocity \( v \) when particle A reaches the table is:
\[ v = \sqrt{4gl} = 2\sqrt{gl} \]

### Final Answer
The velocity with which particle A reaches the table again is \( 2\sqrt{gl} \).

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