A body is hanging from a rigid support. by an inextensible string of length `'1'` . It is struck inelastically by an identical body of mass in with horizontal velocity `v=sqrt(2gl)` the tension in the string increases just after the striking by

Options

A. `mg`
B. `3mg`
C. `2mg`
D. none of these

Explanation

To find the increase in tension in the string just after the inelastic collision, we can follow these steps:

### Step 1: Understand the initial conditions
Initially, we have one body of mass \( m \) hanging from a rigid support by an inextensible string of length \( L \). This body is at rest, so the tension in the string \( T_0 \) is equal to the weight of the body:
\[ T_0 = mg \]

### Step 2: Analyze the collision
An identical body of mass \( m \) strikes the hanging body with a horizontal velocity \( v = \sqrt{2gl} \). Since the collision is inelastic, the two bodies will stick together after the collision.

### Step 3: Apply conservation of momentum
Before the collision, the momentum of the system is:
\[ p_{\text{initial}} = mv \]
After the collision, the two bodies move together with a combined mass of \( 2m \). Let \( v' \) be their final velocity. By conservation of momentum:
\[ mv = (2m)v' \]
Solving for \( v' \):
\[ v' = \frac{v}{2} = \frac{\sqrt{2gl}}{2} = \frac{\sqrt{2gl}}{2} = \sqrt{\frac{gl}{2}} \]

### Step 4: Determine the forces acting on the combined mass
After the collision, the combined mass \( 2m \) will experience two forces:
1. The gravitational force acting downward: \( F_g = 2mg \)
2. The tension \( T \) in the string acting upward.

### Step 5: Apply centripetal force requirement
The tension in the string must provide the centripetal force required for the circular motion of the combined mass. The net force acting towards the center of the circular path is given by:
\[ T - 2mg = \frac{(2m)(v')^2}{L} \]
Substituting \( v' = \sqrt{\frac{gl}{2}} \):
\[ T - 2mg = \frac{(2m)(\frac{gl}{2})}{L} \]
This simplifies to:
\[ T - 2mg = \frac{mgl}{L} \]

### Step 6: Solve for the tension \( T \)
Rearranging the equation gives:
\[ T = 2mg + \frac{mgl}{L} \]
Since \( L \) is the length of the string, we can replace \( \frac{gl}{L} \) with \( g \) (as \( l \) is equal to \( L \)):
\[ T = 2mg + mg = 3mg \]

### Step 7: Calculate the increase in tension
The initial tension was \( T_0 = mg \). The increase in tension \( \Delta T \) is:
\[ \Delta T = T - T_0 = 3mg - mg = 2mg \]

### Final Answer
The increase in tension in the string just after the collision is:
\[ \Delta T = 2mg \]

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