A particle traveling horizontally with speed u collides and sticks with a particle of equal mass hanging at rest at the end of a light inextensible string of length 21. If the string rotates through an angle of 60° before its velocity becomes zero, then u is

Options

A. `sqrt(4gl)`
B. `sqrt(8gl)`
C. `sqrt(2gl)`
D. `sqrt(gl)`

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Determine the vertical distance moved by the hanging mass.
The hanging mass moves through an angle of 60° when the string rotates. The vertical distance \( r \) moved by the mass can be calculated using the formula for vertical displacement in terms of the length of the string \( l \):

\[ r = 2l - 2l \cos(60^\circ) \]

Since \( \cos(60^\circ) = \frac{1}{2} \):

\[ r = 2l - 2l \cdot \frac{1}{2} = 2l - l = l \]

### Step 2: Apply conservation of momentum.
Before the collision, the horizontal particle has momentum \( mu \) (where \( m \) is the mass and \( u \) is the speed). After the collision, both masses stick together and move with a common velocity \( v \). By conservation of momentum:

\[ mu = (m + m)v \]

This simplifies to:

\[ mu = 2mv \]

Dividing both sides by \( m \):

\[ u = 2v \quad \Rightarrow \quad v = \frac{u}{2} \]

### Step 3: Apply conservation of energy.
The initial kinetic energy of the system is due to the moving particle, and the potential energy of the hanging mass is zero since it is at the lowest point. After the collision, the system has both kinetic and potential energy. The conservation of energy can be expressed as:

\[ \text{Initial Kinetic Energy} + \text{Initial Potential Energy} = \text{Final Kinetic Energy} + \text{Final Potential Energy} \]

Initially, the potential energy is zero, and the kinetic energy is:

\[ \text{Initial K.E.} = \frac{1}{2} m u^2 \]

At the highest point, the kinetic energy is:

\[ \text{Final K.E.} = \frac{1}{2} (2m) v^2 = mv^2 \]

The potential energy at the highest point is:

\[ \text{Final P.E.} = 2mg r = 2mg l \]

Setting the initial energy equal to the final energy:

\[ \frac{1}{2} m u^2 = mv^2 + 2mg l \]

### Step 4: Substitute \( v \) and simplify.
Substituting \( v = \frac{u}{2} \):

\[ \frac{1}{2} m u^2 = m \left(\frac{u}{2}\right)^2 + 2mg l \]

This simplifies to:

\[ \frac{1}{2} m u^2 = m \frac{u^2}{4} + 2mg l \]

Multiplying through by 4 to eliminate the fractions:

\[ 2mu^2 = mu^2 + 8mgl \]

### Step 5: Solve for \( u^2 \).
Rearranging gives:

\[ 2mu^2 - mu^2 = 8mgl \quad \Rightarrow \quad mu^2 = 8mgl \]

Dividing both sides by \( m \):

\[ u^2 = 8gl \]

### Step 6: Take the square root to find \( u \).
Taking the square root:

\[ u = \sqrt{8gl} \]

This is the final expression for \( u \).

### Final Answer:
\[ u = \sqrt{8gl} \]