A bullet of mass m passes through a pendulum bob of mass M with velocity v and comes out of it with a velocity `(v)/(2)`. The minimum value of v so that the bob completes one revolution in the vertical circle is (l = length of pendulum)

Options

A. `(M)/(m)sqrt(2gl)`
B. `(2M)/(m)sqrt(2lg)`
C. `(2M)/(m)sqrt(5lg)`
D. `(M)/(m)sqrt(5lg)`

Answer

To solve the problem, we need to analyze the situation step-by-step:

### Step 1: Understand the situation
A bullet of mass \( m \) passes through a pendulum bob of mass \( M \) with an initial velocity \( v \) and exits with a velocity of \( \frac{v}{2} \). We need to find the minimum value of \( v \) such that the pendulum bob can complete a full revolution in a vertical circle of radius \( l \).

### Step 2: Apply conservation of momentum
When the bullet passes through the bob, we can apply the principle of conservation of momentum. The momentum before the bullet enters the bob is equal to the momentum after it exits.

- Initial momentum: \( mv \) (bullet) + \( 0 \) (bob) = \( mv \)
- Final momentum: \( m \cdot \frac{v}{2} \) (bullet) + \( M \cdot u \) (bob)

Setting these equal gives us:
\[ mv = m \cdot \frac{v}{2} + Mu \]

### Step 3: Solve for \( u \)
Rearranging the equation:
\[ mv - m \cdot \frac{v}{2} = Mu \]
\[ \frac{mv}{2} = Mu \]
\[ u = \frac{mv}{2M} \]

### Step 4: Determine the minimum velocity \( u \) for the bob to complete a vertical circle
For the bob to complete a vertical circle, at the top of the circle, the tension in the string can be zero. The only force providing the centripetal force is the weight of the bob.

Using the centripetal force equation at the top of the circle:
\[ Mg = \frac{M v'^2}{l} \]
Where \( v' \) is the velocity of the bob at the top of the circle.

From this, we can express \( v' \):
\[ v'^2 = gl \]
\[ v' = \sqrt{gl} \]

### Step 5: Apply conservation of energy
Using conservation of energy from the lowest point to the highest point:
\[ \text{Initial kinetic energy} + \text{Initial potential energy} = \text{Final kinetic energy} + \text{Final potential energy} \]
\[ \frac{1}{2}Mu^2 + 0 = \frac{1}{2}Mv'^2 + Mg(2l) \]

Substituting \( v' = \sqrt{gl} \):
\[ \frac{1}{2}Mu^2 = \frac{1}{2}M(gl) + Mg(2l) \]
\[ \frac{1}{2}Mu^2 = \frac{1}{2}Mgl + 2Mgl \]
\[ \frac{1}{2}Mu^2 = \frac{5}{2}Mgl \]
\[ u^2 = 5gl \]
\[ u = \sqrt{5gl} \]

### Step 6: Substitute \( u \) back to find \( v \)
Now substituting \( u \) back into the equation for \( v \):
\[ u = \frac{mv}{2M} \]
\[ \sqrt{5gl} = \frac{mv}{2M} \]
\[ v = \frac{2M}{m} \sqrt{5gl} \]

### Final Result
Thus, the minimum value of \( v \) required for the bob to complete one full revolution in the vertical circle is:
\[ v = \frac{2M}{m} \sqrt{5gl} \]