A pendulum of length $$2$$ m consists of a wooden bob of mass $$50$$ g. A bullet of mass $$75$$ g is fired towards the stationary bob with a speed $$v$$. The bullet emerges out of the bob with a speed $$\frac{v}{3}$$ and the bob just completes the vertical circle. The value of $$v$$ is ______ m s$$^{-1}$$
(if $$g = 10$$ m s$$^{-2}$$).

A bullet of mass 75 g is fired at a stationary wooden bob of mass 50 g on a pendulum of length 2 m. The bullet emerges with speed v/3 and the bob just completes the vertical circle.

First, for the bob to just complete the vertical circle, the minimum speed at the bottom must satisfy the energy condition. At the top of the circle, the minimum speed requires:

$$mg = \frac{mv_{top}^2}{L} \implies v_{top}^2 = gL$$

Using energy conservation from bottom to top (height = 2L):

$$\frac{1}{2}mv_{bob}^2 = \frac{1}{2}mv_{top}^2 + mg(2L)$$

$$v_{bob}^2 = gL + 4gL = 5gL$$

Substituting values:

$$v_{bob}^2 = 5 \times 10 \times 2 = 100$$

$$v_{bob} = 10 \text{ m/s}$$

Next, applying conservation of momentum for the bullet-bob collision gives:

$$m_{bullet} \cdot v = m_{bullet} \cdot \frac{v}{3} + m_{bob} \cdot v_{bob}$$

Substituting the masses and speeds,

$$75 \times 10^{-3} \cdot v - 75 \times 10^{-3} \cdot \frac{v}{3} = 50 \times 10^{-3} \times 10$$

$$75 \times 10^{-3} \cdot \frac{2v}{3} = 0.5$$

$$50 \times 10^{-3} \cdot v = 0.5$$

$$v = \frac{0.5}{0.05} = 10 \text{ m/s}$$

Therefore, the value of v is 10 m/s.