1. Given data

• Length of pendulum: $L = 2\,\text{m}$
• Mass of wooden bob: $M = 50\,\text{g} = 0.05\,\text{kg}$
• Mass of bullet: $m = 75\,\text{g} = 0.075\,\text{kg}$
• Initial speed of bullet: $v$
• Final speed of bullet after emerging: $\dfrac{v}{3}$
• $g = 10\,\text{m/s}^2$

The bob is initially at rest.

2. Speed needed by bob to just complete a vertical circle

For a pendulum/string to just complete a vertical circle, the minimum speed at the lowest point is

$u = \sqrt{5gL}$

Substituting $g=10$ and $L=2$:

$u = \sqrt{5 \cdot 10 \cdot 2} = \sqrt{100} = 10\,\text{m/s}$

So, immediately after collision, the bob must acquire speed

$U = 10\,\text{m/s}$

3. Apply conservation of linear momentum during collision

During the short collision, external impulse in horizontal direction is negligible, so linear momentum along bullet direction is conserved.

Initial momentum:

$p_i = mv$

Final momentum:

• Bullet momentum = $m\left(\dfrac{v}{3}\right)$
• Bob momentum = $MU$

Thus,

$mv = m\frac{v}{3} + MU$

Substitute values:

$0.075v = 0.075\cdot \frac{v}{3} + 0.05\cdot 10$

$0.075v = 0.025v + 0.5$

$0.05v = 0.5$

$v = 10\,\text{m/s}$

4. Final answer

$\boxed{10}$

5. Comparison with stored correct answer

Stored correct answer = $10$

This matches the derived answer.