1. Condition for the pendulum bob to complete a vertical circle

For a bob attached to a string to just complete a vertical circle, the minimum speed at the lowest point must be such that at the topmost point the tension is zero:

$v_{\text{top}}^2 = gL$

Using energy conservation from bottom to top:

$\frac12 m u^2 = \frac12 m v_{\text{top}}^2 + mg(2L)$

Substitute $v_{\text{top}}^2 = gL$:

$\frac12 m u^2 = \frac12 m(gL) + 2mgL$

$u^2 = 5gL$

So the minimum speed needed by the bob immediately after collision is

$u_{\min} = \sqrt{5gL}$

Given $g=10\,\text{m/s}^2$ and $L=0.5\,\text{m}$,

$u_{\min} = \sqrt{5 \cdot 10 \cdot 0.5} = \sqrt{25} = 5\,\text{m/s}$

2. Apply conservation of linear momentum during collision

Let

• mass of bullet $m = 10\,\text{g} = 0.01\,\text{kg}$
• mass of bob $M = 1\,\text{kg}$
• initial speed of bullet $= v$
• bob initially at rest
• bullet recoils with speed $100\,\text{m/s}$ in the opposite direction

Take the initial direction of bullet as positive.

Then after collision, bullet velocity is $-100\,\text{m/s}$. Let bob speed just after collision be $u$.

By conservation of momentum:

$mv = Mu + m(-100)$

$0.01v = 1\cdot u - 0.01\cdot 100$

$0.01v = u - 1$

$u = 0.01v + 1$

For minimum $v$, take minimum required bob speed $u=5$ m/s:

$5 = 0.01v + 1$

$0.01v = 4$

$v = 400\,\text{m/s}$

3. Final answer

$\boxed{400}$

The derived answer matches the stored correct answer.