As shown in the figure, a bullet of mass m

Question

and speed v passes completely through a
pendulum bob of mass M. The bullet
emerges with a speed of v/2. The
pendulum bob is suspended by a stiff rod
of length ℓ and negligible mass.
2.1 Derive an expression for the minimum value of v such that the pendulum bob
will barely swing through a complete vertical circle. (6)
2.2 Assuming the same value for v, obtain an expression for the centripetal
acceleration of the bob as it passes through the angular position +/2 radians
beyond its starting point.

Answer 1

momentum released to bob: mv/2
at impact, momentum
mv/2=(M )V so initial V is
V=m/(M ) v/2

so to make it to the top...
1/2 (M)((m/M)^2 v^2/4)=Mg L

then for min speed, solve for v