Answer
(a) h=1.87m
Work Step by Step
Mass of the two balls (m) = 5kg
Mass of the rod (M) = 8kg
Length of the rod (l) = 4m
Height from where the ball is dropped (H) = 12m
Let the height reached by the second ball be h.
Applying energy conservation,
mgH=\frac{1}{2}mv^{2}
v is the velocity of ball just before it strikes the rod
v=\sqrt (2gh)
Let \omega be the angular velocity of the system just after the ball strikes the rod
Conserving angular momentum
mv\frac{l}{2} = (\frac{Mx^{l}}{12} + 2\timesx^{\frac{l}{2}})\omega
Now substituting value of v from above eqn we get
\omega = 3.03 rad/sec
v = 6.06 m/sec
Now applying energy conservation
mgh = \frac{1}{2} mx^{2}
h = x^{2}/2g
h = 1.87m
