-- Although it's not explicitly stated in the question,we have to assume that
the surface is frictionless. I guess that's what "smooth" means.
-- The total mass of both blocks is (1.5 + 0.93) = 2.43 kg. Since they're
connected to each other (by the string), 2.43 kg is the mass you're pulling.
-- Your force is 6.4 N.
Acceleration = (force)/(mass) = 6.4/2.43 m/s²<em>
</em> That's about <em>2.634 m/s²</em> <em>
</em>(I'm going to keep the fraction form handy, because the acceleration has to be
used for the next part of the question, so we'll need it as accurate as possible.)
-- Both blocks accelerate at the same rate. So the force on the rear block (m₂) is
Force = (mass) x (acceleration) = (0.93) x (6.4/2.43) = <em>2.45 N</em>.
That's the force that's accelerating the little block, so that must be the tension
in the string.
The angular momentum is defined as,

Acording to this text we know for conservation of angular momentum that

Where
is initial momentum
is the final momentum
How there is a difference between the stick mass and the bug mass, we define that
Mass of the bug= m
Mass of the stick=10m
At the point 0 we have that,

Where l is the lenght of the stick which is also the perpendicular distance of the bug's velocity
vector from the point of reference (O), and ve is the velocity
At the end with the collition we have

Substituting




Applying conservative energy equation we have


Replacing the values and solving

Substituting
l=\frac{13}{0.54(9.8)}
