physic question!?

Question

Two 2.00 kg masses are connected by a 50.0 cm long rod of negligible mass. This rod can rotate in the vertical plane around a horizontal rotation axis through its middle. Initially assume the rod is balanced and sitting along the horizontal as shown below. Suddenly a 50.0 g spider drops onto one of the masses with a speed of 3.00 m/s. (a) What is the angular speed of the system just after the spider lands? (b) What is the fraction of kinetic energy after the spider lands to just before it lands? (c) Through what angle will the system rotate before coming to rest?

Answer 1

This problem satisfies the conservatin of momentum law but not conservation of kinetic energy since it is an inelastic collision.

Solving
m1*v1=m2*v2
m1*v1/m2=v2
3*.05/4.05
=.0371m/s
divide by the radius for radians/sec
=.0371/.25 m/s

For the angular displacement, conservation of energy is held, so the rotation down and back to horizontal will result in the same energy as when the spider landed, so the spider's weight times vertical displacement will be the height of the spider before equilibrium:

1/2*m*v^2=m(spider)*g*h
.5*4.05*.0371^2/(.05*9.8)=h
h=.0057 m

with this we can compute the angle above horizontal
sin(theta)=.0057/.025
theta =13.18 degrees
so the spider rotated through
90+13.18
=103.18 degrees

Sorry for the previous incorrect answer using conservation of energy for the collision.

j