damn physics 2.... help is highly appreciated?

Question

Two small blocks each of mass M, are connected by a string of constant length of 4h and negligible mass. Block A is placed on a smooth tabletop, and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is then released from rest at a distance h above the floor at time T=0.

Q's:

Determine the acceleration of block B as it descends.

Block B strikes the floor and does not bounce. Determine the time at which block B strikes the floor.

There are many more questions but I do not understand the ones above. Please help :)

Actually this one also....

Determine the distance between the landing points of the two blocks.

Answer 1

> Determine the acceleration of block B as it descends.

By Newton's 2nd Law:
a = Fnet / M

The net force (Fnet_b) on Block B is a combination of two forces:
1. The block's weight pulling down (= Mg)
2. The rope's tension, pulling up (call it "T")

So, the net force on B is: Fnet_b = Mg - T. Substitute that into the previous equation:
(1): a = (Mg-T)/M

The problem is, we don't know the value of "T". But here's where Block "A" comes in. The acceleration of Block "A" must be the same as the acceleration of Block "B" (since the two are connected by a rope). And "A"'s acceleration is due only to the tension T (there are no other unbalanced forces acting on "A"). So, for Block "A" we can write:

(2): a = Fnet_a / M = T/M

From which:
T = Ma

Now substitute this value of T back into (1):

a = (Mg-Ma)/M
which gives:
a = g-a
2a = g
a = g/2

> Determine the time at which block B strikes the floor.

The block has to fall a distance "h" before it hits the floor. Use this equation for elapsed time when accelerating from rest:

t = sqrt(2ah)

Plugging in the value "g/2" for "a" (as figured above):

t = sqrt(2(g/2)h) = sqrt(gh)

Notice that you can't turn that into an actual number without knowing the value of "h" (which they don't give you).

Answer 2

Why should we do your homework?