Please help me with my physics homework!?

Question

(I've tried and tried but I simply don't know how to solve this. Someone please help me.)

Illustration: http://i115.photobucket.com/albums/n305/jlsilo/phy.jpg

Two small blocks, each of massm, are connected by a string of constant length 4h and negligible mass. Block A is placed on a smooth tabletop as shown above, 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 t=0. Express all algebraic answers in terms of h, m, and g.
a) Determine the acceleration of block B as it descends.
b) Block B strikes the floor and does not bounce. Determine the time t, at which block B strikes the floor.
c) Describe the motion of block A from time t=0 to the time when block B strikes the floor.
d) Describe the motion of block A from the time block B strikes the floor to the time block A leaves the table.
e) Determine the distance between the landing points of the two blocks.

a) b) c) d) are all individual questions corresponding to the description above.

Answer 1

The force accelerating block B is m*g. That force must also accelerate A. Therefore, from f=Ma, a = f/M, and here M is both masses = 2m. So the acceleration is a =m*g/2m = g/2

B. The distance an accelerating object moves is given by s=.5*a*t^2. Here, s = h, a=g/2, solve for t

C. Block A and B move together at the same rate. Block B moved a distance h at an acceleration of g/2. Block A must do the same. Since the string length is 4h, it is long enough so that block A does not fall off the edge of the table (yet).

D After block B reaches the floor, there is no more force pulling block A. It therefore stops accelerating, but it doesn't stop moving. It will move at a constant velocity, the velocity it had when B hit the floor. V = a*t. You know a (-g/2) and t (calculated above) so you can compute A's velocity along the table.

E) When A reaches the end of the table, it keeps its horizontal velocity (computed above), but starts accelerating at rate g towards the floor. Using the formula s=.5*g*t^2, knowing s = 2h (the table height), you can figure t. The block will travel horizontally a distance v*t before it lands, and that's how far it will be from block B.

Answer 2

Since we are assuming the table to be frictionless, there is no resistance to the motion of A. Till the mass B hits the ground, the string will drive A. Lets say that the acceleration of the bodies be 'a'. Hence, for body A, T=ma , where T = tension in the string).
now for B, mg-T=ma, equationg two equations, a=g/2.(Answer part a)
Using s=1/2at^2; h=1/2(g/2)t^2 (note a=g/2 here). Hence t=2*sqrt(h/g) (Answer b)

Answer c: Uniform acceleration at a=g/2 units.
Answer d: uniform velocity in x-direction (since there no tenson in the string). To get velocity use
v^2=u^2+2as (for part c) ; (u=0); v=sqrt(2*g/2*h)=sqrt(gh)

Answer e) find time to fall freely and multiply it with u.

Answer e)

Answer 3

The Physics communicate board, as presently constituted, is plenty greater effective than the Astronomy and area communicate board. A&S has a tendency to be somewhat delicate. somewhat some i. grade college homework, ii. hoaxes (2012 stuff), and iii. trolls searching for interest. even nevertheless homework help is stressful, a minimum of it relatively is severe. you could popular self-filter out to discover questions that grant somewhat challenge and a questioner who's appreciative. there has been communicate among the aficionados of bobbing up a severe Physics and Math communicate board. however the final consensus replaced into that it may nonetheless be crammed with homework questions and unsolicited mail trolls, and Y!A does no longer be prepared on the assumption because of the fact what Y!A needs is site visitors.

Answer 4

b