two small blocks, each of mass, m, are connected by a string of constant length, 4h and negligible mass. Bloct

Question

two small blocks, each of mass, m, are connected by a string of constant length, 4h and negligible mass. Block A is placed on a smooth tabletop and block B hangs over the edge of the table. The table top is a distance of 2h above of the floor. Block A is then released from rest at a distance h above the floor at time 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 t1, 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

Answer 1

I assume the table is horizontal and no friction.

Use FBD of the blocks from t=0 to t= B impacts floor
T is the tension in the string

Block B

m*g-T=m*a

Block A
T=m*a

a) a=g/2 downward

b) For Block B y(t)=h-(g*t^2)/4
Block B strikes the floor when y(t1)=0
t1=2*sqrt(h/g)

c) Block A has the same acceleration as B until B strikes the floor
v(t)=g*t/2
x(t)=g*t^2/4
when x(t1)=h
t1=2*sqrt(h/g)

d) at t=t1, the speed of block A is
v(t1)=g*t1
=2*SQRT(g*h)

So for t>t1 before the block falls off the table
the velocity is constant
and the distance form the edge is 2*h
x(t)=2*sqrt(g*h)*t

e) when block A leaves the table it has
y(t)=2*h-.5*g*t^2
and x(t)=2*sqrt(g*h)*t
as long as sqrt(y^2+x^2)<4*h
this is maximal when Block A strikes the floor
which is
t2=2*sqrt(h/g)
and
x(t2)=4*h

j

Answer 2

for e, the answer is 2sqrt(2)*h, because the horizontal velocity of the block leaving the table is sqrt(h*g)(j had this part wrong, he put that it was 2*sqrt(h*g). so you take sqrt(h*g) times the time(2sqrt(2h/g)) and simplify and get 2h*sqrt(2)