Smith, Mary asks:

CATAGORY: Mechanics QUESTION: Willy obtains an inclined plane with theta= 36.9 degrees and length s=1.00 m. He releases block A from rest at the top of the plane at the same instant he relases and identical block B from te bottom with an initial speed Vo. Willy observes that just before the blocks collide, their speeds are equal and opposite. The coefficient of kinetic friction between blocks and plane is Uk=0.3. How far up the plane in cm did this collision occur?

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This is a "way COOL" question(!) because it illustrates that friction 
works against the motion so for one block it works up the plane and for
another it works down the plane. 

Lets start by taking the convention that the bottom of the ramp is
the zero position and the top is at s = 1.00m (along the diagonal
of the ramp).  We have two equations of motion, describing position
and velocity along the plane of the ramp, of block A and block B.

(1)  block A

           s(A) = 1.0m + (0.5) a(A) t^2
           v(A) = a(A) t
     block B
           s(B) = Vo t + (0.5) a(B) t^2
           v(B) = Vo + a(B) t

Where a(A) is the acceleration of block A and a(B) is the acceleration
of block B. v(A) is the velocity of block A and v(B) is the velocity
of block B.  Also t is time and Vo is the initial velocity up the inclined
ramp.  From the problem description we know that the blocks meet when
they have the same speed but are moving in OPPOSITE directions.  Since
block A is always going down the plane the two blocks must meet while
block B is still going up the plane.  This is important because the
accelerations of the two blocks will be different since in the case of
block A friction is working against gravity and ind the case of block
B friction is working with gravity. So first we need to find a(A) and

The force on block A is given by the component of gravity trying to
accelerate the block down the plane (-mgsin(w)) plus the frictional
force that hinders it from accelerating down the plane which is given
by the coefficient of friction Uk times the normal force (mgcos(w)).

(2)       m a(A) = -mg sin(w) + Uk mg cos(w)

Where w is the angle of the inclined plane from horizontal (36.9 degrees)
Dividing through by m gives

(3)        a(A) = -g sin(w) + Uk g cos(w)             
The force on block B is given by the component of gravity trying to
accelerate the block down the plane (-mgsin(w)) plus the frictional
force that (here) helps gravity to slow down block B even more
than gravity would by itself.

(4)       m a(B) = -mg sin(w) - Uk mg cos(w)

Dividing through by m gives

(5)        a(B) = -g sin(w) - Uk g cos(w)

When they meet 
(6)                  v(A) = - v(B)

                      s(A) = s(B)

Substitution of the velocity expressions in (1) into (6) gives                                                            
(7)             a(A) T = -(Vo + a(B) T)

Solving for Vo gives

(8)         a(A) T - a(B) T = Vo

Factoring out t gives

(9)          (a(A) - a(B)) T = Vo 

where T = the time of contact. Substituting (9) and s(A) and s(B) from
(1) into s(A) = s(B) gives us

(10)     1.0m + (0.5) a(A) T^2 = Vo T + (0.5) a(B) T^2


(11)    1.0m + (0.5) a(A) T^2 = [(a(A) - a(B)) T] T + (0.5) a(B) T^2 
Which reduces further (12) ---> (14)

(12)    1.0m + (0.5) a(A) T^2 = a(A)T^2 - a(B)T^2 + (0.5) a(B) T^2 

(13)    1.0m + (-0.5) a(A) T^2 = (-0.5) a(B) T^2 

(14)    1.0m =  (-0.5) a(B) T^2 - (-0.5) a(A) T^2 

Factoring T^2 and -0.5 gives

(15)    1.0m =  -0.5(a(B)-a(A))T^2

substituting for a(A) and a(B) from  (3) and (5) gives

(16)    a(B) - a(A) = -g sin(w) - Uk g cos(w) - [-g sin(w) + Uk g cos(w)]

                    = -g sin(w) - Uk g cos(w) + g sin(w) - Uk g cos(w)
                    = -2 Uk g cos(w)
This is a beautiful result.  Since the gravitational components of 
acceleration act in the same way they do not contribute to the 
relative seperation of the two blocks.  Only the acceleration due to
friction acts between the two and since they are equal and opposite
one another the magnitude of the resulting relative acceleration
is twice the frictional acceleration of one of the blocks.  Substitution
into (15) gives

(17)       1.0m = 0.5(2 Uk g cos(w))T^2 

Solving fot T

(18)         T^2 = 2.0m/2 Uk g cos(w)
                 = 1.0m/ Uk g cos(w)
(19)           T = sqrt(1.0m / Uk g cos(w))

We can substitute (18) into s(A) (or (19) and the expression 
for Vo into s(B)) to find the position the blocks meet. We use s(A)
because it is easier to work with.

(20)  s(A) =  1.0 + (0.5) a(A) T^2                 (units of meters)

       =  1.0 + (0.5)(-g sin(w) + Uk g cos(w))T^2
       =  1.0 + (0.5)(-g sin(w) + Uk g cos(w))(1.0m/ Uk g cos(w))
       =  1.0 -0.5 g sin(w)/ Uk g cos(w) + 0.5 Uk g cos(w)/Uk g cos(w) 

       =  1.0 -0.5 tan(w) + 0.5         since sin(w)/cos(w) = tan(w)
Letting Uk = 0.3 and w = 36.9 degrees gives

(21)    s(A) = 1.0 -0.5 tan(36.9) +0.5
             = .249 meters
             = 24.9 cm

              *                                 *             
              *   The two blocks meet 24.9cm    *
              *   from the bottom of the plane. *                                      
              *                                 *
 Thanks for a great question!

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