A 3 kg block (block A) is released from rest at the top of a 20 m long frictionless ramp that is 5 m high. At the same time, an identical block (block B) is released next to the ramp so that it drops straight down the same 5 m. Find the values for each of the following for the blocks just before they reach ground level.

Question

(A) gravitational potential energy
      Block A	 
           ---J
      Block B	 
        -------J

(B) kinetic energy
    Block A	 
    -----J
    Block B	 
    -----J

(c) speed
  Block A	 
   ----m/s
  Block B	 
    ----m/s

(d) momentum
    Block A	 
    ----kg·m/s
    Block B	 
   ----kg·m/s

Damon · Answer

This is a trick question.

(a) Since there is no friction, the potential energy at the TOP (at 5 meters), m g h = 3 * 9.8 * 5
becomes ZERO AT THE BOTTOM if you define ground level as zero potential. Note the potential difference depends only on altitude. The ramp changes nothing.

(b) The potential drop in the 5 meter change in altitude results in kinetic energy at the bottom (1/2) m v^2 = m g h. (c) Again the ramp means nothing but a change in direction. The speeds are the same.
s = sqrt (2 g h) = sqrt(2*9.8*5)
there is a force component up from the ramp, so the vertical component of velocity will be less and it will take longer to reach the bottom, but the speed will be the same.
(d), now we finally have a difference, momentum is a vector and now direction matters
call u = horizontal velocity component
call w = vertical down velocity component
then u^2 + w^2 = speed^2 where
speed is our speed, call it s, from part (c)
u = s cos T
w = s sin T
where
cos T = 5 sqrt 15 /20
and sin T = 5/20

the magnitude of momentum in both
cases is m s = 3 s
however in the direct fall
horizontal momentum = 3 u = 0
vertical momentum = 3 w = 3 s

in the slide down the ramp
horizontal momentum =3 u = 3s *5sqrt (15)/20
and
vertical momentum = 3 w = 3s*5/20

Anonymous · Answer

thank you very mach