Velocity of the wedge

  • Thread starter mighty2000
  • Start date
  • #1
39
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I think this may be correct I just need a second set of eyes on it to look it over. Thanks in advance.

The question is:

A small block of mass m1 =0.500kg is released from rest at the top of a curve shaped frictionless wedge of mass m2=3.00kg, which sits on a frictionless horizontal surface. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right

a) what is the velocity of the wedge after the block reaches the horizontal surface

b) what is the height h of the wedge

anyway

a)
Pwedge=-Pblock
Mwedge*Vwedge=-Mblock*Vblock

Vwedge=(-mBlockVblock)/mWedge = -(0.500)*(4)/3
Vwedge= 0.667 m/s

b)
Ki+Ugi=Kf+Ugf

(0+Mblock*g*h=1/2*Mblock*V^2block + 1/2*Mwedge*v^2wedge+0)

h= (1/2*Mblock*V^2block + 1/2*Mwedge*v^2wedge)/gMblock

h=(4+.668)/4.9
h=0.953 m

any input?
 

Answers and Replies

  • #2
508
0
Looks OK to me. Better use 2/3 than .668
 
  • #3
39
0
naa, this is a physics class, not a Calculus class. Exact numbers rounded to 3 places is best.

Thanks for the check.

Peace out bro

M2k
 

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