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Problem dealing with momentum

  • Thread starter Spectre32
  • Start date
133
0
ALright i seem to be a little stuck on a Homework problem. Basicly there is a downward curving slope that level so for a few meters> throuhgout a good chunk of the problem the track is frictionless. And the end the coeff of kenetic friction is .50. They want to know how far the complelty inelastic system travels. For the problem set up there is a block of mass M1 sliding down the slide. At the bottem there is another block of Mass 2.00M1. This is all the information i'm given.
 

jamesrc

Science Advisor
Gold Member
476
1
So, is m2 on the rough part of the track? If you know how high m1 started at, you can use the work-energy theorem to find how fast it is going on the level part of the frictionless track before it hits m2. For the inelastic collision, you know that momentum is conserved and that the two masses get stuck together, effectively becoming a single object with mass = m1+m2 (this means the new velocity is m1*vi/(m1+m2)). After that, you have a constant acceleration problem with a known initial velocity, which you can use to find the total distance traveled (I'm thinking along the lines of vf^2 - vo^2 = 2*a*x, where a<0 (constant deceleration from the kinetic friction), vf = 0, and vo is found from the momentum conservation above.
 
133
0
Well m2 isn;t on the friction surface there both on the non friction part of the track and then they slide on to the friction part. Also no velocity is givin.
 

jamesrc

Science Advisor
Gold Member
476
1
Well, the blocks will slide together at a constant velocity until they hit the friction section, so if you call that distance D and the distance they travel with friction h, the total distance traveled together is D+h. You can solve for h in terms of H, the initial height of the first mass (I did it quickly and got h = 2H/9, but you should work it out to see what you get). That's the best I can think of; maybe someone else on the forum can help more if that's not what your teacher's looking for.
 

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