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Mastering Physics Help, Conservation of Momentum

  1. Nov 10, 2009 #1
    1. The problem statement, all variables and given/known data
    A 238U nucleus is moving in the x direction at 4.5*10^5 m/s when it decays into an alpha particle 4He and a 234Th nucleus.

    1.) If the alpha particle moves off at 34degrees above the x axis with a speed of 2.0*10^7 m/s, what is the speed of the thorium nucleus?

    2.)What is the direction of the motion of the thorium nucleus?

    2. Relevant equations
    I know that momentum is conserved since this is a totally inelastic collision in reverse.
    M1(V1)+M2(V2)=(M1+M2)VFinal

    3. The attempt at a solution
    1.)Im using the atomic numbers for the masses.

    238(4.5*10^5)=4(2.0*10^7m/s)+234(V2)
    Solving for V2 I got the answer 1.158*10^5 or 1.2*10^5. Mastering physics says this is wrong and nothing else. I believe I might have the wrong formula or something.

    2.) I have no ideo how to start this part of the problem. Any help/tips would be greatly appreciated
     
  2. jcsd
  3. Nov 10, 2009 #2

    Doc Al

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    Staff: Mentor

    Since momentum is a vector, apply conservation of momentum to each component (x and y) separately.
     
  4. Nov 10, 2009 #3
    Okay, I understand that I need to look at the conservation of momentum for both the x and y directions. Here was my next attempt.

    M1(V1x)+M2(V2x)=(M1+M2)Vfinal(x)
    M1(V1y)+M2(V2y)=(M1+M2)Vfinal(y)

    Using the angle 34 as given I calculated the initial velocities of the 4He particle.
    2.0*10^7(sin34)=V1x=1.12*10^7
    2.0*10^7(cos34)=V2y=1.66*10^7

    I substituted those values into the equations above again using the atomic number as the masses.

    4(1.12*10^7m/s)+234(V2x)=238(4.5*10^5m/s) Solving for V2x I got 2.66*10^5
    4(1.66*10^7m/s)+234(V2y)=238(4.5*10^5m/s) Solving for V2y I got 1.74*10^5

    Next I used the Pythagorean Theorem to calculate the final velocity using the x and y components. (2.66*10^5)^2+(1.74*10^5)^2=Vfinal^2
    Solving for Vfinal I got 3.18*10^5

    This is still wrong according to mastering physics and I received no feedback from the program on how it was wrong. Any help again would be greatly appreciated.
     
  5. Nov 10, 2009 #4
    If my math doesn't fail me velocity in x direction should be vcos34 and vsin34 for y direction.
    Initially the particle is moving in the x direction so second equation is wrong the particle has no velocity in the y direction.
     
  6. Nov 10, 2009 #5
    Okay, this makes sense if you are talking about the 238U particle. I see my simple error in calculating the x-velocity. Here was my next attempt.

    V1x=2.0*10^7(cos34)=1.66*10^7m/s
    V1y=2.0*10^7(sin34)=1.12*10^7m/s

    Since there is no velocity in the y direction for the initial particle I revised my formula.

    M1(V1y)+M2(V2y)=(M1+M2)VFinal(y)
    M1(V1y)+M2(V2y)=0
    M1(V1y)=-M2(V2y)

    4(1.12*10^7m/s)=-234(V2y) Solving for V2y I got -1.91*10^5m/s. This negative sign indicates the particle is moving in the negative y direction.

    So again using the Pythagorean Theorem I calculated the final velocity of the 234Th particle.

    (2.66*10^5m/s)^2+(-1.91*10^5)^2=(V2Final)^2 Solving for V2Final I got 3.30*10^5m/s.

    I submitted it into Mastering Physics and again it was wrong with no feedback. I'm getting a little concerned because I only have one more attempt to get the answer right. Again any help/tips will be greatly appreciated.
     
  7. Nov 10, 2009 #6
    By the way I have to go to work so I'll check back in around 6 hours. Thanks again.
     
  8. Nov 10, 2009 #7
    Dude, work it out again i see velocity in x direction 2.66 which is the same in the previous post.
     
  9. Nov 10, 2009 #8
    Sorry, In my rush to complete the problem I got ahead of myself and forgot to correct my previous mistake.

    Well I fixed my error and found the new V1x and used it to calculate the V2y. Then I used the Pythagorean Theorem and found a velocity of 2.6*10^5 and it was correct.

    Now onto the second part of the question.

    2.) What is the direction of the motion of the thorium nucleus?

    This part was simple. I drew a right triangle and placed the velocities on their corresponding sides. I used sin(theta)=(-1.9*10^5m/s)/(2.6*10^5m/s) and took the inverse sin of that answer to get a angle of 47 degrees which was correct.

    Thanks for the help people.
     
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