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Two blocks of wood colliding

  1. Mar 31, 2009 #1
    A block of wood of mass m1 = 0.15 kg slides to the right on a frictionless horizontal table with an unknown speed. It collides with a second piece of wood of mass m2 = 0.30 kg which is stationary and attached to a spring with a spring constant of k = 159.00 N/m. The two pieces of wood stick together, and the spring compresses by an amount of 0.45 m.


    (a) What is change in total momentum of the two masses as a result of the collision,
    (before the spring compresses)?




    ANSWER: __________________ _____

    (b) What is the velocity of each mass just after collision (and before the spring
    compresses)?





    ANSWER: v1f = _______________________ _____
    ANSWER: v2f = _______________________ _____


    (c) What is the kinetic energy of each mass before collision?




    ANSWER: K1i = _______________________ _____
    ANSWER: K2i = _______________________ _____

    (d) What is the change in total kinetic energy of the two masses as a result of the
    collision (before the spring compresses)?




    ANSWER: __________________ _____

    Answers are suppose to be: A1) 0 A2) 8.46 m/s, 8.46 m/s A3) 48.4 J, 0J A4) -32.3 J
     
  2. jcsd
  3. Mar 31, 2009 #2

    hage567

    User Avatar
    Homework Helper

    Re: Collisions

    Show your work please. We are not going to do your homework for you.
     
  4. Mar 31, 2009 #3
    Re: Collisions

    Its not Homework. It test review but i understand what you are saying. I am just trying to get a starting point since i looked at these problems for about 10 min each. But i have three other tests to study for and all four of them are on the same day so i was trying to get a boost or something. Ill reply in a little with my attempt.
     
  5. Mar 31, 2009 #4
    Re: Collisions

    refresh. Just putting the forums on my subscription. not trying to get people to do it for me
     
  6. Mar 31, 2009 #5
    Re: Collisions

    Ok can anyone help me with part B with finding the initial velocity of mass 1.

    I have the formula to solve for the speeds.

    M1V1 + M2V1 = M1V2 + M2V2

    which the gets simplified to

    .15 kg (v1) +0 = ( .15+.30)V2
     
  7. Apr 1, 2009 #6
    Re: Collisions

    refresh
     
  8. Apr 1, 2009 #7
    Re: Collisions

    refresh
     
  9. Apr 1, 2009 #8
    Re: Collisions

    Someone please help
     
  10. Apr 1, 2009 #9
    Re: Collisions

    Ok to solve B i did.

    W = 1/2 K (X2-X1) = .5 (159.00)(.45)=35.775

    w = (change in K) = .5 (ma + Mb)V2^2

    V2 = (35.775/ (1/2 * .45))^(1/2)

    however i got 12.6 m/s^2

    what was i suppose to do instead?
     
  11. Apr 1, 2009 #10
    Re: Collisions

    ok i forgot the ^2 in theW = 1/2 K (X2-X1)^2

    so i did get the right answer but how do i go about getting part C first part.

    The second part is clearly zero because it doesn;t have any velocity so there for no kinetic energy.
     
  12. Apr 1, 2009 #11
    Re: Collisions

    ok wow i keep finding how to do stuff.

    Ok to get part C. I realized that i had solved for Vf.

    So therefore you can use this to find Vi

    M1V1+M2V1= (M1+M2)V2

    then you can use

    K = 1/2 * m1 * V1^2

    = 48.4J
     
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