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Another One Just like It

  1. Feb 25, 2008 #1
    Two blocks, each with a mass 0.40 {\rm kg}, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed v = 1.2 {\rm m/s}; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.60 {\rm m/s}.

    If the maximum compression of the spring is 1.9 {\rm cm}, what is its force constant?

    So, I just posted a question but this one seems just like it. I guess I don't understand what effect each force has on one another and where to begin calculating it. Should I be thinking about the work energy theorm? How would I set this problem up? Thanks in advance to any that can tackle this. This by the way is nothing close to what we have been learning but I guess they are trying to challenge us...and i'm very curious
  2. jcsd
  3. Feb 25, 2008 #2


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    Staff: Mentor

    It's an inelastic collision since the spring absorbs some of the energy.

    Block 1 will decelerate and block 2 will accelerate, and meanwhile some energy will be stored in spring. But, one is told that maximum deflection occurs when the speed of both blocks is v/2 (0.6 m/s).

    So what is the KE of block 1 before the collision, and what is the KE of both blocks at v/2? What is the significance of the difference?
  4. Feb 25, 2008 #3
    well the kinetic energy of both blocks moving at the same velocity is exactly half of the kinetic energy of the first block in motion. which means? i have the same problem and i'm still lost. i'll take a guess though. since both the velocity and the kinetic energy of the first block are halved, can i assume that potential energy of the first block is halved as well?
    Last edited: Feb 26, 2008
  5. Feb 26, 2008 #4
    yes i can :D thanks for the help. i hope the original poster gets this
  6. Feb 26, 2008 #5

    Hey Sillybean, you have a message! :redface:
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