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Homework Help: Momentum with string

  1. Nov 1, 2013 #1
    Hello, PhysicsForum,
    I haven't posted a physics question, mainly because I don't have mechanics classes anymore :(
    But, studying on my own, I found this problem on the web with the solution:

    «61. A 2.0-kg box is attached by a string to a 5.0-kg box. A compressed spring is placed between them. The two boxes are initially at rest on a friction-free track. The string is cut and the spring applies an impulse to both boxes, setting them in motion. The 2.0-kg box is propelled backwards and moves 1.2 meters to the end of the track in 0.50 seconds. Determine the time it takes the 5.0-kg box to move 0.90 meters to the opposite end of the track.

    Answer: 0.94 s

    For the sake of the discussion, the 2-kg box will be referred to as Box 1 and the 5-kg box will be referred to as box 2.

    Given: mbox 1 = 2.0 kg; mbox 2 = 5.0 kg; dbox 1 = 1.2 m; tbox 1 = 0.50 s; dbox 2 = 0.90 m

    Find: tbox 2 = ???

    The two boxes are initially at rest. The total system momentum is initially 0. After the cutting of the string and the impulse of the spring, the total system momentum must also be 0. Thus, Box 1's backward momentum must be equal to the Box 2's forward momentum. The distance and time for Box 1 must be used to determine its velocity.

    v = d/t = (1.2 m) / (0.5 s) = 2.4 m/s

    Now the principle of momentum conservation can be used to determine Box 2's velocity.

    mbox 1 • vbox 1 = mbox 2 • vbox 2
    (2 kg) • (2.4 m/s) = (5 kg) • vbox 2

    vbox 2 = (2 kg) • (2.4 m/s) / (5 kg) = 0.96 m/s

    The velocity of Box 2 can be used to determine the time it takes it to move a distance of 0.90 meters.

    vbox 2 = dbox 2 / time
    Time = dbox 2 / vbox 2 = (0.90 m) / (0.96 m/s) = 0.9375 s = ~0.94 s »

    Now, what I don't understand from this whole process, is why we consider the speed of the blocks to be uniform. I think that the string, not being specified, could have been long and could therefore provide an acceleration throughout the movement of the blocks. This would make the calculation of the velocity erroneous. Is this correct?

    Also, I want to know how the spring separates energy. Let's say the compression of the spring gathered 1000 J of energy. Would each block get 500J? How does the separation of the energy work?
  2. jcsd
  3. Nov 1, 2013 #2


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    Gold Member

    It's true that during the time interval over which the spring provides the impulse, the blocks will be accelerating from rest, not moving at a constant speed. However, as soon as the spring loses contact with the blocks, there is no longer any force, and so the blocks continue at a constant speed that depends on the total impulse imparted. The point is, this time interval over which the impulse is provided is assumed to be short, as is the distance that the blocks move while accelerating. If you want, for the purpose of this problem, you can take the 1.2 m that block 1 moves to be the distance that it moved after being accelerated up to its final constant speed.

    This is ultimately determined by the one constraint we have on the system, that momentum must be conserved. Conservation of momentum determines the final velocity, and hence the final speed, of each block. Since kinetic energy depends on speed, this also determines the amount of kinetic energy each block will have in the end.
  4. Nov 1, 2013 #3
    Awesome. Finally some mental physics cogitation. :) I had thought that my bio courses had taken that away from me. :(
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