OCR Momentum

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  1. Julii92's Avatar
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    OCR Momentum
    A block of mass 3kg travelling at 5ms-1 catches up another block of mass 7kg travelling at 2 ms-1 along the same line and the same direction.

    The difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds before the collision. Deduce that the speed of the first block after the collision must satisfy the inequalities 0.8<=v1<=2.9

    I had v2=(29-3v1)/5, and v1-3<=v2<=v1+3, which eventually leads me to 14<=8v<=44. This clearly isn't right, but I really don't know what to do. Help appreciated.
  2. TheMagicMan's Avatar
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    Re: OCR Momentum
    (Original post by Julii92)
    A block of mass 3kg travelling at 5ms-1 catches up another block of mass 7kg travelling at 2 ms-1 along the same line and the same direction.

    The difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds before the collision. Deduce that the speed of the first block after the collision must satisfy the inequalities 0.8<=v1<=2.9

    I had v2=(29-3v1)/5, and v1-3<=v2<=v1+3, which eventually leads me to 14<=8v<=44. This clearly isn't right, but I really don't know what to do. Help appreciated.
    Define a number e such that e= \frac{speed of separation}{speed of approach}. Work from there
  3. Julii92's Avatar
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    Re: OCR Momentum
    (Original post by TheMagicMan)
    Define a number e such that e= \frac{speed of separation}{speed of approach}. Work from there
    Thnanks, but I'm still not sure how to proceed from there. Can you tell me any more?
  4. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: OCR Momentum
    (Original post by Julii92)
    I had v2=(29-3v1)/5
    Why do you have it over 5?

    You also need to know that the speed of separation must be >=0.
  5. Julii92's Avatar
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    Re: OCR Momentum
    (Original post by ghostwalker)
    Why do you have it over 5?

    You also need to know that the speed of separation must be >=0.
    m1u1 + m2u2=m1v1 + m2v2
    3x5 + 7x2 = 3v1 + 5v2
    29=3v1+5v2
    29-3v1=5v2
    v2=(29-3v1)/5

    How does the speed of separation being >=0 help?
  6. ghostwalker's Avatar
    • Outcast of Imrryr
    • Location: CA13
    Re: OCR Momentum
    (Original post by Julii92)
    m1u1 + m2u2=m1v1 + m2v2
    3x5 + 7x2 = 3v1 + 5v2
    29=3v1+5v2
    29-3v1=5v2
    v2=(29-3v1)/5
    m_2=7

    How does the speed of separation being >=0 help?
    The inequality you have, although it gives two limits (if I recall), only one of them matches the restriction you're aiming for.

    You need the additional bit to get the second limit.
    Last edited by ghostwalker; 28-06-2012 at 16:36.
  7. Julii92's Avatar
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    Re: OCR Momentum
    (Original post by ghostwalker)
    m_2=7
    Ah. Now I feel silly.


    (Original post by ghostwalker)
    The inequality you have, although it gives two limits (if I recall), only one of them matches the restriction you're aiming for.

    You need the additional bit to get the second limit.
    Okay, I'll have a go at using this.
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