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Homework Help: Tension in a wire

  1. Dec 8, 2013 #1
    1. The problem statement, all variables and given/known data
    a 8kg ball is hanging from a 1.35m string. A 2 kg ball moving horizontally at 5m/s hits the hanging 8kg ball in an elastic collision. What is the tension in the string just after the collision?

    2. Relevant equations

    3. The attempt at a solution
    Initially there is only a gravitational force of 8*9.8 = 78.4 N.

    Using conservation of momentum, we can find that the ball has a velocity of 1.25 m/s just after the collision.

    so the centripetal acceleration is v^2/r = 1.1574 m/s^2.
    That centripetal acceleration is due to a centripetal force point in the opposite direction as the gravitation force

    so F_g - M*a_c = answer
    78.4 - (8)*(1.1574) =69.14 Newtons

    The back of the book says the answer is 102 Newtons.

    I feel like I worked this out right as far as my understanding of physics goes, but then I think about it and I feel like the tension should increase.

    Even if I add the centripetal force to the gravitational force I come up short of 102 N.
  2. jcsd
  3. Dec 8, 2013 #2

    Simon Bridge

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    What does it mean that the collision was elastic?

    When you did the conservation of momentum calculation, what did you use for the final speed of the 2kg ball?
    Looks like you guessed the ball comes to rest - how do you now it comes to rest?

    So the tension in the string has to supply the centripetal force as well as enough to cancel out the weight of the hanging mass.
    Last edited: Dec 8, 2013
  4. Dec 9, 2013 #3
    Elastic means no kinetic energy was lost in the collision. I guess the way I did it is not correct then. How do I find the final energy of the first mass? I thought in an elastic collision it would transfer all it's energy completely to the second mass.
  5. Dec 9, 2013 #4

    Simon Bridge

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    That's not what it means... and it is also not what you did: check

    EK (first mass) before collision: ##K_i= \frac{1}{2}(2\text{kg})(5\text{m/s})^2=25\text{ J}##
    EK (transferred to second mass) after the collision: ##K_f= \frac{1}{2}(8\text{kg})(1.25\text{m/s})^2=6.25\text{ J}##

    "Conserved" means it doesn't change at all - the total kinetic energy should be unchanged.

    After the collision you have two unknown velocities.
    You also have two equations - one for conservation of momentum, and the other for conservation of energy.
    These equations have to hold simultaneously.
    Last edited: Dec 9, 2013
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