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Two balls of different mass elastically collide

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A tennis ball of mass m_t is held just above a basketball of mass m_b. With their centers vertically aligned, both are released from rest at the same moment, so that the bottom of the basketball falls freely through a height h and strikes the floor. Assume an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down because the balls have a very small amount of separation while falling.
    m_(b)asketball ≈ 11m_(t)ennis ball

    (Q1.) The two balls meet in an elastic collision. To what height does the tennis ball & basketball rebound?
    (Q2.) How do you account for the height being larger than h? Does that seem like a violation of conservation of energy?

    2. Relevant equations
    (1.) v_1f (basket ball) = [(m_1-m_2)/(m_1+m_2)]v_1i + [(2*m_2)/(m_1+m_2)]v_2i

    (2.) V_2f (tennis ball) = [(2*m_1)/(m_1+m_2)]v_i + [(m_2-m_1)/(m_1+m_2)]

    (3.) v_1i (basketball) = -v_2i (tennis ball) →
    v_1i = + √(2gh)
    v_2i = -√(2gh)

    *Note: Recall, m_basketball ≈ 11*m_tennis ball


    3. The attempt at a solution

    *I already have the correct solution and answer to the first questions but I don't understand it. I need help with understanding the story behind the mathwork.

    -Solution 1a:
    Answer: Tennis ball will rise up to ~7 times of its original height approximately.

    -Solution 1b:
    (i) v_1f (basket ball) = [(m_1-m_2)/(m_1+m_2)](√(2gh)) + [(2*m_2)/(m_1+m_2)](-√(2gh))
    (ii) =[(m_b-3*m_t)/(m_b+m_t)](√(2gh))
    (iii) =(8/12)(√(2gh))

    ****how does this (direclty below, iv) relate to the last part of my above solution (iii):
    (iv) (1/2)(m_b*v^2_1f)=m_b*g*h_1f)
    (v) h_1f = (v^2_1f)/(2*g) = h*(2/3)^2 ≈ 0.44*h
    Answer: Basketball will rise up to ~.44 times of its original height approximately.

    My Questions:
    A. I can see that the equations given:
    (1.) v_1f (basket ball) = [(m_1-m_2)/(m_1+m_2)]v_1i + [(2*m_2)/(m_1+m_2)]v_2i
    (2.) V_2f (tennis ball) = [(2*m_1)/(m_1+m_2)]v_i + [(m_2-m_1)/(m_1+m_2)]

    relates to the conservation of momentum (which I'm guessing will be the equation that I am supposed to always think of first when it comes to elastic collisions) but how and why did we subtract mass in the equations???? instead of just using m_1*v_1i + m_2*v_2i = m_1*v_1f + m_2*v_2f

    Note: I am really struggling with Physics, helping me understand the material and learning it...I would really appreciate it. I sit at one problem for hours, which eats up my time in studying my other course material.
     
    Last edited: May 15, 2012
  2. jcsd
  3. May 15, 2012 #2
    Hi, Regnum, I am not sure about what your question actually is
    You want to know how to derive the equations that relate masses, initial and final velocities in an elastic one dimensional collision ?
    Conservation of momentum is only half of it and it is not really what you should first think about when told about an elastic collision. Elastic collision tells you that the kinetic energy is conserved.
    Together with the conservation of momentum (or of the speed of the center of mass) you get to those formulas.
    Have a look at this
    there is a nice video of Walter Lewin that should help you a lot
     
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