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Thread: Conservation of energy

  1. #1
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    Conservation of energy

    Conservation of energy

    show that when Mass (m1) and kinetic energy (Q1) collide with a stationary particle (m2) it transfers an energy. Using conservation of energy and momentum

    Q2 = (4M1M2)/(M1 + M2)^2 x Q1

    Cheers
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  2. #2
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    Quote Originally Posted by j-lee00 View Post
    Conservation of energy

    show that when Mass (m1) and kinetic energy (Q1) collide with a stationary particle (m2) it transfers an energy. Using conservation of energy and momentum

    Q2 = (4M1M2)/(M1 + M2)^2 x Q1

    Cheers
    I presume you simply mean find the kinetic energy of particle m2 after the collision.

    $\displaystyle m_1v_{10} = m_1v_1 + m2v_2$
    where $\displaystyle v_{10}$ is the velocity of m1 before the collision and $\displaystyle v_1, v_2$ are the final velocities.

    $\displaystyle \frac{1}{2}m_1v_{10}^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$

    Solve the momentum equation for $\displaystyle v_2$ and plug it into the energy equation. After a few steps:
    $\displaystyle m_1m_2v_{10}^2 = m_1^2v_{10}^2 - 2m_1^2v_{10}v_1 + (m_1^2 + m_1m_2)v_1^2$

    Solve this for $\displaystyle v_1$. Hint: Use the quadratic formula.
    $\displaystyle v_1 = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_{10}$

    So use this to find $\displaystyle v_2$ and find $\displaystyle Q_2$.

    -Dan
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  3. #3
    Member Danshader's Avatar
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    Hello there,

    you need to consider fractional lost in kinetic energy to obtain that equation:

    fractional lost in Kinetic Energy
    = (lost in Kinetic Energy)/(initial Kinetic Energy) → (1)

    given for mass 1:

    mass = $\displaystyle m_1$
    velocity before collision = $\displaystyle u_1$
    velocity after collision = $\displaystyle v_1$

    for mass 2:

    mass = $\displaystyle m_2$
    velocity before collision = $\displaystyle u_2$
    velocity after collision = $\displaystyle v_2$

    using principle of conservation of momentum:
    $\displaystyle
    m_1 u_1= m_1 v_1 + m_2 v_2 \longrightarrow (2)
    $

    from velocity of approach = velocity of separation
    $\displaystyle
    u_1 - u_2 = v_2 -v_1
    $
    $\displaystyle
    u_1 -0 = v_2 -v_1
    $
    $\displaystyle
    u_1 = v_2 -v_1
    $
    or
    $\displaystyle
    v_2 = u_1 +v_1
    $

    Substituting $\displaystyle
    v_2 = u_1 +v_1
    $ into (2):
    $\displaystyle
    m_1 u_1= m_1 v_1 + m_2 (u_1 +v_1)
    $
    $\displaystyle
    m_1 u_1 - m_2 u_1 = m_1 v_1 + m_2 v_1
    $
    $\displaystyle
    (m_1 - m_2) u_1 = (m_1 + m_2) v_1
    $
    $\displaystyle
    v_1=\frac {(m_1 - m_2)} {(m_1 + m_2)}u_1 \longrightarrow(3)
    $


    From equation (1):
    = (lost in Kinetic Energy)/(initial Kinetic Energy)
    $\displaystyle
    =\frac{ \frac{1}{2}m_1 {u_1}^2 - \frac{1}{2}m_1 {v_1}^2}{\frac{1}{2}m_1 {u_1}^2}
    $
    $\displaystyle
    =\frac{ \frac{1}{2}m_1 ({u_1}^2 - {v_1}^2)}{\frac{1}{2}m_1 {u_1}^2}
    $
    $\displaystyle
    =\frac{ {u_1}^2 - {v_1}^2}{{u_1}^2}
    $
    $\displaystyle
    =\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}
    $
    $\displaystyle
    =\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}
    $
    $\displaystyle
    =\frac{ ({u_1} - (\frac{m_1 - m_2}{m_1 + m_2}){u_1})({u_1} + (\frac{m_1 - m_2}{m_1 + m_2}){u_1})}{{u_1}^2}\longrightarrow using(3)
    $
    $\displaystyle
    =\frac{ ({u_1} (\frac{m_1 + m_2 - m_1 + m_2}{m_1 + m_2})({u_1} (\frac{m_1 + m_2 + m_1 - m_2}{m_1 + m_2}))}{{u_1}^2}
    $
    $\displaystyle
    =\frac{2m_1 \cdot 2m_2}{({m_1 +m_2})^2}
    $
    $\displaystyle
    =\frac{4m_1 m_2}{({m_1 +m_2})^2}
    $

    lost in kinetic energy is gained by the stationary mass hence we can rewrite the equation as:

    kinetic energy of mass m2/ initial kinetic energy
    $\displaystyle
    =\frac{4m_1 m_2}{({m_1 +m_2})^2}
    $

    therefore;
    kinetic energy of mass m2 = (initial kinetic energy)$\displaystyle (\frac{4m_1 m_2}{({m_1 +m_2})^2})
    $
    Q2 = $\displaystyle (\frac{4m_1 m_2}{({m_1 +m_2})^2})
    $Q1

    There is another way to get the answer but i think this should be sufficient unless you really want to know more xD cheers!
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