Results 1 to 3 of 3

Math Help - Conservation of energy

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    14

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,079
    Thanks
    375
    Awards
    1
    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.

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

    \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 v_2 and plug it into the energy equation. After a few steps:
    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 v_1. Hint: Use the quadratic formula.
    v_1 = \frac{m_1 - m_2}{m_1 + m_2} \cdot v_{10}

    So use this to find v_2 and find Q_2.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member Danshader's Avatar
    Joined
    Mar 2008
    From
    http://en.wikipedia.org/wiki/Malaysia now stop asking me where is malaysia...
    Posts
    158
    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 = m_1
    velocity before collision = u_1
    velocity after collision = v_1

    for mass 2:

    mass = m_2
    velocity before collision = u_2
    velocity after collision = v_2

    using principle of conservation of momentum:
    <br />
m_1 u_1= m_1 v_1 + m_2 v_2 \longrightarrow (2)<br />

    from velocity of approach = velocity of separation
    <br />
u_1 - u_2 = v_2 -v_1<br />
    <br />
u_1 -0 = v_2 -v_1<br />
    <br />
 u_1 = v_2 -v_1<br />
    or
    <br />
 v_2 = u_1 +v_1<br />

    Substituting <br />
 v_2 = u_1 +v_1<br />
  into (2):
    <br />
m_1 u_1= m_1 v_1 + m_2 (u_1 +v_1) <br />
    <br />
m_1 u_1 - m_2 u_1 = m_1 v_1 + m_2 v_1 <br />
    <br />
(m_1 - m_2) u_1 = (m_1 + m_2) v_1 <br />
    <br />
 v_1=\frac {(m_1 - m_2)} {(m_1 + m_2)}u_1 \longrightarrow(3)<br />


    From equation (1):
    = (lost in Kinetic Energy)/(initial Kinetic Energy)
    <br />
=\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}<br />
    <br />
=\frac{ \frac{1}{2}m_1 ({u_1}^2 - {v_1}^2)}{\frac{1}{2}m_1 {u_1}^2}<br />
    <br />
=\frac{ {u_1}^2 - {v_1}^2}{{u_1}^2}<br />
    <br />
=\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}<br />
    <br />
=\frac{ ({u_1} - {v_1})({u_1} + {v_1})}{{u_1}^2}<br />
    <br />
 =\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)<br />
    <br />
 =\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}<br />
    <br />
=\frac{2m_1 \cdot 2m_2}{({m_1 +m_2})^2}<br />
    <br />
=\frac{4m_1 m_2}{({m_1 +m_2})^2}<br />

    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
    <br />
=\frac{4m_1 m_2}{({m_1 +m_2})^2}<br />

    therefore;
    kinetic energy of mass m2 = (initial kinetic energy) (\frac{4m_1 m_2}{({m_1 +m_2})^2})<br />
    Q2 = (\frac{4m_1 m_2}{({m_1 +m_2})^2})<br />
 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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Need help on the conservation of energy Eq
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 20th 2011, 08:52 AM
  2. Conservation of Energy
    Posted in the Advanced Applied Math Forum
    Replies: 4
    Last Post: March 28th 2010, 08:44 AM
  3. Conservation of Energy
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: November 13th 2009, 03:02 AM
  4. conservation of energy
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 28th 2008, 05:38 PM
  5. Energy Conservation question
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: March 11th 2008, 06:24 PM

Search Tags


/mathhelpforum @mathhelpforum