Elastic Collisions

Key Questions

  • billiard balls
    Newton's cradle
    curling rocks
    shuffleboard pucks
    bowling ball and pins

    Always take time to think because in real-life situations, most collisions are a combination of elastic and inelastic.

  • m_1, m_2 be the two bodies
    u_1, v_1 be initial and final velocities of m_1
    u_2,v_2 be initial and final velocities of m_2

    let the initial velocity of m_2 = o. So, u_2 = 0.

    As the collision is elastic, Kinetic Energy and Momentum are conserved.

    Initial Momentum is m_1u_1
    Final Momentum is m_1v_1+m_2v_2

    So, m_1u_1 = m_1v_1+m_2v_2
    rArr m_1u_1-m_1v_1=m_2v_2
    rArr m_1(u_1-v_1)=m_2v_2 -> Equation'1'

    Initial Kinetic Energy is 1/2m_1u_1^2
    Final Kinetic Energy is 1/2m_1v_1^2+1/2m_2v_2^2

    So, 1/2m_1u_1^2 = 1/2m_1v_1^2+1/2m_2v_2^2
    rArr m_1u_1^2 = m_1v_1^2+m_2v_2^2
    rArr m_1u_1^2-m_1v_1^2 = m_2v_2^2
    rArr m_1(u_1^2-v_1^2) = m_2v_2^2
    rArr m_1(u_1-v_1)(u_1+v_1) = m_2v_2^2

    From Equation '1', m_1(u_1-v_1)=m_2v_2

    rArr m_2v_2(u_1+v_1)=m_2v_2^2
    rArr u_1+v_1=v_2

    Now, v_2=u_1+v_1

    From Equation '1', m_1(u_1-v_1)=m_2v_2

    So, m_1(u_1-v_1)=m_2(u_1+v_1)
    rArr m_1u_1-m_1v_1=m_2u_1+m_2v_1
    rArr v_1= (u_1(m_1-m_2))/(m_1+m_2)

    now, solve the equation '1' using v_1=v_2-u_1 to get v_2

    I think, v_2=(2m_1u_2)/(m_1+m_2)

    Final momentum of of m_1 is m_1v_1 and Final momentum of m_2 is m_2v_2

    :D

    PS : Sorry if the math is clumsy, I'm new here.

  • Answer:

    Elastic collision is the collision where there occurs no loss in net kinetic energy as the result of collision.

    Explanation:

    Total Kinetic energy before the collision= Total kinetic energy after the collision

    For example,

    Bouncing back of a ball from the floor is an example of elastic collision.

    Some other examples are:-
    =>collision between atoms
    =>collision of billiard balls
    =>balls in the Newton's cradle... etc.

Questions