Elastic Collisions
Key Questions
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billiard balls
Newton's cradle
curling rocks
shuffleboard pucks
bowling ball and pinsAlways take time to think because in real-life situations, most collisions are a combination of elastic and inelastic.
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m_1, m_2 be the two bodies
u_1, v_1 be initial and final velocities ofm_1
u_2,v_2 be initial and final velocities ofm_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 ism_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 is1/2m_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 getv_2 I think,
v_2=(2m_1u_2)/(m_1+m_2) Final momentum of of
m_1 ism_1v_1 and Final momentum ofm_2 ism_2v_2 :D
PS : Sorry if the math is clumsy, I'm new here.
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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.