What is elastic collision.Derive expression for the final velocity of colliding bodies in elastic collision.
Consider two bodies A and B of masses m1 and m2 respectively, moving along the same straight line, in the same direction. Let be their respective velocities, such that
The two bodies will collide after some time.
During collision, the bodies will be deformed at the region of contact. So, a part of the kinetic energy will be converted into potential energy. The bodies will regain their original shape due to elasticity. The potential energy will be reconverted into kinetic energy. The bodies will separate and continue to move along the same straight line, in the same direction, but with different velocities.
Applying the law of conservation of momentum,
Total momentum before collision = total momentum after collision
m1v1i + m2v2i = m1v1f + m2v2f (in magnitude)
or m1 (v1i - v1f) = m2 (v2f - v2i) --------------- (i)
Since the collision is elastic, the kinetic energy will be conserved.
Kinetic energy before collision = Kinetic energy after collision
Dividing (ii) by (i), we get
v1i + v1f = v2f + v2i
or v1i - v2i = v2f - v1f ------------ (iii)
(v1i - v2i) is the magnitude of the relative velocity of A w.r.t B.
(v2f - v1f) is the magnitude of the relative velocity of B w.r.t A.
It may be noted that the direction of relative velocity is reversed after the collision.
From equation (iii)
v2f = v1i - v2i + v1f
From equation (i)
m1 (v1i - v1f) = m2 (v1i - v2i + v1f - v2i)
or m1v1i - m1v1f = m2v1i - 2m2v2i + m2v1f
or -m1v1f - m2v1f = -m1v1i + m2v1i - 2m2v2i
or (m1 + m2) v1f = (m1 - m2) v1i + 2m2v2i
Again, from equation (iii),
v1f = v2f - v1i + v2i
Substituting this value in equation (i) and simplifying, we get
Hope this clarifies your doubt.