Question

Find out the expression of the final velocities of body in perfectly head on collision

Answer 1

*
It is that elastic collision in which the colliding bodies move along the same straight line before n after th collision......

Consider two bodies A and B of masses m1 and m2respectively, 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