Question

In head on collision of two bodies, derive the expressions for the velocities of the bodies in terms of their masses and velocities before collision.

Answer 1

Consider two bodies with masses m1 and m2

Let the masses be travelling with velocities v1 and v2

Let their initial (i.e., before collision) velocities be v1i and v2i respectively

Similarly, let their final (i.e., after collision) velocities be v1f and v2f respectively

According to the law of conservation of linear momentum,

Total momentum Pi before collision = Total momentum Pf after collision

In a two body collision, this implies,

 p1i + p2i = p1f + p2f

∵ the collision is one-dimensional (along x-axis), the vectors of the equation become unidirectional

∴ substituting p = mv in the above equation becomes,

  m1v1i + m2v2i = m1v1f +m2v2f

Now, in terms of parameters before collision, the equation becomes,

  m1v1i +m1v1f = m2v2i + m2v2f

  m1 (v1i + v1f) = m2 (v2i + v2f)

Special Case: If the one of the bodies, say the second body, is at rest before collision, then

         v2i = 0

Then the equation becomes,  m1 (v1i + v1f) = m2v2f

Answer 2

in head on collision velocity of first body is directed just opposite to velocity of 2nd body. or you can say that, when two bodies move in just opposite direction , and finally they collide . this type of collision is known as head on collision.

Let a body of mass m1, moves with speed v1 in east direction . another body of mass m2 moves with speed v2 in west direction. after interval of time, bodies collide at point.

from law of conservation of linear momentum, we know, if there is no any external force acts on system of bodies , linear momentum remains constant.
e.g., Initial linear momentum = final linear momentum

m1v1 - m2v2 = Lf

if we assume after collision , velocity of first body is v3 and velocity of 2nd body is v4
then, m1v1 - m2v2 = m1v3 + m2v4 ........(1)

[ here you should include sign of velocity ]

from work energy theorem,
initial kinetic energy = final kinetic energy
1/2m1v1² + 1/2m2v2² = 1/2m1v3² + 1/2mv4² .....(2)

here we have two unknown and two equations we can get v3 and v4 easily to solve them.