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Elastic Collision in One Dimension

Let us consider a collision between two bodies moving along the same straight line before and after the collision.

Let us consider two bodies A and B of masses m1 and m2 moving along the same straight line with velocities u1 and u2 respectively.

Let us assume that u1 is greater then u2. A and B suffer a head-on collision when they meet and continue moving along the same straight line with new velocities v1 and v2 respectively.

Let us further assume that the collision is elastic, i.e. in addition to momentum, the kinetic energy is also conserved.

From the law of conservation of linear momentum, we have 

One-dimensional elastic collision between two bodies (u1 > u2)

Since the kinetic energy of the bodies is also conserved during this collision, we have

where u1, u2, υ1and and and υ2 are the magnitudes of u1, u2, v1 and v2 respectively. Since the direction of all these velocities is the same, we can write in terms of magnitudes as

The above equation shows that in an elastic one-dimensional collision, the relative velocity with which the two bodies approach each other before collision is equal to the relative velocity with which they recede from each other after the collision.
To find υ1 and υ2
substituting for υ2


Similarly eliminating υ1 from eqs. (6.15) and (6.16), we get

Now let us consider the following special cases:
  1. If both bodies have the same mass, then
    m1 = m2 = m
    In this case, eqs. (6.17) and (6.18) respectively give
    υ1 = u2
    and υ2 = u1
    This means that in a one-dimensional elastic collision between two bodies of equal mass, the bodies merely exchange their velocities after the collision.
  2. If one of he bodies, say m2, is initially at rest, then
    u2 = 0
    If, in addition, m1 = m2 = m, these equations give
    υ1 = 0 = 0 = 0 υ2 = u1
Thus, if a body suffers a one-dimensional elastic collision with another body of the same mass at rest, the first body is stopped dead, but the second begins to move with the velocity of the first.

However, if the body at rest, namely B, is much more massive than the colliding body A, i.e. m2 » m1, such that m1 is negligibly small, 
υ1 = -u1
υ2 0

Thus, if a very light body suffers an elastic collision with a very heavy body at rest, the velocity of the lighter body is reversed on collision, while the heavier body remains practically at rest.  

A common example of this type of collision is the dropping of a hard steel ball on a hard concrete floor. The ball rebounds and regains its original height from where it was dropped while the much more massive ground remains at rest.

Finally, if the body at rest is much lighter than the colliding body, i.e. if m2 « m1, we have
υ1 = u1 
υ2 = 2u1 

i.e. the velocity of the massive body remains practically unchanged on collision with the lighter body at rest and the lighter body acquires nearly twice the initial velocity of the massive body.

In a reactor, fast neutrons are produced in the fission of uranium atoms. These neutrons have to be slowed down so that a sustained reaction is possible. They are, therefore, made to collide with hydrogen. 

The hydrogen nucleus (i.e. proton) has nearly the same mss as a neutron and hence in a head-on collision with a hydrogen nucleus at rest, the neutrons are nearly stopped dead or considerably slowed down. If, instead of hydrogen, a more massive nucleus like that of lead were used as the target, the neutrons would simply bounce back from the target (an electron, for example), the neutrons would continue to move with their original velocity.

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