# 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 m

_{1}and m

_{2}moving along the same straight line with velocities

**u**

_{1}and

**u**

_{2}respectively.

Let us assume that

**u**

_{1}is greater then

**u**

_{2}. A and B suffer a head-on collision when they meet and continue moving along the same straight line with new velocities v

_{1}and v

_{2}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 havewhere u

_{1}, u

_{2}, Ï…

_{1}and and and Ï…

_{2 }are the magnitudes of

**u**

_{1},

**u**

_{2}, v

_{1}and v

_{2}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 Ï…

_{1 }and Ï…

_{2},

substituting for Ï…

_{2 }

Similarly eliminating Ï…

_{1 }from eqs. (6.15) and (6.16), we get

Now let us consider the following special cases:

- If both bodies have the same mass, then

m_{1}= m_{2}= m

In this case, eqs. (6.17) and (6.18) respectively give

Ï…_{1 }= u_{2 }

and Ï…2 = 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. - If one of he bodies, say m
_{2}, is initially at rest, then

u_{2}= 0

and

If, in addition, m_{1}= m_{2}= m, these equations give

Ï…_{1}= 0 = 0 = 0 Ï…_{2}= u_{1 }

However, if the body at rest, namely B, is much more massive than the colliding body A, i.e. m

_{2 }Â» m

_{1}, such that m

_{1}is negligibly small,

Ï…

_{1}= -u

_{1}

Ï…

_{2 }→ 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 m

_{2 }Â« m

_{1}, we have

Ï…

_{1 }= u

_{1 }

Ï…

_{2}

_{ }= 2u

_{1 }

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.