Collisions in One Dimension

Consider first a completely inelastic collision in one dimension. Then, in Fig. 1,

$\theta_1=\theta_2$

$m_1v_{1i}=(m_1+m_2)v_f$  (momentum conservation)

$v_f=\frac{m_1v_{1i}}{m_1+m_2}$

Fig.1: Collision of mass $m_1$, with a stationary mass $m_2$.

The loss in kinetic energy on collision is

$\Delta K=\frac{1}{2}m_1v_{1i}^2-\frac{1}{2}(m_1+m_2)v_f^2$

       $=\frac{1}{2}m_1v_{1i}^2-\frac{1}{2}\frac{m_1^2}{(m_1+m_2)}v_{1i}^2$ using Eq. (6.23)]

       $=\frac{1}{2}m_1v_{1i}^2\left[1-\frac{m_1}{m_1+m_2}\right]$

       $=\frac{1}{2}\frac{m_1}{m_1+m_2}v_{1i}^2$

which is a positive quantity as expected.

Consider next an elastic collision. Using the above nomenclature with θ$_1$ = θ$_2$ = 0, the momentum and kinetic energy conservation equations are

$m_1v_{1i}=m_1v_{1f}+m_2v_{2f}$

$m_1v_{1i}^2=m_1v_{1f}^2+m_2v_{2f}^2$

From Eqs. (6.24) and (6.25) it follows that,

$m_1v_{1i}(v_{2f}-v_{1i})=m_1v_{1f}(v_{2f}-v_{1f})$

or $v_{2f}(v_{1i}-v_{1f})=v_{1i}^2-{1f}^2$

$v_{2f}(v_{1i}-v_{1f})=(v_{1i}-v_{1f})(v_{1i}+v_{1f})$

Hence, ∴ $v_{2f}=v_{1i}+v_{1f}$

Substituting this in Eq. (6.24), we obtain

$v_{1f}=\frac{m_1-m_2}{m_1+m_2}v_{1i}$

and $v_{2f}=\frac{2m_1}{m_1+m_2}v_{1i}$

Thus, the ‘unknowns’ ($v_{1f}$ , $v_{2f}$) are obtained in terms of the ‘knowns’ ($m_1$, $m_2$, $v_{1i}$). Special cases of our analysis are interesting.

Case I : If the two masses are equal

$v_{1i}=0$

$v_{2f}=v_{1i}$

The first mass comes to rest and pushes off the second mass with its initial speed on collision.

Case II : If one mass dominates, e.g. $m_2$ > > $m_1$ 

$v_{1f}\simeq -v_{1i}$

$v_{2f} \simeq 0$

The heavier mass is undisturbed while the lighter mass reverses its velocity.

Example 1: Slowing down of neutrons: 

In a nuclear reactor a neutron of high speed (typically $10^7$ m/s) must be slowed to $10^3$ m/s so that it can have a high probability of interacting with isotope $^{235}_{92}U$ and causing it to fission. Show that a neutron can lose most of its kinetic energy in an elastic collision with a light nuclei like deuterium or carbon which has a mass of only a few times the neutron mass. The material making up the light nuclei, usually heavy water (D$_2$O) or graphite, is called a moderator.

Answer:

The initial kinetic energy of the neutron is

$K_{1i}=\frac{1}{2}m_1v_{1i}^2$

while its final kinetic energy from Eq. (6.27)

$K_{1f}=\frac{1}{2}m_1v_{1f}^2=\frac{1}{2}m_1\left(\frac{m_1-m_2}{m_1+m_2}\right)^2v_{1i}^2$

The fractional kinetic energy lost is

$f_1=\frac{K_{1f}}{K_{1i}}=\left(\frac{m_1-m_2}{m_1+m_2}\right)^2$

while the fractional kinetic energy gained by the moderating nuclei $K_{2f} /K_{1i}$ is

$f_2 = 1 − f_1$ (elastic collision)

$f_2=\frac{4m_1m_2}{(m_1+m_2)^2}$

One can also verify this result by substituting from Eq. (6.28).

For deuterium $m_2 = 2m_1$ and we obtain $f_1$ = 1/9 while $f_2$ = 8/9. Almost 90% of the neutron’s energy is transferred to deuterium. For carbon $f_1$ = 71.6% and $f_2$ = 28.4%. In practice, however, this number is smaller since head-on collisions are rare.

If the initial velocities and final velocities of both the bodies are along the same straight line, then it is called a one-dimensional collision, or head-on collision. In the case of small spherical bodies, this is possible if the direction of travel of body 1 passes through the centre of body 2 which is at rest. In general, the collision is twodimensional, where the initial velocities and the final velocities lie in a plane.