In Progress
Lesson 13, Topic 2
In Progress

# Collisions in One Dimension

Lesson Progress
0% Complete

Consider first a completely inelastic collision in one dimension. Then, in Fig. 10,

θ1 = θ2 = 0

m1v1i = (m1 +m2)vf (momentum conservation)

vf = [m1 /(m1+m2)]v1i ……………………………………………………..(23)

The loss in kinetic energy on collision is

ΔK = 1/2m1(v1i)2-1/2 (m1+m2)(v1f)2

=1/2m1(v1i)2-1/2 (m12/m1+m2)(v1i)2 using Eq. (23)

=1/2m1(v1i)2[1-m1 /(m1+m2)]

= 1/2 [(m1m2)/(m1+m2)](v1i)2

which is a positive quantity as expected.

Consider next an elastic collision. Using the above nomenclature with θ1 = θ2 = 0, the momentum and kinetic energy conservation equations are

m1v1i = (m1 v1f+m2v2f ) ……………………………………………………….(24)

m1(v1i)2 = (m1 (v1f)2+m2 (v2f)2) …………………………………………….(25)

From Eqs. (24) and (25) it follows that,

m1v1i (v2f – v1i ) = m1 v1f (v2f – v1f )

Or, v2f (v1i – v1f )= (v1i )2 – (v1f )2

= (v1i – v1f )((v1i + v1f )

Hence, v2f = (v1i – v1f ) ………………………………………………………………………..(26)

Substituting this in Eq. (24), we obtain

v1f = [(m1-m2)/(m1+m2)]v1i …………………………………………………………………(27)

v2f = 2m1v1i/(m1+m2) …………………………………………………………………(28)

Thus, the ‘unknowns’ {v1f , v2f } are obtained in terms of the unknowns {m1, m2, v1i}. Special cases of our analysis are interesting.

Case I : If the two masses are equal

v1f = 0

v2f = v1f

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. m2 » m1

v1f ≅ -v1i

v2f ≅ 0

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

Example 12
Slowing down of neutrons: In a nuclear reactor a neutron of high speed (typically 107 m s -1) must be slowed to 103 m s -1 so that it can have a high probability of interacting with isotope and causing it to fission. Show that 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 (D2O) or graphite, is called a moderator.

Ans.: The initial kinetic energy of the neutron is

K1i = 1/2 m1 (v1i)2

while its kinetic energy from Eq. (27)

K1f = 1/2 m1 (v1f)2=1/2m1 [(m1 – m2)/(m1 + m2)]2(v1i)2

The fractional kinetic energy lost is

f1 = K1f / K1i = [(m1 – m2)/(m1 + m2)]2

while the fractional kinetic energy gained by the moderating nuclei K2f / K1i is

f2 = 1 – f1 (elastic collision)

= (4m1m2)/(m1 + m2 )2

One can also verify this result by substituting from Eq. (28).

For deuterium m2 = 2m1 and we obtain f1 = 1/9 while f2 = 8/9. Almost 90% of the neutron’s energy is transferred to deuterium. For the carbon f1 = 71.6% and f2 = 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 two dimensional, where the initial velocities and the final velocities lie in a plane. 