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WORK, ENERGY AND POWER 131

to 10 3 m s –1 so that it can have a high 6.12.3 Collisions in Two Dimensions
probability of interacting with isotope 235 U Fig. 6.10 also depicts the collision of a moving
92 with the stationary mass m . Linear
and causing it to fission. Show that a mass m 1 2
neutron can lose most of its kinetic energy momentum is conserved in such a collision.
in an elastic collision with a light nuclei Since momentum is a vector this implies three
like deuterium or carbon which has a mass equations for the three directions {x, y, z}.
of only a few times the neutron mass. The Consider the plane determined by the final
material making up the light nuclei, usually velocity directions of m and m and choose it to
1
2
heavy water (D O) or graphite, is called a be the x-y plane. The conservation of the
2
moderator. z-component of the linear momentum implies
that the entire collision is in the x-y plane. The
Answer The initial kinetic energy of the neutron x- and y-component equations are
is m v = m v cos θ + m v cos θ (6.29)
1 1i 1 1f 1 2 2f 2
1
K = m v 2 0 = m v sin θ − m v sin θ (6.30)
1i 1 1i 1 1f 1 2 2f 2
2
One knows {m , m , v } in most situations. There
while its final kinetic energy from Eq. (6.27) 1 2 1i
are thus four unknowns {v , v , θ and θ }, and
1f 2f 1 2
1 2 1  m − m  2 2 only two equations. If θ 1 = θ 2 = 0, we regain

K = m v = m 1 2 v Eq. (6.24) for one dimensional collision.
1 f 1 1 f 1   i 1
2 2  m + m 
1
2
If, further the collision is elastic,
1 2 1 2 1 2
The fractional kinetic energy lost is m v = m v + m v (6.31)
1 1f
2 2 f
1 1i
2 2 2
K  m − m  2
f = 1 f =  1 2  We obtain an additional equation. That still
1
K i 1  m + m  leaves us one equation short. At least one of
1
2
while the fractional kinetic energy gained by the the four unknowns, say θ , must be made known
1
moderating nuclei K /K is for the problem to be solvable. For example, θ 1
2f 1i
can be determined by moving a detector in an
f = 1 − f (elastic collision)
2 1 angular fashion from the x to the y axis. Given
4m m {m , m , v , θ } we can determine {v , v , θ }
1f
2f
2
1
1
1i
2
= 1 2 2 from Eqs. (6.29)-(6.31).
(m + m 2 )
1
Example 6.13 Consider the collision
t
One can also verify this result by substituting depicted in Fig. 6.10 to be between two
from Eq. (6.28). billiard balls with equal masses m = m .
2
1
For deuterium m = 2m and we obtain The first ball is called the cue while the
2 1
f = 1/9 while f = 8/9. Almost 90% of the second ball is called the target. The
1 2
neutron’s energy is transferred to deuterium. For billiard player wants to ‘sink’ the target
carbon f = 71.6% and f = 28.4%. In practice, ball in a corner pocket, which is at an
1 2
however, this number is smaller since head-on angle θ = 37°. Assume that the collision
2
collisions are rare. t is elastic and that friction and rotational
If the initial velocities and final velocities of motion are not important. Obtain θ .
1
both the bodies are along the same straight line, Answer From momentum conservation, since
then it is called a one-dimensional collision, or the masses are equal
head-on collision. In the case of small spherical
bodies, this is possible if the direction of travel v = v + v 2f
1i
1f
of body 1 passes through the centre of body 2
or 2 )
⋅ v
which is at rest. In general, the collision is two- v 1i = v + v 2 f ) ( 1 f + v 2 f
( 1f
dimensional, where the initial velocities and the = 2 + 2 +
final velocities lie in a plane. v 1 f v 2 f 2v 1 f .v 2 f
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