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352 CHAPTER 11 Collisions

In an atomic collision experiment, or “scattering” experiment,
EXAMPLE 8
a helium ion of mass m 4.0 u with speed v 1200 m s
1 1
strikes an oxygen (O ) molecule of mass m 32 u which is initially at rest (see
2 2
Fig. 11.10a). The helium ion exits the collision at 90 from its incident direction
with one-fourth of its original kinetic energy.What is the recoil speed of the oxygen
molecule? What fraction of the total kinetic energy is lost during the collision?
[This energy is lost to the internal (vibrational and rotational) motions of the
oxygen molecule.]
SOLUTION: In the absence of external forces, momentum is always conserved. If
we choose the direction of incident motion along the x axis, then for 90 scatter-
ing, we can choose the direction in which the helium ion exits (the direction of v' )
1
to be along the y axis (see Fig. 11.10b). Conservation of momentum in the two
directions then requires

for x direction: m v m v'
1 1 2 2x
for y direction: 0 m v' m v'
1 1 2 2y
Since the helium ion exits with one-fourth of its initial kinetic energy,
1 2 1 1 2
2 m v' m v
4
2
1 1
1 1
or
1
v' v
1 2 1
Substituting this v' and the given m 8m into the x and y components of the
1 2 1
momentum gives for the velocity of the oxygen molecule:
m 1 1 1
v' v v 1200 m s 150 m s
2x m 1 1
2 8 8
m 1 1 1 1
v' v' v 1200 m/s 75 m/s
2y m 1 1
2 8 2 16
The speed of recoil of the oxygen molecule is thus
2 2 2 2
v' 2v' v' 2(150 m s) ( 75 m s) 170 m s
2 2x 2y

(a) (b)
Helium ion
v' 1 exits at 90°.

y To conserve momentum,
velocity of the oxygen
molecule must have +x
and –y components.
He + x O O x
v 1
Helium ion is v'
a projectile. Oxygen molecule is 2 x
a stationary target. v' 2 y
v' 2

FIGURE 11.10 (a) A helium ion with velocity v v i is moving toward a stationary oxygen molecule.
1x
1
(b) After the collision, the helium ion exits perpendicular to its incident direction with velocity v' , while the oxygen
1
molecule acquires a velocity v' v' i v' j.
2 2x 2y