Environmental Engineering Reference
In-Depth Information
(b) For the case where the cars stick together we work in the centre-of-mass
frame. The centre-of-mass velocity, as measured in the lab, is (using Eq. (3.34))
1
2
m
(mv
v
2
.
V
=
+
m
×
0
)
=
Hence
v
2
v
2
,
v
1
c
=
v
−
=
v
2
=−
v
2
.
v
2
c
=
0
−
If the cars stick together then all of the kinetic energy in the centre-of-mass frame is
lost (presumably into deforming the material that sticks the cars together) and we
have
v
1
c
=
v
2
c
=
0
.
Thus
1
2
2
m
−
2
m
v
2
1
v
2
1
4
mv
2
.
Q
=
0
−
+
=−
2
Example 3.3.2
An experiment measures the elastic scattering of a beam of particles
from a stationary target. The beam and target particles have equal mass. Show that
the angle between the final velocity vectors is
90
◦
, as long as the final velocities are
both non-zero.
Solution 3.3.2
The initial state is similar to that described in the previous example,
although we must be careful with the vector nature of the velocities. Momentum
conservation now gives
v
1
+
v
2
=
v
,
where
v
is the velocity of the beam particles and
v
1
and
v
2
are the final velocity
vectors. Taking the scalar product of each side with itself (“squaring”) gives
v
1
2
v
1
v
2
cos
φ
v
2
v
2
,
+
+
=
where φ is the angle between the final velocity vectors. Invoking energy conservation
for an elastic collision we have
v
1
v
2
v
2
+
=
and therefore
2
v
1
v
2
cos
φ
=
0
.
Since both final speeds are non-zero we must have
cos
φ
0
and we conclude that
the final velocity vectors form a right-angle in the lab frame.
=
