Example 7-7 Drama in the Skies
The photograph that opens this chapter shows a Cooper’s hawk attacking a pigeon from behind. Each bird has a mass of
0.600 kg. The pigeon is gliding due west at 8.00 m>s at a constant altitude, and the Cooper’s hawk is diving on the pigeon
at a speed of 12.0 m>s at an angle of 30.0° below the horizontal. Just after the Cooper’s hawk grabs the pigeon in its talons,
how fast and in what direction are the two birds moving?
Set Up
This is a completely inelastic collision
because the two birds move together afterward. Unlike Example 7-6, this collision takes
place in two dimensions, so we will need to
consider more than one component of Equavf of the
tion 7-18 to find the final velocity s
two birds after the collision.
Momentum conservation in a
before
completely inelastic collision:
mpigeon s
vpigeon, i + mhawk s
vhawk, i
sf = 1mpigeon + mhawk 2v
(7-18)
hawk
30.0°
vhawk,i = 12.0 m/s
pigeon
vpigeon,i = 8.00 m/s
after
hawk + pigeon
O=?
vf = ?
Solve
We choose the positive x axis to be horizontal
and to the west, and we choose the positive y
axis to be upward. We write the components of
the initial velocities of the hawk and pigeon.
Pigeon is initially moving in the positive x direction,
so
vpigeon, ix = vpigeon, i = +8.00 m>s
vpigeon, iy = 0
y
x
Hawk is initially moving in the positive x direction
and the negative y direction, so
vhawk, ix = vhawk, i cos 30.0 = 112.0 m>s2 cos 30.0
= +10.4 m>s
vhawk, iy = -vhawk, i sin 30.0 = - 112.0 m>s2 sin 30.0
= -6.00 m>s
Use conservation of momentum in component
form to find the components of the final
velocity of the two birds.
Equation for conservation of x momentum in the collision:
mpigeon vpigeon, ix + mhawk vhawk, ix = (mpigeon + mhawk)vfx
Solve for the final x velocity of the two birds:
vfx =
=
mpigeon vpigeon, ix + mhawk vhawk, ix
mpigeon + mhawk
10.600 kg2 1 +8.00 m>s2 + 10.600 kg2 1 +10.4 m>s2
= +9.20 m>s
0.600 kg + 0.600 kg
Equation for conservation of y momentum in the collision:
mpigeon vpigeon, iy + mhawk vhawk, iy = (mpigeon + mhawk)vfy
Solve for the final y velocity of the two birds:
vfy =
=
mpigeon vpigeon, iy + mhawk vhawk, iy
mpigeon + mhawk
10.600 kg2 10 m>s2 + 10.600 kg2 1 -6.00 m>s2
= -3.00 m>s
0.600 kg + 0.600 kg
vpigeon,i
30.0°
vhawk,i
Given the components of the final velocity
vf , use trigonometry to find the
vector s
vf .
magnitude and direction of s
y
Speed after collision = magnitude of final
velocity vector:
vf = 2v 2fx + v 2fy
2
= 21 +9.20 m>s2 + 1 -3.00 m>s2
= 9.67 m>s
2
x
O = –18.1°
vf = 9.67 m/s
Direction of final velocity vector:
vfy
-3.00 m>s
= -0.326
tan u =
=
vfx
+ 9.20 m>s
u = arctan(20.326) = 218.1°
After the collision, the two birds move at 9.67 m>s
in a direction 18.1° below the horizontal.
Reflect
The impact causes the hawk to slow down from 12.0 m>s to 9.67 m>s, and causes the pigeon to speed up from 8.00 m>s
to 9.67 m>s. This is just what we would expect when a slow-moving pigeon is hit from behind by a fast-moving hawk.