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7-6 Inelastic Collisions
With inelastic collisions, some of
the initial kinetic energy is lost to
thermal or potential energy. It
may also be gained during
explosions, as there is the
addition of chemical or nuclear
energy.
A completely inelastic collision is
one where the objects stick
together afterwards, so there is
only one final velocity.
Example 7-9
For the completely inelastic collision of two railroad cars that we considered in
Example 7-3, calculate how much of the initial kinetic energy is transformed to
thermal or other forms of energy.
Before the collision,
the total kinetic energy is
1
1
mA v 2A = (10,000. kg)(24.0 m/s) 2 = 2.88x10 6 J
2
2
After the collision, the total kinetic energy is
1
1
(mA + mB )v' 2 = (20,000. kg)(12.0 m/s) 2 = 1.44x10 6 J
2
2
The energy transformed is KE i - KE f
2.88x10 6 J -1.44x10 6 J = 1.44x10 6 J, half of the initial KE

Example 7-10
The ballistic pendulum is a device used to measure the speed of a
projectile, such as a bullet. The projectile, of mass m, is fired into a large
block (of wood or other material) of mass M, which is suspended like a
pendulum. (Usually, M is somewhat greater than m.) As a result of the
collision, the pendulum and projectile together swing up to a maximum
height h. Determine the relationship between the initial horizontal speed of
the projectile, v, and the maximum height h.
(a) total p before impact = total p after impact
mv = (m +M)v'
(b) E just after impact = E at max height
1
(m +M)v' 2 +0 = 0 +(m +M)gh
2
v' = 2gh
We insert this value for v' into our equation in
part (a) to get
m +M
m +M
v=
v' =
2gh
m
m
7-7 Collisions in Two or Three Dimensions
Conservation of energy and momentum can also
be used to analyze collisions in two or three
dimensions, but unless the situation is very
simple, the math quickly becomes unwieldy.
Here, a moving object
collides with an object
initially at rest. Knowing
the masses and initial
velocities is not enough;
we need to know the
angles as well in order to
find the final velocities.
7-7 Collisions in Two or Three Dimensions
Problem solving:
1. Choose the system. If it is complex,
subsystems may be chosen where one or
more conservation laws apply.
2. Is there an external force? If so, is the
collision time short enough that you can
ignore it?
3. Draw diagrams of the initial and final
situations, with momentum vectors labeled.
4. Choose a coordinate system.
7-7 Collisions in Two or Three Dimensions
5. Apply momentum conservation; there will be
one equation for each dimension.
6. If the collision is elastic, apply conservation
of kinetic energy as well.
7. Solve.
8. Check units and magnitudes of result.
Example 7-11
Billiard ball A moving with speed vA=3.0 m/s in the +x direction strikes an
equal-mass ball B initially at rest. The two balls are observed to move off
at 45 degrees to the x axis, ball A above the x axis and ball B below.
What are the speeds to the two balls after the collision?
for x : mv A = mv 'Acos(45) + mv 'Bcos(-45)
for y : 0 = mvsin(45) + mvsin(-45)
From the y equation, we get
'
' sin(45)
' sin45
v B = -vA
= -vA
= v'A
sin(-45)
-sin45
Using this in the x equation, we have
v A = v'Acos45 + v'Bcos45 = 2v'Acos45
vA
3.0 m/s
v'A = v'B =
=
= 2.1 m/s
2cos45 2(0.707)
7-8 Center of Mass
In (a), the diver’s motion is pure translation; in (b)
it is translation plus rotation.
There is one point that moves in the same path a
particle would
take if subjected
to the same force
as the diver. This
point is called the
center of mass
(CM).
7-8 Center of Mass
The general motion of an object can be
considered as the sum of the translational
motion of the CM, plus rotational, vibrational, or
other forms of motion about the CM.
7-8 Center of Mass
For two particles, the center of mass lies closer
to the one with the most mass:
where M is the total mass.
Example 7-12
Three people of roughly equal masses m on a lightweight (air-filled)
banana boat sit along the x axis at positions xA=1.0 m, xB=5.0 m, and
xC=6.0 m, measured from the left-hand end. Find the position of the CM.
Ignore the mass of the boat.
mx A + mx B + mx C m(x A + x B + x C )
=
m +m +m
3m
(1.0 m + 5.0 m + 6.0 m) 12.0 m
x CM =
=
= 4.0 m
3
3
The CM is 4.0 m from the left - hand end of the boat.
x CM =
7-8 Center of Mass
The center of gravity is the point where the
gravitational force can be considered to act. It is
the same as the center of mass as long as the
gravitational force does not vary among different
parts of the object.
7-8 Center of Mass
The center of gravity can be found experimentally
by suspending an object from different points.
The CM need not be within the actual object – a
doughnut’s CM is in the center of the hole.
7-9 CM for the Human Body
The x’s in the small diagram mark the CM of
the listed body segments.
7-9 CM for the Human Body
The location of the center of
mass of the leg (circled) will
depend on the position of
the leg.
7-9 CM for the Human Body
High jumpers have
developed a technique
where their CM actually
passes under the bar as
they go over it. This allows
them to clear higher bars.
Example 7-13
Determine the position of the CM of a whole leg (a) when stretched out,
and (b) when bent at 90 degrees. Assume the person is 1.70 m tall.
(21.5)(9.6) + (9.6)(33.9) + (3.4)(50.3)
= 20.4 units
21.5 + 9.6 + 3.4
So the CM is 20.4 units from the hip joint, or 52.1 - 20.4
= 31.7 units from the foot. The person is 1.70 m tall,
this
is (1.70 m)(31.7/100) = 0.54 m above the bottom of the foot.
(21.5)(9.6) + (9.6)(23.6) + (3.4)(23.6)
(b) x CM =
= 14.9 units
21.5 + 9.6 + 3.4
(3.4)(1.8) + (9.6)(18.2) + (21.5)(28.5)
y CM =
= 23.0 units
21.5 + 9.6 + 3.4
So the CM is (1.70 m)(14.9/100) = 0.25 m from the hip joint
and (1.70 m)(23.0/100) = 0.39 m above the floor.
(a) x CM =

7-10 Center of Mass and Translational Motion
The total momentum of a system of particles is
equal to the product of the total mass and the
velocity of the center of mass.
The sum of all the forces acting on a system is
equal to the total mass of the system multiplied
by the acceleration of the center of mass:
(7-11)
7-10 Center of Mass and Translational Motion
This is particularly useful in the analysis of
separations and explosions; the center of
mass (which may not correspond to the
position of any particle) continues to move
according to the net force.
Summary of Chapter 7
• Momentum of an object:
• Newton’s second law:
•Total momentum of an isolated system of objects is
conserved.
• During a collision, the colliding objects can be
considered to be an isolated system even if external
forces exist, as long as they are not too large.
• Momentum will therefore be conserved during
collisions.
Summary of Chapter 7, cont.
•
• In an elastic collision, total kinetic energy is
also conserved.
• In an inelastic collision, some kinetic energy
is lost.
• In a completely inelastic collision, the two
objects stick together after the collision.
• The center of mass of a system is the point at
which external forces can be considered to
act.
Homework - Ch.7
• Questions #’s 3, 4, 5, 7, 12, 15, 19
• Problems #’s 3, 5, 7, 11, 15, 17, 23, 25,
27, 31, 35, 41, 49, 57