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Momentum Conservation
Conservation of Linear Momentum
The net force acting on an object is the rate of
change of its momentum:
If the net force is zero, the momentum does not
change:
Conservation of Linear Momentum
Internal Versus External Forces:
Internal forces act between objects within the system.
As with all forces, they occur in action-reaction pairs. As all
pairs act between objects in the system, the internal forces
always sum to zero:
Therefore, the net force acting on a system is the sum of
the external forces acting on it.
Conservation of Linear Momentum
Furthermore, internal forces cannot change the momentum of
a system.
However, the momenta of components of the system may
change.
Conservation of Linear Momentum
An example of internal forces moving
components of a system:
Canoe Example
Two groups of canoeists meet in the
middle of a lake. After a brief visit, a
person in Canoe 1 (total mass 130
kg) pushes on Canoe 2 (total mass
250 kg) with a force of 46 N to
separate the canoes.
Find the momentum of each canoe
after 1.20 s of pushing.
a1, x
F


1, x
m1

46 N
 0.354 m/s 2
130 kg
a2, x
F


v1, x  a1, xt  (0.354 m/s2 )(1.20 s)  0.425 m/s
2, x
m2

46 N
 0.184 m/s 2
250 kg
v2, x  a2, xt  (184 m/s2 )(1.20 s)  0.221 m/s
p1, x  m1v1, x  (130 kg)(0.425 m/s)  55.3 kg  m/s
p2, x  m2 v2, x  (250 kg)(0.221 m/s)  55.3 kg  m/s
Net momentum = 0
Example: Velocity of a Bee
A honeybee with a mass of 0.150 g lands on one end of a floating 4.75 g
popsicle stick. After sitting at rest for a moment, it runs to the other end of
the stick with a velocity vb relative to still water. The stick moves in the
opposite direction with a velocity of 0.120 cm/s.
Find the velocity vb of the bee.
ps  ms vs
Conservation of x-axis momentum: ps  pb  0
pb  mb vb   ps  ms vs
ms
(4.75 g)
vb   vs  
(0.120 cm/s)  3.80 cm/s
mb
(0.150 g)
Inelastic Collisions
Collision: two objects striking one another.
Time of collision is short enough that external
forces may be ignored.
Inelastic collision: momentum is conserved but
kinetic energy is not: pf = pi but Kf ≠ Ki.
Completely inelastic collision: objects stick
together afterwards: pf = pi1 + pi2
Inelastic Collisions
A completely inelastic collision:
m1v1,i  m2v2,i  (m1  m2 )v f
Example: Goal-Line Stand
On a touchdown attempt, a 95.0 kg
running back runs toward the end
zone at 3.75 m/s. A 111 kg line backer
moving at 4.10 m/s meets the runner
in a head-on collision and locks his
arms around the runner.
(a) Find their velocity immediately
after the collisions.
(b) Find the initial and final kinetic energies and
the energy DK lost in the collision.
m1v1,i  m2v2,i (95.0 kg)(3.75 m/s)  (111.0 kg)( 4.10 m/s)
vf 

 0.480 m/s
m1  m2
(95.0 kg)  (111.0 kg)
Ki  12 m1v1,2i  12 m2v2,2 i  12 (95.0 kg)(3.75 m/s)2  12 (111.0 kg)(4.10 m/s)2  1600 J
K f  12  m1  m2  v2f  12 (95.0 kg)  (111.0 kg) (0.480 m/s)2  23.7 J
DK  1576 J
Inelastic Collisions
Ballistic pendulum: the height h can be found using
conservation of mechanical energy after the object
is embedded in the block.
Example: Ballistic Pendulum
A projectile of mass m
is fired with an initial
speed v0 at the bob of a
pendulum. The bob has
mass M and is suspended
by a rod of negligible mass.
After the collision the projectile and bob stick together and swing at speed
vf through an arc reaching height h.
Find the height h.
Momentum Conservation: mv0   m  M  v f
Energy Conservation: E  12  m  M  v2f   m  M  gh
Inelastic Collisions
For collisions in two dimensions, conservation of
momentum is applied separately along each axis:
Example: A Traffic Accident
A car of mass m1 = 950 kg and a speed v1,i = 16
m/s approaches an intersection. A minivan of
mass m2 = 1300 kg and speed v2,i = 21 m/s
enters the same intersection. The cars collide
and stick together.
Find the direction q and final speed vf of the
wrecked vehicles just after the collision.
x-momentum: m1v1  (m1  m2 )v f cos q
y-momentum: m2 v2  (m1  m2 )v f sin q
m2 v2 (m1  m2 )v f sin q sin q


 tan q
m1v1 (m1  m2 )v f cos q cos q
m2 v2
(1300 kg)(21 m/s)
 arctan
 61
m1v1
(950 kg)(16 m/s)
m1v1
(950 kg)(16 m/s)
vf 

 14 m/s
(m1  m2 ) cos q (950 kg)  (1300 kg) cos 61
q  arctan
Explosions
An explosion in which the particles of a system move apart from each
other after a brief, intense interaction, is the opposite of a collision. The
explosive forces, which could be from an expanding spring or from
expanding hot gases, are internal forces. If the system is isolated, its total
momentum will be conserved during the explosion, so the net momentum of
the fragments equals the initial momentum.
Elastic Collision
31.
• A 722-kg car stopped at an intersection is rear-ended by a 1620-kg
truck moving with a speed of 14.5 m/s. If the car was in neutral and its brakes
were off, so that the collision is approximately elastic, find the final speed of both
vehicles after the collision.
Pbefore  Pafter
0
mc vc  mt vt  mc vc'  mt vt'
K before  K after