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Momentum
Momentum—Inertia in Motion
• Momentum = mass x velocity or momentum = mv
22 km/h
5225 kg
75 km/h
1540 kg
Which has greater momentum?
Impulse Changes Momentum
The greater the force acting on an object, the
greater its velocity and momentum.
Time is also a factor in determining momentum.
How long is force applied?
Impulse = force x time interval or impulse = Ft
Impulse = change in momentum
Note: The difference between impulse and impact force
involves the time the force acts.
Ft  (mv)
Egg drop challenge
Increasing Momentum
To increase momentum, force must be applied
as hard as possible for as long as possible.
This is why follow through is important in a golf
swing or in baseball.
Large force applied for a long time = maximum impulse
Impact measured in N. Impulse measured in N-s
Decreasing Momentum
To decrease the force of impact, the impact time
(time during which momentum is brought to
zero) must be lengthened.
Decreasing Momentum
To decrease the force of impact, the impact time
(time during which momentum is brought to
zero) must be lengthened.
• A 0.50-kg cart (#1) is pulled with a 1.0N force for 1 second; another 0.50 kg
cart (#2) is pulled with a 2.0 N-force
for 0.50 seconds.
– Which cart (#1 or #2) has the greatest
acceleration? Explain.
– Which cart (#1 or #2) has the greatest
impulse? Explain.
– Which cart (#1 or #2) has the greatest
change in momentum? Explain.
Decreasing Momentum
Decreasing Momentum
Decreasing Momentum
Decreasing Momentum
Momentum (mv) is what is gained by the
Jumper until the chord begins to stretch.
Decreasing Momentum
Momentum (mv) is what is gained by the
Jumper until the chord begins to stretch.
Ft is the impulse the cord supplies to reduce
the momentum to zero.
Because of the long time it takes the cord to
stretch, the average force (F) on the jumper
is minimal.
Decreasing Momentum
HELP!!!
Bouncing
Impulses are greater when
an object bounces.
Flower pot on the head
Pelton Wheel
Conservation of Momentum
• Law of conservation of momentum:
In the absence of an external force, the momentum of a system
remains unchanged.
The bullet gains momentum and so does
the rifle, but the rifle-bullet system gains none.
Collisions
• When objects collide in the absence of external forces,
the net momentum of both objects before the collision
equals the net momentum of both objects after the
collision.
net momentum
before collision
= net momentum
after collision
Elastic collision—when objects collide without being
deformed permanently or generating heat.
Inelastic collision—when colliding objects become tangled
or coupled together
Elastic Collisions
• The sum of the momentum vectors are
the same before and after the collisions.
Inelastic Collisions
• If mass is equal, the momentum is shared
equally after the collision by each of the
objects.
V = 12 m/s
V=0
V = 6 m/s
Inelastic Lunch
5 kg
v=0
3 m/s
20 kg
Inelastic Lunch
(net mv) before = (net mv) after
(20 kg)(3 m/s)+(5 kg)(0 m/s) = (25 kg)(v after)
(60 kg m/s)+(0) = (25 kg)(v after)
(60 kg m/s) = (25 kg)(v after)
2.4 m/s = (v after)
5 kg
3 m/s
25 kg
2.4 m/s
20 kg
Inelastic Lunch
(net mv) before = (net mv) after
(20 kg)(3 m/s)+(5 kg)(-2 m/s) = (25 kg)(v after)
(60 kg m/s)+(-10 kg m/s) = (25 kg)(v after)
(50 kg m/s) = (25 kg)(v after)
2 m/s = (v after)
5 kg
2 m/s
3 m/s
20 kg
Inelastic Lunch
(net mv) before = (net mv) after
(20 kg)(3 m/s)+(5 kg)(-2 m/s) = (25 kg)(v after)
(60 kg m/s)+(-10 kg m/s) = (25 kg)(v after)
(50 kg m/s) = (25 kg)(v after)
2 m/s = (v after)
5 kg
v=0
2 m/s
25 kg
Inelastic Lunch
(net mv) before = (net mv) after
(20 kg)(3 m/s)+(5 kg)(-10 m/s) = (25 kg)(v after)
(60 kg m/s)+(-50 kg m/s) = (25 kg)(v after)
(10 kg m/s) = (25 kg)(v after)
.4 m/s = (v after)
5 kg
10 m/s
3 m/s
20 kg
Inelastic Lunch
(net mv) before = (net mv) after
(20 kg)(3 m/s)+(5 kg)(-10 m/s) = (25 kg)(v after)
(60 kg m/s)+(-50 kg m/s) = (25 kg)(v after)
(10 kg m/s) = (25 kg)(v after)
.4 m/s = (v after)
5 kg
v=0
.4 m/s
25 kg
Inelastic Lunch
Solve this problem on a sheet of loose
leaf paper.
Show all work. Write the answer at the
bottom along with
the appropriate velocity vector.
50 kg
10 kg
7 m/s
5 m/s
Momentum Vectors
2  length of a side
(diagonal of a square)