The net force acting on an object is the
rate of change of its momentum:
The General form of Newton’s 2nd law
If the net force is zero, the total momentum does not change: Newton’s 1st Law
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 (Newton’s 3rd law). 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 ONLY External forces acting on it.
Conservation of Linear Momentum
Internal forces cannot change the momentum of a system.
The momenta of individual components in the system
may change, however this change will be exactly
balanced by other components in the system.
OK, so what is the change in momentum?
(a) a 0.10‐kg beanbag bear is dropped to the floor, where it hits with a speed of 4.0 m/s and sticks. (b) 0.10‐kg rubber ball also hits the floor with a speed of 4.0 m/s, but in this case the ball bounces upward off the floor. Assuming an ideal rubber ball, its initial upward speed is 4.0 m/s. Momentum is a vector quantity, possessing a direction as
well as a magnitude.
Crash Course in Momentum
Change in momentum example illustrated by this cartoon of a car crash test
Case 2
Case 1
-15.0 m/s
-17.0 m/s
0.0 m/s
3.0 m/s
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Just as the Work Done to an object transforms Energy,
the Impulse to an object can transfer momentum
Impulse is a vector, in the same direction as the average force.
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Momentum allows us to characterize collisions into two categories. Inelastic
Elastic
KE is not conserved What kind of collisions do you usually have here? KE is conserved Oh, we got both kinds. We got elastic *and* inelastic. Inelastic Collisions
0
Before
m1
m2
What is the final velocity?
After
m1
m2
total momentum cannot change
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Before
m1
m2
What is the final velocity?
After
m1
m2
Elastic Collisions
Both kinetic energy and momentum are conserved.
One‐dimensional elastic collision:
Elastic Collisions
We have two equations (conservation of momentum and conservation of kinetic energy) and two unknowns (the final speeds). Solving for the final speeds:
Elastic Collisions in 2D
Two‐dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object:
A proton collides elastically with
another proton that is initially at rest.
v1i
The incoming proton has an initial
speed of 3.5x105 m/s and makes a
glancing collision with the 2nd proton.
After the collision, one proton moves off at an angle of θ= 37 degrees to the original direction of motion and the 2nd deflects at an angle of φ to the same axis. y
x
Find the final speeds of the two protons and the angle φ.
Conservation of momentum in 2D
Conservation of energy in 2D
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Find the final speeds of the two protons and the angle φ.
v1i
Conservation of momentum in 2D
0
0
v1fx = v1f cosθ
v2fx = v2f cosφ
in the Y direction
Conservation of energy in 2D
0
v1fy = v1f sinθ
v2fx = v2f sinφ
Conservation of momentum in the X direction
0
Conservation of energy 14
Rearrange the momentum equations
Square both equations and add them
One possible solution is a head‐on collision in which the first proton stops. That is not the solution we want. So we divide both sides by V1f
Energy conservation relationship
Substitute this relationship into the energy equation
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Reference copy from previous slides
To determine φ we use the Y component of momentum equation
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