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Transcript
Elastic Collisions in One Dimension
β€’ For an elastic collision, both momentum and mechanical
energy are conserved.
β€’ In one dimension,
π‘šπ΄ 𝑣𝐴1π‘₯ + π‘šπ΅ 𝑣𝐡1π‘₯ = π‘šπ΄ 𝑣𝐴2π‘₯ + π‘šπ΅ 𝑣𝐡2π‘₯
1
2
2π‘šπ΄ 𝑣𝐴1π‘₯
2
2
2
+ 12π‘šπ΅ 𝑣𝐡1π‘₯
= 12π‘šπ΄ 𝑣𝐴2π‘₯
+ 12π‘šπ΅ 𝑣𝐡2π‘₯
β€’ Given the masses and initial velocities, we can solve for the final
velocities.
β€’ We find that the relative velocities before and after the
collision are equal magnitude but opposite sign:
𝑣𝐴1π‘₯ βˆ’ 𝑣𝐡1π‘₯ = βˆ’ 𝑣𝐴2π‘₯ βˆ’ 𝑣𝐡2π‘₯
Elastic Collisions in One Dimension
β€’ For the special case where one
body is initially at rest, the
velocities after the collision
reduce to:
π‘šπ΄ βˆ’ π‘šπ΅
𝑣𝐴2π‘₯ =
𝑣𝐴1π‘₯
π‘šπ΄ + π‘šπ΅
𝑣𝐡2π‘₯
2π‘šπ΄
=
𝑣𝐴1π‘₯
π‘šπ΄ + π‘šπ΅
Collision Example #2
Center of Mass
β€’ For a collection of particles, we can define the center of
mass as the mass-weighted average position of the
particles.
π‘š1 π‘₯1 + π‘š2 π‘₯2 + π‘š3 π‘₯3 + β‹―
𝑖 π‘šπ‘– π‘₯𝑖
π‘₯cm =
=
π‘š1 + π‘š2 + π‘š3 + β‹―
𝑖 π‘šπ‘–
π‘š1 𝑦1 + π‘š2 𝑦2 + π‘š3 𝑦3 + β‹―
𝑖 π‘šπ‘– 𝑦𝑖
𝑦cm =
=
π‘š1 + π‘š2 + π‘š3 + β‹―
𝑖 π‘šπ‘–
β€’ In vector form,
π‘š1 𝒓1 + π‘š2 π’“πŸ + π‘š3 𝒓3 + β‹―
𝒓cm =
=
π‘š1 + π‘š2 + π‘š3 + β‹―
β€’ The sum of the masses is the total mass 𝑀.
𝑀=
π‘šπ‘–
𝑖
𝑖 π‘šπ‘– 𝒓𝑖
𝑖 π‘šπ‘–
Center of Mass Example
Center of Mass of Solid Bodies
β€’ To find the center of
mass of a solid body, we
replace the sums with
integrals.
β€’ For uniform bodies, the
center of mass is the
geometric center.
β€’ The center of mass will
lie on an axis of
symmetry, if there is one.
β€’ The center of mass does
not have to be within the
body.
Motion of the Center of Mass
β€’ Taking the time derivative of the
center of mass position gives the
center of mass velocity.
π‘š1 𝒗1 + π‘š2 π’—πŸ + π‘š3 𝒗3 + β‹―
𝒗cm =
π‘š1 + π‘š2 + π‘š3 + β‹―
𝑖 π‘šπ‘– 𝒗𝑖
𝒗cm =
𝑀
β€’ We then see that the total
momentum is the total mass times
the velocity of the center of mass.
𝑀𝒗cm = π‘š1 𝒗1 + π‘š2 π’—πŸ + π‘š3 𝒗3 + β‹―
𝑷 = 𝑀𝒗cm
External Forces and Motion of the Center of Mass
β€’ Similarly, taking the time derivative of the center of mass
velocity gives the center of mass acceleration.
π‘š1 𝒂1 + π‘š2 π’‚πŸ + π‘š3 𝒂3 + β‹―
𝑖 π‘šπ‘– 𝒂𝑖
𝒂cm =
=
π‘š1 + π‘š2 + π‘š3 + β‹―
𝑖 π‘šπ‘–
β€’ The sum of all of the forces, internal and external is equal
to the mass times acceleration.
β€’ By Newton’s third law, for each internal force, there is an equal
but opposite reaction force, so the internal forces cancel out.
Σ𝑭ext = 𝑀𝒂cm
β€’ So, the collection of particles moves as if all of the mass
were located at the center of mass, and were acted on by
only the external forces.
External Forces and Motion of the Center of Mass
β€’ We can also express this in terms of momentum.
𝑑𝑷
Σ𝑭ext =
𝑑𝑑
β€’ Only external forces can change the total momentum.
β€’ This is a restatement of conservation of momentum.
Rocket Propulsion
β€’ In outer space, there is no atmosphere, so how do space
ships maneuver?
β€’ Rocket propulsion
β€’ Conservation of momentum
β€’ Burned fuel is ejected at a high velocity from the rear of
the rocket, propelling it forward.
β€’ The mass of the rocket is therefore decreasing.
Rocket Propulsion
β€’ Consider a rocket in outer space (no gravity or air
resistance) with a mass π‘š and velocity of magnitude 𝑣.
β€’ In time 𝑑𝑑, the rocket ejects burned fuel and its mass changes by
π‘‘π‘š, a negative quantity.
β€’ A positive mass of βˆ’π‘‘π‘š is ejected at a relative velocity of βˆ’π‘£ex
β€’ From conservation of momentum, we can find the thrust,
or force exerted on the rocket, and its acceleration.
β€’ The mass of the rocket is therefore decreasing.
Rocket Propulsion
β€’ The thrust is given by:
π‘‘π‘š
𝐹 = βˆ’π‘£ex
𝑑𝑑
β€’ From Newton’s second law, the
acceleration in the absence of any
other forces, such as gravity, is:
𝑣ex π‘‘π‘š
π‘Ž=βˆ’
π‘š 𝑑𝑑
β€’ Integrating, we find that the change in
velocity is given by:
π‘š0
𝑣 βˆ’ 𝑣0 = 𝑣ex ln
π‘š
Rocket Propulsion Example
Chapter 8 Summary
Momentum, Impulse, and Collisions
β€’ Momentum: 𝒑 = π‘šπ’—
β€’ Newton’s second law:
β€’ Impulse:
𝑭=
𝑑𝒑
𝑑𝑑
𝑑2
𝑱 = 𝒑2 βˆ’ 𝒑1 = 𝑑 Σ𝑭 𝑑𝑑 =
1
𝑭ave Δ𝑑
β€’ Conservation of momentum
β€’ No external forces: 𝑷 = π‘šπ΄ 𝒗𝐴 + π‘šπ΅ 𝒗𝐡 + β‹― = constant
β€’ Collisions – Momentum is conserved
β€’ Elastic – Kinetic energy is conserved
β€’ Inelastic – Loss of kinetic energy
Chapter 8 Summary
Momentum, Impulse, and Collisions
β€’ Center of mass
β€’ 𝒓cm
=
π‘š1 𝒓1 +π‘š2 π’“πŸ +π‘š3 𝒓3 +β‹―
π‘š1 +π‘š2 +π‘š3 +β‹―
𝑑𝑷
β€’ Σ𝑭ext = 𝑀𝒂cm = 𝑑𝑑
β€’ Rocket propulsion
β€’ Thrust: 𝐹 = βˆ’π‘£ex
β€’ 𝑣 βˆ’ 𝑣0 = 𝑣ex ln
π‘š0
π‘š
π‘‘π‘š
𝑑𝑑
=
𝑖 π‘šπ‘– 𝒓𝑖
𝑖 π‘šπ‘–