General Physics IGeneral Physics I

Lecture 6: Conservation of Lecture 6: Conservation of MomentumMomentum

Prof. WAN, Xin （万歆）

xinwan@zju.edu.cnhttp://zimp.zju.edu.cn/~xinwan/

OutlineOutline

● Importance of conservation laws in physics− Conservation of linear momentum

● Applications of the conservation laws− Collisions

● Many-particle systems− Center of mass− Collisions in the center-of mass frame

A Question From YouA Question From You

● We have learn the work-kinetic energy theorem. What happens to it under Galileo transformation?

(1) Nothing happens, we get the exactly same work-kinetic energy theorem.

(2) Since velocity and displacement are reference frame dependent, the work-kinetic energy theorem only works at an inertial frame at rest (w.r.t. fixed stars).

(3) Some additional law will emerge to ensure the work-kinetic energy theorem valid in a different inertial frame.

Galilean Transformation Galilean Transformation

The two inertial observers agree on measurements of acceleration.

Let's Work it outLet's Work it out

W '=F d'=F (d−v0t)=W−v0F t

Assume an object in 1D subject to a total force F for simplicity.

K '=12mv'2=

12m(v−v0)

2=K−v0mv+

12mv0

2

According to the work-kinetic energy theorem, W=K f−K i

W '−(K ' f−K 'i)=W−(K f−K i)−v0F t+v0m(v f−vi)

W '=(K ' f−K 'i) ⇔ F t=mv f−mvi

Impulse-Momentum TheoremImpulse-Momentum Theorem

● The impulse of the force F acting on a particle equals the change in the linear momentum of the particle caused by that force I = Dp.

− Linear momentum

− Impulse

Connection to Newton's LawsConnection to Newton's Laws

● Assume that a single force F acts on a particle and that this force may vary with time

Two-Particle SystemTwo-Particle System

● Consider two particles 1 and 2 that can interact with each other but are isolated from their surroundings.

● Newton's second law

● Newton's third law

Conservation of Linear MomentumConservation of Linear Momentum

● Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant.

● The total momentum of an isolated system at all times equals its initial momentum.

● The only requirement is that the forces must be internal to the system.

A Question for YouA Question for You

● In an isolated system the validity of Galileo invariance and the conservation of kinetic energy (and mass) can lead to the conservation of linear momentum. Can you derive it? Discuss whether your derivation is still valid when some kinetic energy converts into internal energy.

Inelastic scattering involves internal energy change.

m1v1i+m2v2i = m1v1f+m2v2f

12m1v1i

2+12m2v2i

2=

12m1v1f

2+12m2v2f

2+ Δεinternal

Conservation LawsConservation Laws

● A conservation law is usually the consequence of some underlying symmetry in the universe.

● Why are conservation laws powerful tools?− Can assure something impossible. − Applicable even when the force is unknown.− An intimate connection with invariance. − Convenient in solving for the particle motion.

● First use the relevant conservation laws one by one, then differential equations, computers, etc.

Rutherford ScatteringRutherford Scattering

The Structure of AtomThe Structure of Atom

The Standard ModelThe Standard Model

Large Hadron Collider (LHC)Large Hadron Collider (LHC)

Search for Higgs BosonSearch for Higgs Boson

The top event in the CMS experiment shows a decay into two photons (dashed yellow lines and green towers).

The lower event in the ATLAS experiment shows a decay into 4 muons (red tracks).

CollisionCollision

● The event of two particles’ coming together for a short time and thereby producing impulsive forces on each other.

− These forces are assumed to be much greater than any external forces present.

Elastic and Inelastic CollisionElastic and Inelastic Collision

● Momentum is conserved in any collision in which external forces are negligible.

● Kinetic energy may or may not be constant.− Elastic collision between two objects is one in which

total kinetic energy (as well as total momentum) is the same before and after the collision.

− Inelastic collision is one in which total kinetic energy is not the same before and after the collision (even though momentum is constant).

− Kinetic energy is constant only in elastic collisions.

Perfectly Inelastic CollisionsPerfectly Inelastic Collisions

● When the colliding objects stick together after the collision, the collision is called perfectly inelastic.

Ballistic PendulumBallistic Pendulum

Two-Body Collision with a SpringTwo-Body Collision with a Spring

Elastic Collision in 1DElastic Collision in 1D

m1v1i+m2v2i = m1v1f+m2v2f

12m1v1i

2+12m2v2i

2=

12m1v1f

2+12m2v2f

2

Show that

Stress RelieverStress Reliever

Special CasesSpecial Cases

● Particle 2 initially at rest:

● Equal masses: m1=m2

v1f=v2i , v2f=v1i

v2i=0

(1) m1≫m2

(2) m1≪m2

v1f≈v1i , v2f≈2v1i

v1f≈−v1i , v2f≈v2i=0

Elastic Collision in 2DElastic Collision in 2D

We need one more equation to solve.

Example: Proton CollisionExample: Proton Collision

Whenever two equal masses collide elastically in a glancing collision and one of them is initially at rest, their final velocities are always at right angles to each other.

We are going to show:

An Equivalent ProblemAn Equivalent Problem

Conservation of energy

Conservation of momentum

The Center of MassThe Center of Mass

Many-Particle SystemsMany-Particle Systems

Consider a system of many particles in three dimensions

We can think of an extended object as a system containing a large number of particles

The Center of Mass of a RodThe Center of Mass of a Rod

The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry.

● Show that the center of mass of a rod of mass M and length L lies midway between its ends, assuming the rod has a uniform mass per unit length.

xCM =1M

∫0

L

xλdx

=λ L2

2M=

L2

= λM

x2

2 ∣0

L

Nonuniform RodNonuniform Rod

● Suppose a rod is nonuniform such that its mass per unit length varies linearly with x according to

λ=αx

The CM of a Right TriangleThe CM of a Right Triangle

Motion of a Many-Particles SystemMotion of a Many-Particles System

Collision in the CM FrameCollision in the CM Frame

● Perfectly inelastic scattering

vCM =m1v1i+m2v2i

m1+m2

u1i = v1i−vCM =m2

m1+m2(v1i−v2i )

u2i = v2i−vCM =m1

m1+m2(v2i−v1i )

u1f = u2f = 0

In the center-of-mass frame, the total momentum is zero.

m1u1i+m2u2i = m1u1f+m2u2f = 0

Collision in the CM FrameCollision in the CM Frame

● Elastic scattering vCM =m1v1i+m2v2i

m1+m2

u1i = v1i−vCM =m2

m1+m2(v1i−v2i )

u2i = v2i−vCM =m1

m1+m2(v2i−v1i )

u1f = v1f−vCM =m2

m1+m2(v2i−v1i )

u2f = v2f−vCM =m1

m1+m2(v1i−v2i )

12m1u1i

2+12m2u2i

2=

12m1u1f

2+12m2u2f

2 Conservation of kinetic energy

Collision in the CM FrameCollision in the CM Frame