Momentum and Impulse

8.01 W05D1

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Exam 1 Re-grade Policy

If you would like any problems re-graded on your exam, please write a note on the cover sheet indicated which questions you would like re-graded and your reason. Submit the exam for re-grading no later then the following class, W05D2. Some exams have been photocopied so under no circumstances should you change any of your original answers. Any altered exams will automatically be graded zero.

Announcements

Exam 1 Regrade Requests due W05D2 Experiment 2 Report due Week 4 Tuesday 9 pm Week 5 Day 2 Prepset due Wed 8:30 am Sunday Tutoring in 26-152 from 1-5 pm Problem Set 4 due 9 pm Tuesday Week 6

Momentum and Impulse

•  Obeys a conservation law

•  Simplifies complicated motions

•  Describes collisions

•  Basis of rocket propulsion & space travel

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Momentum and Force: Single Particle

•  Momentum

SI units

•  Change in momentum •  Force

m=p v

mΔ = Δp v [kg ⋅m ⋅ s-1] = [N ⋅ s]

F = ma = m dv

dt=

dpdt

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Impulse and Change in Momentum: Single Particle

•  Impulse

•  Average force •  Impulse SI units •  Momentum Law: Impulse equals change in momentum

[N ⋅ s]

I ≡

F dt

ti

t f

∫ =FaveΔt =

Fave(t f − ti )

I ≡

F dt

ti

t f

∫ = dpdt

dtti

t f

∫ = d pti

t f

∫ = Δp ≡ p(tf ) − p(ti )

Fave ≡

1(t f − ti )

F dt

ti

t f

∫

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Concept Question: Pushing Identical Objects

Identical constant forces push two identical objects A and B continuously from a starting line to a finish line. Neglect friction. If A is initially at rest and B is initially moving to the right, 1.  Object A has the larger change in

momentum. 2.  Object B has the larger change in

momentum. 3.  Both objects have the same change in

momentum 4.  Not enough information is given to

decide.

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Demo B61: Impulse of Bouncing Balls

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Group Problem: Bouncing Ball A superball of m, starting at rest, is dropped from a height hi above the ground and bounces back up to a height of hf. The collision with the ground occurs over a time interval Δtc. Ignore air resistance. a)  What are the external forces acting on the ball during the

collision? b)  What is the momentum of the ball immediately before the

collision? c)  What is the momentum of the ball immediately after the collision? d)  What is the external impulse acting on the ball? 9

Group Problem: Bouncing Ball Solution

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!

Choose positive direction up. External forces: Kinematics for falling motion implies: Momentum before collision: Kinematics for rising motion implies: Momentum after collision: Momentum Law for Collision: External Impulse:

vy ,1 = − 2hig

p(t1) = m 2ghi (− j)

vy ,2 = 2hf g

p(t2 ) = m 2ghf j

!Nave − mg j= Δ!p

Δt=!p(t2 )− !p(t1)

Δtc

= m 2ghf + hi

Δtc

j⇒

!Fave =

!Nave − mg j

!I = (!Nave − mgj)Δtc = m 2ghf + 2ghi( )( ) j

Concept Question: Impulse

The figure to the right depicts the paths of two colliding steel balls, A and B. Which of the arrows 1-5 best represents the impulse applied to ball B by ball A during the collision?

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System of Particles: Newton’s Second Law

The momentum of a system of N particles is defined as the sum of the individual momenta of the particles

Force changes the momentum of the system

!F =

!Fi

i=1

N

∑ =d!pi

dti=1

N

∑ ≡d!psys

dt

!psys ≡!pi

i=1

N

∑ = mi!v i

i=1

N

∑

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Internal Force on a System of Particles is Zero

The internal force on the ith particle is sum of the internal interaction forces with all the other particles The sum of the internal forces on each particle in the system is Newton’s Third Law: internal forces cancel in pairs So the sum of the internal forces on the system is zero

Fint =

0

Fi, j = −

Fj ,i

!Fint,i =

!Fj ,i

j=1i≠ j

j=N

∑

!Fint =

!Fint,i

i=1

N

∑ =!Fj ,i

j=1i≠ j

N

∑i=1

N

∑

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Newton’s Second and Third Laws for System of Particles

Newton’s Third Law: The force on a system of particles is the external force because internal forces cancel in pairs

Newton’s Second Law: The total external force is equal to the rate of change of the momentum of the system

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F =Fext +

Fint =

Fext

Fext =

dpsys

dt

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Conservation of Momentum

For a fixed choice of system, if there are no external forces acting on the system then the momentum of the system is constant is constant. This is a vector equation (a separate equation for each direction).

Fext =

0 ⇒Δpsys =

0⇒ psys, f =

psys,i

Strategy: Momentum of a System 1. Choose system 2. Identify initial and final states 3. Identify any external forces in order to determine whether any

component of the momentum of the system is constant or not i) If there is a non-zero total external force:

ii) If the total external force is zero then momentum is constant iii) If the collision time is approximately zero,

Fext = dpsys / dt

psys,i =

psys,f

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0coltΔ

(Fext )aveΔtcol = Δpsys ⇒Δpsys ≅

0

Group Problem: Landing Plane and Sandbag

A light plane of mass m1 makes an emergency landing on a short runway. With its engine off, it lands on the runway at a speed of vi. A hook on the plane snags a cable attached to a sandbag of mass m2 and drags the sandbag along. If the coefficient of friction between the sandbag and the runway is µ, and if the plane’s brakes give an additional retarding force of magnitude F, how far does the plane go before it comes to a stop?

Position of Center of Mass of System of Particles

Mass for collection of N discrete bodies (system) is a constant: Position of center of mass

Rcm ≡ 1

msys

mjrj

j=1

j=N

∑

msys = mj

j=1

j=N

∑

..

.mjrjmsys

Rcm..

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Continuous Bodies: Center of Mass

Infinitesimal mass element Mass for system: Position vector for infinitesimal element Position of center of mass

rj →

r msys = mj

j=1

j=N

∑ → dmbody∫

mj → dm =ρdV , volume element

σ dA , area elementλds , length element

⎧

⎨⎪

⎩⎪

Rcm = 1

msys

mj

rjj=1

j=N

∑ → 1msys

dmrbody∫

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Worked Example: Center of Mass of Uniform Rod

Choose a coordinate system with the rod aligned along the x-axis and origin located at the left end of the rod. Center of mass: Choose an infinitesimal mass element dm located a distance x’ from the origin. In this problem, x’ will be the integration variable. Let the length of the mass element be dx’ . Then dm=(M/L)dx’. The vector . Then the center of mass of the uniform rod is

!

!Rcm = 1

M!r dm

body∫ = 1

L′x d ′x

′x =0

x

∫ i = 12L

′x 2

′x =0

′x =Li = 1

2L(L2 − 0) i = L

2i

Rcm = 1

Mr dm

body∫

r = ′x i

Velocity of center of mass Momentum of system:

Vcm =

dRcm

dt= 1

msys

mj

drj

dtj=1

j=N

∑Vcm = 1

msys

mjv j

j=1

j=N

∑

Velocity and Momentum of Center of Mass

psys = mjv j

j=1

j=N

∑ = msys

Vcm

..

.mj

msys.. v j

Vcmcm

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Group Problem: Center of Mass of Rod and Particle Rod Collision

A slender uniform rod of length d and mass m rests along the x-axis on a frictionless, horizontal table. A particle of equal mass m is moving along the x-axis at a speed v0. At t = 0, the particle strikes the end of the rod and sticks to it. Find a vector expression for the position of the center of mass of the system for (i) t = 0, (ii) t > 0. The figure on the right shows an overhead view of the rod lying on the table.

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Newton’s Second and Third Laws: Center of Mass Motion

The external force is equal to the change in momentum of the system and is proportional to the acceleration of the center of mass.

Fext =

dpsys

dt=

d(msys

Vcm )

dt= msys

Acm

Concept Q.: Center of Mass Motion

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The same force is applied separately to each of the three objects shown above with masses satisfying m1 > m2 > m3. Which of the following statements is true? 1.  In all cases the center of mass of the system has the same acceleration. 2.  In all cases the acceleration of the center of mass of the system are different. 3.  The acceleration of the center of mass of the system is largest in figure (1). 4.  The acceleration of the center of mass of the system is largest in figure (2).

5.  The acceleration of the center of mass of the system is largest in figure (3).

cm1

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F Figure 1

cm1

23F

Figure 2

cm1

23 F

Figure 3

Demo : Center of Mass Trajectory B78

Odd-shaped objects with their centers of mass marked are thrown. The centers of mass travel in a smooth parabola. The objects consist of: a squash racket, a 16” diameter disk weighted at one point on its outer rim, and two balls connected with a rod. This demonstration is shown with UV light.

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http://tsgphysics.mit.edu/front/?page=demo.php&letnum=B 78&show=0

High Jump: Fosbury Flop

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The jumper’s center of mass follows a parabolic trajectory of a point moving in a uniform gravitational field

Video link: Fosbury 1968 Olympics https://www.youtube.com/watch?v=Id4W6VA0uLc

Group Problem: Center of Mass Motion of Exploding Projectile

An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is x0. The projectile breaks into two pieces which fly apart horizontally. The larger piece, m3, has three times the mass of the smaller piece, m2. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station. a)  How far has the center of mass of the system traveled from the

launch when the pieces hit the ground? b)  How far from the launch point has the larger piece traveled when

it first hits the ground? 27