Momentum and Impulse

8.01 W06D2

Associated Reading Assignment:Young and Freedman: 8.1-8.5

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Momentum and Impulse

Obeys a conservation law

Simplifies complicated motions

Describes collisions

Basis of rocket propulsion & space travel

DEFINITION II Newton’s Principia

The quantity of motion is the measure of the same, arising from the velocity and quantity of matter conjointly.

The motion of the whole is the sum of the motions of all the parts; and therefore in a body double in quantity, with equal velocity, the motion is double; with twice the velocity, it is quadruple.

Quantity of Motion

Momentum and Impulse: Single Particle

• Momentum

SI units

• Change in momentum

• Impulse

• SI units

m=p vr r

mΔ = Δp vr r

f

i

t

t

dt≡∫I Fr r

[kg ⋅m ⋅s-1] =[N ⋅s]

[N ⋅s]

Newton’s Second Law“The change of motion is proportional to the motive force impresses, and is made in the direction of the right line in which that force is impressed”,

rI ≡

rFdt

ti

tf

∫ =drp

dtdt

ti

tf

∫ = d rp

ti

tf

∫ =Δrp ≡

rp(tf) −

rp(ti)

rF =

drp

dt=m

drvdt

=m ra

ddt

= =p

F 0r rr

When then

then constant (vector)m= =p vr r

rp(tf ) =

rp(ti)

Concept Question: Pushing Identical Carts

Identical constant forces push two identical objects A and B continuously from a starting line to a finish line. 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.

Concept Question: Pushing Non-identical Carts Kinetic Energy

Consider two carts, of masses m and 2m, at rest on an air track. If you push one cart for 3 s and then the other for the same length of time, exerting equal force on each, the kinetic energy of the light cart is

1) larger than 2) equal to 3) smaller than

the kinetic energy of the heavy car.

Concept Question: Same Momentum, Different Masses

Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the distances needed to stop them compare?

1. It takes a shorter distance to stop the ping-pong ball.

2. Both take the same distance.

3. It takes a longer distance to stop the ping-pong ball.

Demo: Jumping Off the Floor

with a Non-Constant Force

Demo Jumping: Non-Constant Force

• Plot of total external force vs. time for Andy jumping off the floor. Weight of Andy is 911 N.

Demo Jumping: Impulse• Shaded area represents impulse of total force

acting on Andy as he jumps off the floor

1.23 stotal

0.11s

[ , ] ( ) 199 N sf

i

t

i ft

t t t dt=

=

= = ⋅∫I Fr r

total ( ) ( ) gravt t= +F N Fr r r

Demo Jumping: Height• When Andy leaves the

ground, the impulse is

• So the y-component of the velocity is

I y[0.11 s,1.23 s] =199 N ⋅s

Vy (t) =Iy[ti,t] / m =(199 N ⋅s)(9.80 m ⋅s-2 )/(911 N) =2.14 m ⋅s-1

System of Particles: Center of Mass

Position and Velocity of Center of Mass

Total mass for discrete or continuous body (mass density ρ)

Position of center of mass

Velocity of center of mass1sys sys

1 1i N

cm i ii body

m dVm m

r=

=

= =∑ ∫R r rr r r

sys

1sys sys sys

1 1i N

cm i ii body

m dVm m m

r=

=

= = =∑ ∫p

V v vrr r r

sys1

i N

ii body

m m dVr=

=

= =∑ ∫

Table Problem: Center of Mass of Rod and Particle

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 at t = 0.

System of Particles: Internal and External Forces,

Center of Mass Motion

Internal Force on a System of N Particles

• The total internal force on the ith particle is sum of the interaction forces with all the other particles

• The total internal force is the sum of the total internal force on each particle

• Newton’s Third Law: internal forces cancel in pairs

• So the total internal force is zero total

int =F 0rr

, ,i j j i=−F Fr r

int, ,1

j N

i i jji j

=

=≠

=∑F Fr r

int

1total

int, ,1 1

j Ni

i i ji j

i j

==

= =≠

= =∑ ∑F F Fr r r

Total Force on a System of N Particles is the External Force

The total force on a system of particles is the sum of the total external and total internal forces. Since the total internal force is zero

total total total totalext int ext= + =F F F F

r r r r

External Force and Momentum Change

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

Total force changes the momentum of the system

Total force equals total external force

Newton’s Second and Third Laws for a system of particles: The total external force is equal to the change in momentum of the system

systotal total

1 1

i N i Ni

ii i

dddt dt

= =

= =

= = ≡∑ ∑pp

F Frrr r

total totalext=F F

r r

systotalext

ddt

=p

Frr

sys1

i N

ii

=

=

≡∑p pr r

System of Particles: Newton’s Second and Third Laws

The total momentum of a system remains constant unless the system is acted on by an external force

systotalext

ddt

=p

Frr

System of Particles: Translational Motion of the Center of

Mass

Translational Motion of the Center of Mass

Momentum of system

System can be treated as point mass located at center of mass. External force accelerates center of mass

Impulse changes center of mass momentum

sys sys cmm=p Vrr

systotalext sys sys

cmcm

d dm m

dt dt= = =

p VF A

rr rr

totalext sys sys f( ( ) ( ))

f

i

t

cm cm it

dt m t t≡ =Δ = −∫I F p V Vr r r rr

Demo : Center of Mass trajectory B78

http://tsgphysics.mit.edu/front/index.php?page=demo.php?letnum=B%2078&show=0

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.

CM moves as though all external forces on the system act on the CM

so the jumper’s cm follows a parabolic trajectory of a point moving in a uniform gravitational field

Concept Question: Pushing a Baseball Bat Recall Issue with

Phrasing

1 32

The greatest acceleration of the center of mass will be produced by pushing with a force F at1. Position 12. Position 23. Position 34. All the same

Table Problem: Exploding Projectile Center of Mass Motion

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 graveled when it first hits the ground?

System of Particles:Conservation of Momentum

External Forces and Constancy of Momentum Vector

The external force may be zero in one direction but not others

The component of the system momentum is constant in the direction that the external force is zero

The component of system momentum is not constant in a direction in which external force is not zero

Table Problem: Center of Mass of Rod and Particle Post- 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 t > 0.

Strategy: Momentum of a System1. 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 rFext

total =dr

psys

dt

sys,0 sys,f=p pr r

Modeling: Instantaneous Interactions

• Decide whether or not an interaction is instantaneous.

• External impulse changes the momentum of the system.

• If the collision time is approximately zero,

then the change in momentum is approximately zero.

[ , ] ( )colt t

col ext ext ave col syst

t t t dt t+Δ

+ Δ = = Δ =Δ∫I F F pr r r r

systemΔ ≅p 0rr

0coltΔ ;

Table Problem: Landing Plane and Sandbag

A light plane of mass 1000 kg makes an emergency landing on a short runway. With its engine off, it lands on the runway at a speed of 40 ms-1. A hook on the plane snags a cable attached to a sandbag of mass 120 kg and drags the sandbag along. If the coefficient of friction between the sandbag and the runway is μ = 0.4, and if the plane’s brakes give an additional retarding force of magnitude 1400 N, how far does the plane go before it comes to a stop?