lecture 15

Physics 201: Lecture 15, Pg 1

Lecture 15 GoalsGoals

Employ conservation of momentum in 1 D & 2D

Introduce Momentum and Impulse

Compare Force vs time to Force vs distance

Introduce Center-of-Mass

Note: 2nd Exam, Monday, March 19th, 7:15 to 8:45 PM


Comments on Momentum Conservation

More general than conservation of mechanical energy

Momentum Conservation occurs in systems with no net external forces (as a vector quantity)


Explosions: A collision in reverse A two piece assembly is hanging vertically

at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg is ejected horizontally at 30 m/s. The time of the “explosion” is short compared to the swing of the string.

Does the tension in the string increase or decrease after the explosion?

If the time of the explosion is short then momentum is conserved in the x-direction because there is no net x force. This is not true of the y-direction but this is what we are interested in.

Before After


Explosions: A collision in reverse

A two piece assembly is hanging vertically at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 and 20 kg respectively. Suddenly you observe that the 20 kg mass is ejected horizontally at 30 m/s.

Decipher the physics:

1. The green ball recoils in the –x direction (3rd Law) and, because there is no net external force in the x-direction the x-momentum is conserved.

2. The motion of the green ball is constrained to a circular path…there must be centripetal (i.e., radial acceleration)

Before After


Explosions: A collision in reverse

A two piece assembly is hanging vertically at rest at the end of a 20 m long massless string. The mass of the two pieces are 60 & 20 kg respectively. Suddenly you observe that the 20 kg mass is suddenly ejected horizontally at 30 m/s.

Cons. of x-momentum

px before= px after = 0 = - M V + m v

V = m v / M = 20*30/ 60 = 10 m/s

Tbefore = Weight = (60+20) x 10 N = 800 N

Fy = m acy = M V2/r = T – Mg

T = Mg + MV2 /r = 600 N + 60x(10)2/20 N = 900 N

AfterBefore


Exercise Momentum is a Vector (!) quantity

A. Yes

B. No

C. Yes & No

D. Too little information given

A block slides down a frictionless ramp and then falls and lands in a cart which then rolls horizontally without friction

In regards to the block landing in the cart is momentum conserved?



Let a 2 kg block start at rest on a 30° incline and slide vertically a distance 5.0 m and fall a distance 7.5 m into the 10 kg cart

What is the final velocity of the cart?

x-direction: No net force so Px is conserved. y-direction: Net force, interaction with the ground so

depending on the system (i.e., do you include the Earth?)

py is not conserved (system is block and cart only)

5.0 m30°

7.5 m

10 kg

2 kg



Initial Final

Px: MVx + mvx = (M+m) V’x M 0 + mvx = (M+m) V’x

V’x = m vx / (M + m) = 2 (8.7)/ 12 m/s

V’x = 1.4 m/s

x-direction: No net force so Px is conserved y-direction: vy of the cart + block will be zero and

we can ignore vy of the block when it lands in the cart.

5.0 m

30°

7.5 m

N

mg

1) ai = g sin 30°

= 5 m/s2

2) d = 5 m / sin 30°

= ½ ai t2

10 m = 2.5 m/s2 t2

2s = t

v = ai t = 10 m/s

vx= v cos 30°

= 8.7 m/s

i

j

x

y

30°


Impulse (A variable external force applied for a given time)

Collisions often involve a varying force

F(t): 0 maximum 0 We can plot force vs time for a typical collision. The

impulse, II, of the force is a vector defined as the integral of the force during the time of the collision.

The impulse measures momentum transfer


Force and Impulse (A variable force applied for a given time)

F

ptt pddtdtpddtFI )/(

J a vector that reflects momentum transfer

t

ti tf

t

Impulse I = area under this curve !

(Transfer of momentum !)

Impulse has units of Newton-seconds


Force and Impulse Two different collisions can have the same impulse since

I depends only on the momentum transfer, NOT the nature of the collision.

t

F

t

F

tt

same area

t big, FF smallt small, FF big


Average Force and Impulse

t

F

t

F

tt

t big, Fav smallt small, Fav big

Fav

Fav


Exercise Force & Impulse

A. heavier

B. lighter

C. same

D. can’t tell

Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force F acts on each one for exactly 1 second.

Which box has the most momentum after the force acts ?

F F light heavy


A perfectly inelastic collision in 2-D

Consider a collision in 2-D (cars crashing at a slippery intersection...no friction).

vv1

vv2

VV

before after

m1

m2

m1 + m2

If no external force momentum is conserved. Momentum is a vector so px, py and pz


A perfectly inelastic collision in 2-D

vv1

vv2

VV

before after

m1

m2

m1 + m2

x-dir px : m1 v1 = (m1 + m2 ) V cos y-dir py : m2 v2 = (m1 + m2 ) V sin

If no external force momentum is conserved. Momentum is a vector so px, py and pz are conseved


2D Elastic Collisions

Perfectly elastic means that the objects do not stick and, by stipulation, mechanical energy is conservsed.

There are many more possible outcomes but, if no external force, then momentum will always be conserved

Before After


Billiards

Consider the case where one ball is initially at rest.

ppa

ppb

FF PPa

beforeafter

The final direction of the red ball will depend on where the balls hit.

vvcm


Billiards: Without external forces, conservation of both momentum & mech. energy

Conservation of Momentum x-dir Px : m vbefore = m vafter cos + m Vafter cos y-dir Py : 0 = m vafter sin + m Vafter sin

ppafter

ppb

FF PPafter

before after

If the masses of the two balls are equal then there will always be a 90° angle between the paths of the outgoing balls


Center of Mass

Most objects are not point-like but have a mass density and are often deformable.

So how does one account for this complexity in a straightforward way?

Example

In football coaches often tell players attempting to tackle the ball carrier to look at their navel.

So why is this so?


System of Particles: Center of Mass (CM)

If an object is not held then it will rotate about the center of mass. Center of mass: Where the system is balanced !

Building a mobile is an exercise in finding

centers of mass.

m1m2

+m1 m2

+

mobile


System of Particles: Center of Mass

How do we describe the “position” of a system made up of many parts ?

Define the Center of Mass (average position): For a collection of N individual point-like particles whose

masses and positions we know:

(In this case, N = 2) y

x

r2r1

m1m2

RCM

M

rmrmrm

m

rmr N

ii

N

iii

CM

332211

1

1


Momentum of the center-of-mass is just the total momentum

Notice

N

iiiMCM rmr

1

1

)(1

1

N

iiiMdt

dCMdt

d rmr

)(1

1

N

iidt

diMCM rmv

...3211

pppvmpvMN

iiiCMCM

Impulse and momentum conservation applies to the center-of-mass


Sample calculation:

Consider the following mass distribution:

(24,0)(0,0)

(12,12)

m

2m

m

RCM = (12,6)

kji CM CM CM1

CM ZYXM

mN

iii

r

r

XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters

YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters

XCM = 12 meters

YCM = 6 meters


A classic example There is a disc of uniform mass and radius r.

However there is a hole of radius a a distance b (along the x-axis) away from the center.

Where is the center of mass for this object?

)0,0(CMdisk green r

)0,(CM hole br

2rm

22

2

22

22

2

2

//

)()(0

CM abba

rmamrbma

amabmx

02

2

)()(00

CM

amamy


System of Particles: Center of Mass

For a continuous solid, convert sums to an integral.

y

x

dm

rr

where dm is an infinitesimal mass element (see text for an example).

M

dmr

dm

dmr

rCM


Recap

Thursday, Review for exam For Tuesday, Read Chapter 10.1-10.5

lecture 15

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