Physics 207: Lecture 2 Notes

Transcript Physics 207: Lecture 2 Notes

Lecture 15

 Goals  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: 2 nd Exam, Monday, March 19 th , 7:15 to 8:45 PM Physics 201: Lecture 15, Pg 1

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) Physics 201: Lecture 15, Pg 2

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 Before because there is no net x force. This is not true of the y-direction but this is what we are interested in.

After Physics 201: Lecture 15, Pg 3

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 (3 rd 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 Physics 201: Lecture 15, Pg 4

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 p x before = p x after = 0 = - M V + m v V = m v / M = 20*30/ 60 = 10 m/s T before = Weight = (60+20) x 10 N = 800 N Before After S F y = m a cy = M V 2 /r = T – Mg T = Mg + MV 2 /r = 600 N + 60x(10) 2 /20 N = 900 N Physics 201: Lecture 15, Pg 5

Exercise Momentum is a Vector (!) quantity  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

conserved ?

block landing in the cart is momentum A.

Yes B.

No C.

Yes & No D.

Too little information given Physics 201: Lecture 15, Pg 6

Exercise Momentum is a Vector (!) quantity   x-direction: No net force so P x is conserved.

y-direction: Net force, interaction with the ground so depending on the system (i.e., do you include the Earth?) p y is not conserved (system is block and cart only) 2 kg 5.0 m 30 ° 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 7.5 m 10 kg What is the final velocity of the cart?

Physics 201: Lecture 15, Pg 7

Exercise Momentum is a Vector (!) quantity   x-direction: No net force so P x is conserved y-direction: v y of the cart + block will be zero and we can ignore v y of the block when it lands in the cart.

j N i 5.0 m Initial Final P x : MV x + mv x M 0 + mv x V ’ x = (M+m) V ’ x = (M+m) V = m v x ’ x / (M + m) = 2 (8.7)/ 12 m/s V ’ x = 1.4 m/s 30 ° y mg 30 ° 7.5 m x 1) a i = g sin 30 ° = 5 m/s 2 2) d = 5 m / sin 30 ° = ½ a i D t 2 10 m = 2.5 m/s 2 D t 2 2s = D t v = a i D t = 10 m/s v x = v cos 30 ° = 8.7 m/s Physics 201: Lecture 15, Pg 8

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,

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

 The impulse measures momentum transfer Physics 201: Lecture 15, Pg 9

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

a vector that reflects momentum transfer

  



F dt

 

( 

d p

)

 



d p F

Impulse

= area under this curve !

(Transfer of momentum !) Impulse has units of Newton-seconds

t i

t t f

Physics 201: Lecture 15, Pg 10

Force and Impulse 

Two different collisions can have the same impulse since depends only on the

momentum transfer

, NOT the nature of the collision.

F F

same area D

big,

small

t t

small,

big Physics 201: Lecture 15, Pg 11

Average Force and Impulse

F av

big,

F av

small

F av

F t

small,

F av

big Physics 201: Lecture 15, Pg 12

Exercise

Force & Impulse  Two boxes, one heavier than the other, are initially at rest on a horizontal frictionless surface. The same constant force

acts on each one for exactly 1 second .

Which box has the most momentum after the force acts ?

F light F heavy

heavier B.

lighter C.

same can’t tell Physics 201: Lecture 15, Pg 13

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

m 1 + m 2 m 1 m 2

before after  If no external force momentum is conserved.

 Momentum is a vector so p x , p y and p z Physics 201: Lecture 15, Pg 15

A perfectly inelastic collision in 2-D  If no external force momentum is conserved.

 Momentum is a vector so p x , p y and p z are conseved

V v

m 1 + m 2 m 1 m 2

before  x-dir p x : m 1 v 1  y-dir p y : m 2 v 2 = (m 1 = (m 1 + m 2 + m 2 after ) V cos q ) V sin q Physics 201: Lecture 15, Pg 16

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 Physics 201: Lecture 15, Pg 17

Billiards  Consider the case where one ball is initially at rest. after before

F P

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

Physics 201: Lecture 15, Pg 18

Billiards: Without external forces, conservation of both momentum & mech. energy    Conservation of Momentum x-dir P y-dir P x y : : m v before = m v after 0 = m v after cos sin q q + m V after + m V after cos f sin f before after

after

F P

after

f If the masses of the two balls are equal then there will always be a 90 ° angle between the paths of the outgoing balls Physics 201: Lecture 15, Pg 19

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?

Physics 201: Lecture 15, Pg 20

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.

1 +

2 mobile Physics 201: Lecture 15, Pg 21

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

individual point-like particles whose masses and positions we know:

1 (In this case,

N = 2

)

2 

r CM



i N

  1

m i



r i i N

  1

m i



1 

1 

2 



3 

  Physics 201: Lecture 15, Pg 22

Momentum of the center-of-mass is just the total momentum  Notice 

M v CM



d dt



v r



r CM CM

   ( 1

M d dt

M i N

  1

m i r



( 1

M i N

  1

i N

  1

m i m i d dt



r i



r i

) ) 

p CM



i N

  1

m i



v i

 

1  

2  

3  ...

 Impulse and momentum conservation applies to the center-of-mass Physics 201: Lecture 15, Pg 23

Sample calculation:  Consider the following mass distribution:

 CM 

i N

  1

m M i





CM iˆ 

CM jˆ 

CM kˆ

= (m

x 0 + 2m x 12 + m x 24 )/4m meters R CM = (12,6)

= (m

x 0 + 2m x 12 + m x 0 )/4m meters (12,12)

2m X

CM = 12 meters

CM = 6 meters

m m

(0,0) (24,0) Physics 201: Lecture 15, Pg 24

A classic example

 There is a disc of uniform mass and radius

. However there is a hole of radius

a distance

(along the x-axis) away from the center.  Where is the center of mass for this object?

  (  green

hole disk CM   CM 



2 (

, 0 ) 0 , 0 )

CM  0

m m

  0 ( (     ) )  

a a

2 2  0

CM  0



( (     ) )  

a a

2 2  

m bma



2 2 / / 2

r r

2 

 2



2 2 Physics 201: Lecture 15, Pg 25

System of Particles: Center of Mass  For a continuous solid, convert sums to an integral.

r  CM   



dm dm

 



dm M y

where

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

Physics 201: Lecture 15, Pg 26

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

Physics 201: Lecture 15, Pg 27