9.4 Two-Dimensional 9.4 Two-Dimensional CollisionsCollisions

Two-Dimensional Two-Dimensional CollisionsCollisions

The momentum is conserved in all momentum is conserved in all directionsdirections

Use subscripts for identifying the object indicating initial or final values the velocity components

If the collision is elasticcollision is elastic, use conservation conservation of kinetic energyof kinetic energy as a second equation Remember, the simpler equation can only be used

for one-dimensional situations

Two-Dimensional Collision, Two-Dimensional Collision, 22

Qualitative AnalysisQualitative Analysis Physical Principles:Physical Principles: The same as in One-The same as in One-

DimensionDimension 1. Conservation of 1. Conservation of VECTORVECTOR momentum: momentum:

PP1x1x + P + P2x2x = P = P1x1x + P + P2x2x && PP1y1y + P + P2y2y= P= P1y1y + P + P2y2y 2. Conservation of Kinetic Energy2. Conservation of Kinetic Energy

½½mm11vv1122 + ½ + ½mm22vv22

22 = ½ = ½mm11v’v’1122 + ½ + ½mm22v’v’22

22

Two-Dimensional Collision, Two-Dimensional Collision, 33

For a collision of two particlestwo particles in two dimensionstwo dimensions implies that the momentum in each direction xx and yy is conservedis conserved

The game of billiardsgame of billiards is an example for such two dimensional collisions

The equations for conservationequations for conservation of momentum are:

mm11vv11ixix mm22vv22ix ix ≡≡ mm11vv11fxfx mm22vv22fxfx

mm11vv11iyiy mm22vv22iy iy ≡≡ mm11vv11fyfy mm22vv22fyfy

Subscripts represent: (1,2)(1,2) Objects ((i,fi,f)) Initial and final values ((x,yx,y)) Component direction

Two-Dimensional Collision, Two-Dimensional Collision, 44

Particle 1 is moving at velocity vv1i1i and particle 2 is at rest

In the xx-direction-direction, the initial momentum is mm11vv11ii

In the yy-direction-direction, the initial momentum is 00

Two-Dimensional Collision, Two-Dimensional Collision, finalfinal

After the glancingglancing collision, the conservation of momentum in the xx-direction-direction is

mm11vv11ii ≡≡ mm11vv11ff coscos mm22vv22ff coscos(9.24)(9.24)

After the collision, the conservation of momentum in the yy-direction-direction is

0 0 ≡ ≡ mm11vv11ff sinsin mm22vv22ff sinsin(9.25)(9.25)

Active Figure 9.13Active Figure 9.13

Example 9.8Example 9.8 Collision at Collision at an Intersection an Intersection (Example 9.10 (Example 9.10

Text Book)Text Book) Mass of the car mmcc = 1500kg= 1500kg Mass of the van mmvv = 2500kg = 2500kg Find vvff if this is a perfectly

inelastic collision (they stick together).

Before collisionBefore collision The car’s car’s momentummomentum is:

ΣΣppxixi = = mmccvvcc ΣΣppxixi == (1500)(25) = (1500)(25) = 3.75x103.75x1044 kg kg··m/sm/s

The van’svan’s momentummomentum is:

ΣΣppyiyi = = mmvvvvvv ΣΣppyiyi == (2500)(20) = (2500)(20) = 5.00x105.00x1044 kg kg··m/sm/s

Example 9.8Example 9.8 Collision at Collision at an Intersection, 2an Intersection, 2

After collisionAfter collision, bothboth have the same xx-- and yy--components:

ΣΣppxf xf = (= (mmc c + m+ mv v ))vvff cos cosΣΣppyf yf = (= (mmc c + m+ mv v ))vvff sin sin Because the total momentumtotal momentum

is both directions is conserved:is conserved:ΣΣppxf xf = = ΣΣppxixi

3.75x103.75x1044 kg kg··m/s = (m/s = (mmc c + m+ mv v ))vvff cos cos(1)(1)

ΣΣppyf yf = = ΣΣppyiyi

5.00x105.00x1044 kg kg··m/s = (m/s = (mmc c + m+ mv v ))vvff sinsin(2)(2)

Example 9.8Example 9.8 Collision at Collision at an Intersection, finalan Intersection, final

Since ((mmc c + m+ mv v )) = 400kg.

3.75x103.75x1044 kg kg··m/s = m/s = 40004000 vvff cos cos(1)(1)

5.00x105.00x1044 kg kg··m/s = m/s = 40004000vvff sin sin(2)(2) Dividing Eqn (2) by (1)5.00/3.75 =1.33 = tan

= 53.1= 53.1°° Substituting in Eqn (2) or (1) 5.00x105.00x1044 kg kg··m/s = m/s = 40004000vvff sin sin53.153.1° °

vvf f == 5.00x105.00x1044/(4000/(4000sinsin53.153.1° ° ) )

vvf f == 15.6m/s15.6m/s

9.5 The Center of 9.5 The Center of MassMass

There is a special point in a system or object, called the center of mass (CM)center of mass (CM), that moves as if all of the mass of the system is concentrated at that point

The system will move as if an external force were applied to a single particle of mass MM located at the CMCM MM = = ΣΣmmii is the total mass of the system

The coordinatesThe coordinates of the center of masscenter of mass are

(9.28) (9.28) (9.29)(9.29)

CM CM CM

i i i i i ii i i

m x m y m zx y z

M M M

Center of Mass, positionCenter of Mass, position The center of mass can be located by

its position vector, rCM

(9.30)(9.30)

ri is the position of the i th particle, defined by

CM

i ii

m

M r

r

ˆ ˆ ˆi i i ix y z r i j k

Active Figure 9.16Active Figure 9.16

Center of Mass, ExampleCenter of Mass, Example Both masses are on the x-

axis The center of mass (CM) is

on the x-axis One dimension, x-axisOne dimension, x-axis

xxCMCM = = (m(m11xx11 + m + m22xx22)/M)/M

M = mM = m11+m+m22

xxCMCM ≡≡ (m(m11xx11 + m + m22xx22)/(m)/(m11+m+m22))

Center of Mass, finalCenter of Mass, final The center of mass is closer

to the particle with the larger mass

xxCMCM ≡≡ ((mm11xx11 + + mm22xx22)/()/(mm11++mm22)) If: xx11 = 0, = 0, xx22 = = dd & & mm22 = =

22mm11

xxCMCM ≡≡ (0 + 2(0 + 2mm11dd)/(m)/(m11+2+2mm11) )

xxCMCM ≡≡ 22mm11dd/3/3mm11

xxCMCM = 2 = 2dd/3/3

Active Figure 9.17Active Figure 9.17

Center of Mass, Extended Center of Mass, Extended ObjectObject

Up to now, we’ve been mainly concerned with the motion of single (point)single (point) particles.

To treat extended bodiesextended bodies, we’ve approximated the body as a point particle & treated it as ifas if it had all of its mass at a point!

How is this possible?How is this possible? Real, extended bodies have complex motion,

including: translation, rotation, & translation, rotation, & vibration!vibration!

Think of the extended objectextended object as a system containing a large number of particleslarge number of particles

Center of Mass, Extended Center of Mass, Extended Object, CoordinatesObject, Coordinates

The particle separationparticle separation is very small very small, so the mass can be considered a continuous mass distribution:continuous mass distribution:

The coordinates of theCMCM of the object are:

(9.31)(9.31)

(9.32)(9.32)

M

mxx i

ii CM

1

& 1

:Similarly

1lim

CMCM

0CM

zdmM

zydmM

y

xdmMM

mxx i

ii

mi

Center of Mass, Extended Center of Mass, Extended Object, PositionObject, Position

The positionThe position of CMCM can also be found by:

(9.33)(9.33)

The CMCM of any symmetrical symmetrical objectobject lies on an axis of axis of symmetrysymmetry and on any plane plane of symmetryof symmetry

An extended object can be considered a distribution of small mass elements, mm

The CMCM is located at position rrCMCM

CM

1dm

M r r

Example 9.9Example 9.9 Three Guys on Three Guys on a Rafta Raft

A group of extended bodies, each with a known CMCM and equivalent mass mm. Find the CMCM of the group.

xxCMCM = ( = (ΣΣmmiixxii)/)/ΣΣmmii

xxCMCM = = (m(mxx11 + m+ mxx22+ m+ mxx33)/(m+m+m))/(m+m+m)

xxCMCM = = m(m(xx11 + + xx22+ + xx33)/3m)/3m = = (x(x11 + + xx22+ + xx33)/)/3 3

xxCMCM = (1.00m + 5.00m + 6.00m)/3 = 4.00m = (1.00m + 5.00m + 6.00m)/3 = 4.00m

Example 9.10Example 9.10 Center of Center of Mass of a Rod Mass of a Rod (Example 9.14 Text (Example 9.14 Text Book)Book)

Find the CMCM position of a rod of mass MM and length LL

The location is on the x-axis (yCM = zCM = 0)

(A).(A). Assuming the road has a uniform mass per unit length

λλ = M/L = M/L (Linear mass densityLinear mass density) From Eqn 9.31

2

0

2

00CM 22

11L

Mx

Mxdx

Mdxx

Mxdm

Mx

LLL

Example 9.10Example 9.10 Center of Center of Mass of a Rod, 2Mass of a Rod, 2

But λλ = M/L = M/L

(B).(B). Assuming now that the linear mass densitylinear mass density of the road is no uniform: no uniform: λλ = = xx

The CMCM will be:

22

/

222

CM

LL

M

LML

Mx

3

0

2

00CM

3

11

LM

x

dxxM

xdxxM

dxxM

xdmM

x

CM

LLL

Example 9.10Example 9.10 Center of Center of Mass of a Rod, finalMass of a Rod, final

But mass of the rod and are related by:

The CMCM will be:

2

2

0 00

LxdxdxdmM

L LL

LLL

M

Lx

3

2

23

3 2

33

CM

Examples to Read!!!Examples to Read!!! Example 9.12Example 9.12 (page 269) Example 9.13Example 9.13 (page 272) Example 9.16Example 9.16 (page 276)

Homework to be solved in Homework to be solved in Class!!!Class!!! Problems: 43Problems: 43

Material for the FinalMaterial for the Final