9.4 two-dimensional collisions. two-dimensional collisions momentum is conserved in all directions...
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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)
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
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
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
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