9 4 two dimensional collisions
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9.4 Two-Dimensional Collisions. Two-Dimensional Collisions. The momentum is conserved in all directions Use subscripts for identifying the object indicating initial or final values the velocity components

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9.4 Two-Dimensional Collisions

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9 4 two dimensional collisions

9.4 Two-Dimensional Collisions


Two dimensional collisions

Two-Dimensional Collisions

  • The momentum is conserved in all directions

  • Use subscripts for

    • identifying the object

    • indicating initial or final values

    • the velocity components

  • If the collision is elastic, use conservation of kinetic energy as a second equation

    • Remember, the simpler equation can only be used for one-dimensional situations


Two dimensional collision 2

Two-Dimensional Collision, 2

Qualitative Analysis

  • Physical Principles:The same as in One-Dimension

  • 1. Conservation of VECTOR momentum:

    P1x + P2x = P1x + P2x &P1y + P2y= P1y + P2y

  • 2. Conservation of Kinetic Energy

    ½m1v12 + ½m2v22 = ½m1v’12 + ½m2v’22


Two dimensional collision 3

Two-Dimensional Collision, 3

  • For a collision of two particles in two dimensions implies that the momentum in each direction xand yis conserved

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

  • The equations for conservation of momentum are:

    m1v1ix+m2v2ix ≡m1v1fx+m2v2fx

    m1v1iy+m2v2iy ≡m1v1fy+m2v2fy

  • Subscripts represent:

    • (1,2) Objects

    • (i,f) Initial and final values

    • (x,y) Component direction


Two dimensional collision 4

Two-Dimensional Collision, 4

  • Particle 1 is moving at velocity v1iand particle 2 is at rest

  • In the x-direction, the initial momentum is m1v1i

  • In the y-direction, the initial momentum is 0


Two dimensional collision final

Two-Dimensional Collision, final

  • After the glancing collision, the conservation of momentum in the x-direction is

    m1v1i≡m1v1fcosq+m2v2f cosf (9.24)

  • After the collision, the conservation of momentum in the y-direction is

    0 ≡ m1v1fsinq+m2v2f sinf(9.25)


Active figure 9 13

Active Figure 9.13


Example 9 8 collision at an intersection example 9 10 text book

Example 9.8 Collision at an Intersection (Example 9.10 Text Book)

  • Mass of the car mc= 1500kg

  • Mass of the van mv = 2500kg

  • Findvfif this is a perfectly inelastic collision (they stick together).

  • Before collision

    • The car’smomentum is:

      Σpxi= mcvc

      Σpxi= (1500)(25) = 3.75x104 kg·m/s

    • The van’smomentum is:

      Σpyi= mvvv

      Σpyi= (2500)(20) = 5.00x104 kg·m/s


Example 9 8 collision at an intersection 2

Example 9.8 Collision at an Intersection, 2

  • After collision,both have the same x- and y-components:

    Σpxf = (mc + mv )vf cos

    Σpyf = (mc + mv )vf sin

  • Because the total momentum is both directions is conserved:

    Σpxf = Σpxi

    3.75x104 kg·m/s = (mc + mv )vf cos (1)

    Σpyf = Σpyi

    5.00x104 kg·m/s = (mc + mv )vf sin (2)


Example 9 8 collision at an intersection final

Example 9.8 Collision at an Intersection, final

  • Since (mc + mv ) = 400kg. 

    3.75x104 kg·m/s = 4000vf cos (1)

    5.00x104 kg·m/s = 4000vf sin (2)

  • Dividing Eqn (2) by (1)

    5.00/3.75 =1.33 = tan 

     = 53.1°

  • Substitutingin Eqn (2) or (1) 

    5.00x104 kg·m/s = 4000vf sin53.1° 

    vf =5.00x104/(4000sin53.1° ) 

    vf =15.6m/s


9 5 the center of mass

9.5 The Center of Mass

  • There is a special point in a system or object, called the 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 M located at the CM

    • M= Σmiis the total mass of the system

  • The coordinates of the center of mass are

    (9.28) (9.29)


Center of mass position

Center of Mass, position

  • The center of mass can be located by its position vector, rCM

    (9.30)

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


Active figure 9 16

Active Figure 9.16


Center of mass example

Center of Mass, Example

  • Both masses are on the x-axis

  • The center of mass (CM) is on the x-axis

  • One dimension, x-axis

    xCM = (m1x1 + m2x2)/M

    M = m1+m2

    xCM≡(m1x1 + m2x2)/(m1+m2)


Center of mass final

Center of Mass, final

  • The center of mass is closer to the particle with the larger mass

    xCM≡(m1x1 + m2x2)/(m1+m2)

  • If: x1 = 0, x2 = d & m2 = 2m1

    xCM≡(0 + 2m1d)/(m1+2m1) 

    xCM≡2m1d/3m1 

    xCM = 2d/3


Active figure 9 17

Active Figure 9.17


Center of mass extended object

Center of Mass, Extended Object

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

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

    How is this possible?

  • Real, extended bodies have complex motion, including: translation, rotation, & vibration!

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


Center of mass extended object coordinates

Center of Mass, Extended Object, Coordinates

  • The particle separationis very small, so the mass can be considered a continuous mass distribution:

  • The coordinates of theCM of the object are:

    (9.31)

    (9.32)


Center of mass extended object position

Center of Mass, Extended Object, Position

  • The position of CM can also be found by:

    (9.33)

  • The CM of any symmetrical object lies on an axis of symmetry and on any plane of symmetry

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

  • The CM is located at position rCM


Example 9 9 three guys on a raft

Example 9.9 Three Guys on a Raft

A group of extended bodies, each with a known CM and

equivalent mass m.

Find the CM of the group.

xCM = (Σmixi)/Σmi

xCM = (mx1+ mx2+ mx3)/(m+m+m) 

xCM = m(x1+ x2+ x3)/3m = (x1+ x2+ x3)/3 

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


Example 9 10 center of mass of a rod example 9 14 text book

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

  • Find the CM position of a rod of mass M and length L

  • The location is on the x-axis

    (yCM = zCM = 0)

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

    λ = M/L (Linear mass density)

  • From Eqn 9.31


Example 9 10 center of mass of a rod 2

Example 9.10 Center of Mass of a Rod, 2

  • But λ = M/L

  • (B). Assuming now that the linear mass density of the road is no uniform: λ = x

  • The CM will be:


Example 9 10 center of mass of a rod final

Example 9.10 Center of Mass of a Rod, final

  • But mass of the rod and  are related by:

    The CM will be:


9 4 two dimensional collisions

Material for the Final

  • Examples to Read!!!

    • Example 9.12(page 269)

    • Example 9.13(page 272)

    • Example 9.16(page 276)

  • Homework to be solved in Class!!!

    • Problems: 43


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