# 9.4 Two-Dimensional Collisions - PowerPoint PPT Presentation

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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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### 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

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

• 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

• 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

• 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)

### 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

• 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

• 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

• 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

• 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

### 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

• 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

### 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

• 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

• 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

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)

• 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

• 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

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

The CM will be:

Material for the Final