R
E C - T E C

ACCIDENT .
. . . . RECONSTRUCTION
. . . . . .
SOFTWARE

*"**when**
**performance**
**counts"*

George M. Bonnett, J.D.

REC-TEC LLC

**The graphics and
mathematical computations used throughout this article were
generated using REC-TEC Professional - Version V accident
reconstruction software from REC-TEC LLC.**

The reader may find a working knowledge of both linear momentum and vector sum analysis to be beneficial in understanding the concepts presented in this manual. Impact Velocity from Conservation of Linear Momentum and Conservation of Linear Momentum Using Vector Sum Analysis, both by W. A. Kennedy, offer an excellent background for Vector Momentum Analysis. These manuals are available through I.P.T.M., University of North Florida, Jacksonville, Florida 32224-2645.

Copyright © 1993 by George M. Bonnett, J.D. All rights reserved. No part of this publication may be reproduced or transmitted in any form, or any means, electronic or mechanical, including photography, recording, or by any information storage and retrieval system, without prior permission in writing from the copyright holder.

Published and distributed by:

REC-TEC LLC

George M. Bonnett, J.D.

**INTRODUCTION:** Vector Momentum Analysis
is an effective method of solving linear momentum problems for
the post impact data resulting from known, or given, impact
information. This is advantageous during any in depth analysis of
vehicles in collision.

**Linear Momentum**is used to determine the impact velocities of vehicles in collision when the angles and post impact velocities of the vehicles are known.**Vector Sum Analysis**provides a graphical check on the mathematical solution of the linear momentum equations as well as providing additional information, including the change of momentum or impulse, and the principle direction of force (PDOF) involved in the collision.**Vector Momentum Analysis**can be used to determine the post impact data resulting from given impact data. It takes the investigation into the collisions full circle.

**COORDINATE SYSTEMS:** The most widely used
systems for both Cartesian and polar coordinates are the RIGHT
HAND COORDINATE SYSTEM (RHCS) and the LEFT HAND COORDINATE SYSTEM
(LHCS).

**The two (2) systems share several common
attributes:**

- Two (2) perpendicular lines or axes (plural of axis), marked off to a scale that includes positive and negative values;
- An origin where the lines cross at the (0,0) point;
- Angles are measured positively from the X axis.

**Differences between the systems are:**

- The RHCS has the X axis as the vertical axis and the Y axis as the horizontal axis;
- The LHCS has the Y axis as the vertical axis and the X axis as the horizontal axis;
- Angles are measured clockwise from the X axis in the RHCS and counter-clockwise from the X axis in the LHCS.

The LHCS will be used in all future discussions and diagrams. The same scale will be used on both axes. This is for convenience only. It is not required by the LHCS, but will allow placement of a protractor over the diagrams that may assist in comprehending some of the information being presented.

**VECTORS:** A clear understanding of the
concept of vectors and their components is mandatory to an
understanding of accident reconstruction in general, and linear
momentum in particular. A vector is a graphical representation of
a physical action that has both magnitude and direction. A scalar
is an algebraic quantity having magnitude, but not direction.
Speed without direction is a scalar. Scalars can have units, and
perhaps an algebraic sign, but no direction. Speed with direction
(often referred to as velocity) is a VECTOR. In addition to
magnitude and direction, a vector requires a sense (+ or -) to
the direction.

**Scalars** are added mathematically. It is
simple addition. Vectors must be added using the parallelogram
rule. They must be added geometrically and not mathematically.
The vector addition law states that if the arrows representing a
set of vectors are laid tail to head in any order, the vector sum
(the resultant vector) is represented by the arrow that connects
the tail of the first to the head of the last.

While adequately illustrating the concept, it is cumbersome to carry a ruler and protractor in order to add vectors, and often very inaccurate. The method of components is more convenient in that it reduces all vector additions to the addition of perpendicular vectors.

**CONSTANTS:** Everything will be discussed
in reference to the origin (0,0), providing a constant frame of
reference. None of the problems will contain labels for either
weights or speeds (velocities). They have no bearing on the
subject matter, serving only to confuse, and will therefore be
eliminated from all future discussions.

While purists may raise eyebrows, the angles involved in the collision will be referred to as A1, A2, A3, A4, and not by their traditional Greek names. The new names are more easily identified with the associated vector by both the expert and the lay person (judge or jury).

In order to simplify understanding of the concepts involved, the following values will be held constant unless otherwise stated in the problem:

Vehicle #1

- impact velocity (V1) = 30
- weight (W1) = 2000

Vehicle #2

- impact velocity (V2) = 20
- weight (W2) = 3000

**METHODOLOGY:** Vector momentum analysis
combines linear momentum, vector sum analysis, and trigonometry
to solve the post impact data that results from given impact
information. The impact data is used to compute a resultant
vector (Mr) as well as the orientation angle (AO) of the
resultant vector. The resultant vector (Mr) and the two (2) post
impact momentum vectors (M3 and M4) form a triangle. The
resultant vector is a (known) side of this triangle. Two (2)
angles of the triangle (a and b) have a direct correlation to the
relationship between the departure angles (A3 and A4) and the
orientation angle (AO) of the resultant vector.

The Law of Sines and the Law of Cosines will permit the computation of the sides and the angles of a triangle if the following can be determined:

1. Two (2) angles and one (1) side;

2. Two (2) sides and one (1) angle;

3. Three (3) sides.

Once results are obtained using vector momentum analysis, the post impact data can then be used to generate impact velocities using linear momentum. The results obtained from linear momentum should be identical to the input data used for vector momentum analysis. Vector sum analysis may be used as a graphical check on the mathematical results obtained from both vector momentum analysis and linear momentum.

**PROBLEMS:** Three basic problems will be
used in examining the concept of vector momentum analysis. The
problems will consist of collisions with the approach angles
between vehicle #1 and vehicle #2 of: (1) 90 degrees; (2) less
than 90 degrees; and (3) greater than 90 degrees.

1. If vehicle #1 has a known departure angle of 60 degrees (A3), and vehicle #2 has a known departure angle of 30 degrees (A4) then vector M3 would have an angle of 60 degrees (A3) and vector M4 would have an angle of 30 degrees (A4). The vectors M3 and M4 can be added by joining the tail of M4 to the head of M3. M4 and M3 can be also be added by joining the tail of M4 to the head of M3, forming a parallelogram. (Review the rule for vector addition on page 2.) The resultant vector (Mr) divides the parallelogram into two similar triangles that share a common side. This may now be treated as a trigonometry problem. Since two (2) angles (A3 and A4) and a side (Mr) are known, The Law of Sines can be used to solve for sides (M3 and M4). This method will be used in solving Problem 1C.

2. If, instead of angles (A3 and A4), the departure velocities (V3 and V4) are known, the magnitude of the two (2) momentum vectors (M3 and M4) can be obtained by multiplying the velocities (V3 and V4) by their respective weights (W1 and W2). All three (3) sides of the triangle(s) are then known, and by using the Law of Cosines, the respective departure angles (A3 and A4) can be determined. This method will be used in solving Problem 2.

3. If only one (1) of the departure angles (A3 or A4) and one (1) of the departure velocities (V3 or V4) is known, the other two (2) unknowns can be solved by using the Law of Cosines. Some of the combinations may involve the use of a quadratic equation. This results in an ambiguous case that has two (2) possible solutions. There is only one (1) correct solution due to the nature of the triangle involved. This will be discussed further in Problem 3B. These methods (with and without the quadratic) will be used in solving Problems 3A and 3B.

**Problem 1A:** In linear momentum, the post
impact velocities of vehicle #1 (V3) and vehicle #2 (V4) are used
to solve for the respective impact velocities (V1 and V2).

- Velocity of vehicle #1 after impact (V3) = 30
- Velocity of vehicle #2 after impact (V4) = 20
- Weight of vehicle #1 (W1) = 2000
- Weight of vehicle #2 (W2) = 3000
- Approach angle of vehicle #1 (A1) = 180 degrees (ALWAYS)
- Approach angle of vehicle #2 (A2) = 270 degrees
- Departure angle of vehicle #1 (A3) = 90 degrees
- Departure angle of vehicle #2 (A4) = 0 degrees

V1, and therefore M1,is always aligned with the X axis. It never has a vertical (Y axis) component of velocity. The sine of angle A1 (180 degrees) is always zero (0) and the cosine is always -1. Solving for V2 first eliminates a variable and makes the solution possible. This alignment makes it possible to solve a linear momentum problem that is not collinear (inline).

**Conservation of Linear
Momentum**

**M1*V3 + M2*V4 = M1*V1 + M2*V2**

M = W/g ; substitute for M:

**(W1*V1)/g + (W2*V2)/g =
(W1*V3)/g + (W2*V4)/g**

Multiply by g:

**W1*V1 +W2*V2 = W1*V3 + W2*V4**

**Solving for V2:**

W1*V1*sin(A1) + W2*V2*sin(A2) = W1*V3*sin(A3) + W2*V4*sin(A4)

V2 = (W1*V3*sin(A3) + W2*V4*sin(A4) - W1*V1*sin(A1)) / W1*sin(A2)

V2 = 2000*20*-1 + 3000*30*0 - 2000*V1*0 / 2000*-1

V2=(-40000 + 0 - 0) / -2000

**V2 = 20**

**Solving for V1:**

W1*V1*cos(A1) + W2*V2*cos(A2) = W1*V3*cos(A3) + W2*V4*cos(A4)

V1 = (W1*V3*cos(A3) + W2*V4*cos(A4) - W2*V2*cos(A2)) / W1*cos(A1)

V1 = 2000*20*0 + 3000*30*0 + 3000*20*-1 / 2000*-1

V1=(-60000 + 0 - 0) / -2000

**V1 = 30**

**Problem 1B:** This is basically the same
problem with a few minor differences. Impact velocities and
angles are given as well as the fact that the two vehicles remain
together from impact until final rest.

- Velocity of vehicle #1 at impact (V1) = 30
- Velocity of vehicle #2 at impact (V2) = 20
- Weight of vehicle #1 (W1) = 2000
- Weight of vehicle #2 (W2) = 3000
- Approach angle of vehicle #1 (A1) = 180 degrees (ALWAYS)
- Approach angle of vehicle #2 (A2) = 270 degrees

If the vehicles remain together from impact until final rest, the departure angles for vehicle #1 (A3) and vehicle #2 (A4) are identical as are the post impact velocities.

Total momentum (Mr) at impact = SQR(Mx^2 + My^2)

Mr = SQR((2000*30)^2 + (3000*20)^2 = 84,852.81

**The conservation of
linear momentum states that the total momentum after collision
must equal the total momentum before collision.**

Total momentum after collision (Mr) = 84,852.81

The resultant vector has a direction. The orientation angle of the vector is the arctangent of the Y component divided X component.

arctangent(60,000/60,000) = atn(1) = 45 degrees

**The departure angle for both vehicles is 45
degrees.**

Since the vehicles remained together, the velocity is the resultant momentum (Mr) divided by the weight of both vehicles.

Velocity = Mr/(W1+W2) = 84852.81/(2000+3000)

**V3 and V4 = 16.9705**

The resultant momentum vectors must be equal in both magnitude and direction for both impact and post impact. This is true even when the exit vehicles do not have identical angles and/or velocities.

**Problem 1C: **This problem will be
basically the same as Problem 1B, except that the departure
angles and velocities are (for the moment) unknown.

- Velocity of vehicle #1 at impact (V1) = 30
- Velocity of vehicle #2 at impact (V2) = 20
- Weight of vehicle #1 (W1) = 2000
- Weight of vehicle #2 (W2) = 3000
- Approach angle of vehicle #1 (A1) = 180 degrees (ALWAYS)
- Approach angle of vehicle #2 (A2) = 270 degrees
- Departure angle of vehicle #1 (A3) = UNKNOWN
- Departure angle of vehicle #2 (A4) = UNKNOWN

Based on the formulae in problem 1B , the resultant momentum vector has an orientation angle of 45 degrees and a magnitude of 84,852.81.

Angle A3 (60 degrees) and angle A4 (30 degrees) will be treated as given. This information will be used to solve for M3 and M4 using the Law of Sines. The velocities (V3 and V4) can be obtained by dividing by the respective weights.

**Law of Sines: sin(angle a)/A = sin(angle
b)/B = sin(angle c)/C**

The following is known:

side A = M3

side B = M4

side C = Mr = 84,852.81

angle a = (60 - 45) = 15 degrees

angle b = (45 - 30) = 15 degrees

angle c = 180 - angle a - angle b = 150 degrees

**Solving for M3 and M4:**

M4 = Mr * sin(a) / sin(c) = 43,923.05

M3 = M4 * sin(b) / sin(a) = 43,923.05

**V3 = M3 / W1 = 21.9615**

**V4 = M4 / W2 = 14.6410**

**Problem 2:** Post impact velocities of
both vehicles are given. Post impact angles will be determined
using the Law of Cosines.

**Law of Cosines:**

**A^2 = B^2 + C^2 - 2 * B * C *
cos(angle a)**

**B^2 = A^2 + C^2 - 2 * A * C *
cos(angle b)**

**C^2 = A^2 + B^2 - 2 * A * B *
cos(angle c)**

- Velocity of vehicle #1 at impact (V1) = 30
- Velocity of vehicle #2 at impact (V2) = 20
- Weight of vehicle #1 (W1) = 2000
- Weight of vehicle #2 (W2) = 3000
- Approach angle of vehicle #1 (A1) = 180 degrees (ALWAYS)
- Approach angle of vehicle #2 (A2) = 250 degrees
- Departure velocity of vehicle #1 (V3) = 25
- Departure velocity of vehicle #2 (V4) = 17

Using the formulae in problem 1B, the length of the resultant vector can be determined.

The length of all three (3) sides of the triangle are known.

**Solving for the angles (A3 and A4):**

My = sin(250)*60,000 = -56,381.56

Mx = cos(180)*(2000*30) + cos(250)*(3000*20) = -80,521.21

Mr = SQR(Mx^2 + My^2) = 98,298.25

M3 = V3 * W1 = 25*2000 = 50,000

M4 = V4 * W2 = 17*3000 = 51,000

**Orientation angle = angle c =
atn(My/Mx) = 35**

ANGLE a = inverse cos((M3^2 + Mr^2 - M4^2) / (2*M3*Mr)) = 13.42

ANGLE b = inverse cos((M4^2 + Mr^2 - M3^2) / (2*M4*Mr)) = 13.15

**A3 = angle a + angle c = 48.42 degrees**

**A4 = angle c - angle b = 21.85 degrees**

**Problem 3A:** The post impact velocity
(V3) and departure angle (A3) of vehicle #1 are given. The post
impact velocity (V4) and departure angle (A4) of vehicle #2 will
be determined using the Law of Cosines.

- Velocity of vehicle #1 at impact (V1) = 30
- Velocity of vehicle #2 at impact (V2) = 20
- Weight of vehicle #1 (W1) = 2000
- Weight of vehicle #2 (W2) = 3000
- Approach angle of vehicle #1 (A1) = 180 degrees (ALWAYS)
- Approach angle of vehicle #2 (A2) = 290 degrees
- Departure velocity of vehicle #1 (V3) = 25
- Departure angle of vehicle #1 (A3) = 70

Using the formulae in problem 1B, the length of the resultant vector can be determined.

The length of two (2) sides and one (1) angle of the triangle are known.

**Solving for the side M4:**

My = sin(290)*60,000 = -56,381.56

Mx = cos(180)*(2000*30) + cos(290)*(3000*20) = -39,478.79

Mr = SQR(Mx^2 + My^2) = 68,829.17

M3 = V3 * W1 = 25*2000 = 50,000

**Orientation angle = angle c = atn(My/Mx) =
55**

ANGLE a = angle A3 - angle c = 15 degrees

M4 = SQR(M3^2 + Mr^2 - (2 * M3 * Mr * cos (a))) = 24,270.71

**V4 = M4/W2 = 24,270.71/3000 = 8.09**

Knowing both M3 and M4 the solution for angle b is solved as it was in Problem 2 above using the Law of Cosines.

ANGLE b = inverse cos((M4^2 + Mr^2 - M3^2) / (2*M4*Mr)) = 32.22

**A4 = angle c - angle b = 22.78 degrees**

**Problem 3A - Alternative solution:**
Vector component addition can also be used to solve problem 3A.

**The total vertical
momentum at impact must equal the total vertical momentum after
impact.**

Solving for the total vertical component of momentum after impact (Py):

Py = M1y + M2y = M3y + M4y

The vertical component of momentum for vehicle #1 at impact is zero (0).

Py = V2 * W2 * cos(A2) = -60,000 * sin(290) = 56,318.56

V1 * W1 * sin(A1) + V2 * W2 * sin(A2) = M3y + M4y

0 + (-30) * 2000 * sin(A2) = V3 * W1 * sin(A3) + M4y

-60000 * sin(290) = 25 * 2000 * sin(70) + M4y

M4y = 56,381.56 - 46,984.63

M4y = 9396.95

Solving for the total horizontal component of momentum after impact (Px):

Px = M3x + M4x = M1x + M3x

Px = V1 * W1 * cos(A1) + V2 * W2 * cos(A2)

Px = -30 * 2000 * cos(180) + (-20) * 3000 * cos(290)

Px = 60,000 - 20,521.21 = 39,478.79

M3x = M3 * cos(A3) = V3 * W1 * cos(A3)

M3x = 25 * 3000 * cos(70) = 17,101.07

M4x = P4x - M3x = 39,478.79 - 20,521.21 = 22,377.78

M4 = SQR(M4x^2 + M4y^2) = 24,270.72

V4 = M4 / W2 = 24,270.72 / 3000

**V4 = 8.09**

A4 = atn(M4y/M4x) = atn(9396.95 / 22,377.78)

**A4 = 22.78 degrees**

**Problem 3B: **The post impact velocity of
vehicle #1 (V3) and departure angle of vehicle #2 (A4) are given.
The post impact velocity of vehicle #2 (V4) and the departure
angle of vehicle #1 (A) will be determined using the Law of
Cosines.

- Velocity of vehicle #1 at impact (V1) = 30
- Velocity of vehicle #2 at impact (V2) = 20
- Weight of vehicle #1 (W1) = 2000
- Weight of vehicle #2 (W2) = 3000
- Approach angle of vehicle #1 (A1) = 180 degrees (ALWAYS)
- Approach angle of vehicle #2 (A2) = 290 degrees
- Departure velocity of vehicle #1 (V3) = 25
- Departure angle of vehicle #2 (A4) = 20

Using the formulae in problem 1B, the length of the resultant vector can be determined.

Two (2) sides and one (1) angle of the triangle are known.

**Solving for the third side (M4):**

My = sin(290)*60,000 = -56,381.56

Mx = cos(180)*(2000*30) + cos(290)*(3000*20) = -39,478.79

Mr = SQR(Mx^2 + My^2) = 68,829.17

M3 = V3 * W1 = 25*2000 = 50,000

**Orientation angle = angle c =
atn(My/Mx) = 55**

ANGLE b = angle c - angle A4 = 35 degrees

BB = 2 * Mr * cos (b) = 112,763.11

CC = Mr^2 - M3^2 = 2,237,454,968

Q1 = (BB - SQR(BB^2 - 4 * CC)) / 2 = 25,698.91

Q2 = (BB + SQR(BB^2 - 4 * CC)) / 2 = 87,064.21

**Two (2)
possible solutions, or roots, result from the quadratic formula.
Both roots must be checked in order to determine which is the
correct solution to the problem.**

**DETERMINATION OF THE CORRECT ROOT OF THE
QUADRATIC:**

If M4 = Q2 = 87.064.21 then:

V4 = M4/W2 = 87.064.21/3000 = 29.02

ANGLE a = inverse cos((M3^2 + Mr^2 - M4^2) / (2*M3*Mr)) = 92.15

A3 = angle c + angle a = 147.85 degrees

This angle is outside of the cone of departure (0 to 110 degrees) that is dictated by the approach angles(A1 = 180 and A2 = 290), therefore:

**The correct root of the quadratic is:**

M4 = Q1 = 25,698.91

**V4 = M4/W2 = 25,698.9/3000 = 8.57**

Knowing both M3 and M4 the solution for angle b is solved as it was in Problem 2 above.

ANGLE a = inverse cos((M3^2 + Mr^2 - M4^2) / (2*M3*Mr)) = 17.15

**A3 = angle c + angle a = 72.15 degrees**

**Having completed vector
momentum analysis, the reconstructionist should have a better
understanding of linear momentum and vector sum analysis as well
as a greater appreciation of the forces involved in the
collision.**

**The graphics and
mathematical computations used throughout this article were
generated using REC-TEC Professional - Version V accident
reconstruction software from REC-TEC LLC.**

**Copyright (C) 1993 George M.
Bonnett, J.D. All Rights Reserved**

**Voice or Fax 1-321-639-7783**

**REC-TEC LLC P.O. BOX 561031
ROCKLEDGE, FL 32956 USA**

**Copyright © George
M. Bonnett, JD**

**Last edited on Monday, 16 March 2009 03:33:55 PM -0500
**