Lesson 9: Linear Kinetics: Linear Impulse – Momentum Applications
Assignment
Read Chapter 5.
Complete the review questions.
Take the quiz for Lesson 9.
Objectives
 Define horizontal impulse-momentum.
 Define vertical impulse-momentum.
 Define braking and propulsion impulse.
 Solve impulse-momentum problems.
Impulse – Momentum Relationship
Impulse can be computed either by measuring the force and the time of force application, or by
measuring the change in velocity before and after the force is applied and computing the change in
momentum.
Average Force times Time = Change in Linear Momentum
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
The Area under the Force – Time Curve = Change in Linear Momentum
𝑡1
∫ 𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝑡0
The two equations above are general equations for computing impulse. They do not differentiate
between horizontal and vertical impulse. In this lesson we will differentiate between horizontal and
vertical impulse and then apply the above equations to data sets on running, walking and the jumping.
Horizontal Force and Horizontal Impulse in Running
Braking Impulse
The figure below shows horizontal force and horizontal velocity for a runner during ground contact. In
addition, the free-body diagram is shown for three phases of the motion: heel-strike, mid-stance and
toe-off. Begin by examining the horizontal force from t = 0 to t = .112 s, notice that the force is negative.
In the FBD for heel-strike the horizontal force (Rx) is shown as negative. Since this force is opposing the
runner’s motion it will cause the runner to decelerate, also notice that the Rx force causes the runner’s
center of mass to have a negative horizontal acceleration (ax). In biomechanics, it is customary to refer
to the impulse caused by this force which opposes motion as a braking impulse. The result of this
braking impulse is to decrease the horizontal velocity of the runner’s center of mass from Vx = 3.0 m/s
at heel-strike to Vx = 2.79 m/s at mid-stance. A propulsion impulse is defined as an impulse where the
direction of the impulse is the same as the direction of motion. In the horizontal force graph below the
force Rx is positive from t = .113 to t = .235 s, the propulsion impulse from this positive force causes the
runner’s horizontal velocity to increase from mid-stance to toe-off.
Example: Use the average force and area under the curve impulse-momentum formulas to compute the
Vxf for the runner below.
Average Force times Time = Change in Linear Momentum
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
̅̅̅̅ (∆𝑡) = 𝑚𝑉𝑥𝑓 − 𝑚𝑉𝑥𝑖
𝑅𝑥
−132.2 𝑁 (. 112 𝑠) = 70𝑘𝑔(𝑉𝑥𝑓 ) − 70𝑘𝑔(3.0
𝑚
)
𝑠
𝑉𝑥𝑓 = 2.79 𝑚/𝑠
The Area under the Force – Time Curve = Change in Linear Momentum
𝑡1
∫ 𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝑡0
.112
∫
𝑅𝑥 𝑑𝑡 = 70𝑘𝑔(𝑉𝑥𝑓 ) − 70𝑘𝑔(0.0)
0
−14.7 𝑁 · 𝑠 = 70𝑘𝑔(𝑉𝑥𝑓 ) − 70𝑘𝑔(0.0)
𝑉𝑥𝑓 = 2.79 𝑚/𝑠
After experiencing the braking impulse the runner’s center of mass has slowed down to 2.79 m/s. Next
we will now compute the impulse for the propulsion phase.
Propulsion Impulse
The propulsion phase begins at mid-stance at t = .113 s. At the start of propulsion the horizontal
velocity is 2.79 m/s, so the final velocity from the previous step (braking phase) becomes the initial
velocity for propulsion.
Average Force times Time = Change in Linear Momentum
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
̅̅̅̅
𝑅𝑥 (∆𝑡) = 𝑚𝑉𝑥𝑓 − 𝑚𝑉𝑥𝑖
115.58 𝑁 (. 122 𝑠) = 70𝑘𝑔(𝑉𝑥𝑓 ) − 70𝑘𝑔(2.79
𝑚
)
𝑠
𝑉𝑥𝑓 = 2.99 𝑚/𝑠
The Area under the Force – Time Curve = Change in Linear Momentum
𝑡1
∫ 𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝑡0
.235
∫
𝑅𝑥 𝑑𝑡 = 70𝑘𝑔(𝑉𝑥𝑓 ) − 70𝑘𝑔(2.79)
.113
14.1 𝑁 · 𝑠 = 70𝑘𝑔(𝑉𝑥𝑓 ) − 70𝑘𝑔(2.79)
𝑉𝑥𝑓 = 2.99 𝑚/𝑠
{IM-App-0.gif}
Vertical Force and Vertical Impulse in Running
When computing vertical impulse it is necessary to account for the vertical impulse caused by the force
of gravity. The figure below shows the free-body diagram for a runner. Substituting the kinematic
formula for vertical velocity into Newton’s second law of motion gives the vertical impulse-momentum
equation. Similar to the general equation for impulse-momentum, vertical impulse-momentum can be
computed using either the average vertical force or integrating to get the area under the vertical forcetime curve.
{IM-App-1.gif}
The figure below shows vertical force and vertical velocity for a runner during ground contact. In
addition, the free-body diagram is shown for three phases of the motion: heel-strike, mid-stance and
toe-off. The vertical force (Ry) is positive throughout ground contact. At heel-strike the vertical velocity
of the center of mass (Vy) is negative. Since the force Ry is opposing the runner’s motion it will cause
the vertical velocity of the runner’s center of mass to change to change from negative (Vyi = −.6 m/s) at
heel-strike to positive (Vyf = .48 m/s) at toe-off.
Example: Use the average force and area under the curve impulse-momentum formulas to compute the
Vyf for the runner below.
Average Force times Time = Change in Linear Momentum
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
̅̅̅̅̅̅̅̅̅̅̅̅
𝑅𝑦 + 𝑚𝑔(∆𝑡) = 𝑚𝑉𝑦𝑓 − 𝑚𝑉𝑦𝑖
321.1 𝑁 (. 235 𝑠) = 70𝑘𝑔(𝑉𝑦𝑓 ) − 70𝑘𝑔(−.6
𝑚
)
𝑠
𝑉𝑦𝑓 = 0.48 𝑚/𝑠
The Area under the Force – Time Curve = Change in Linear Momentum
𝑡1
∫ 𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝑡0
.235
∫
𝑅𝑦 + 𝑚𝑔 𝑑𝑡 = 70𝑘𝑔(𝑉𝑦𝑓 ) − 70𝑘𝑔(−.6)
0
75.5 𝑁 · 𝑠 = 70𝑘𝑔(𝑉𝑦𝑓 ) − 70𝑘𝑔(−.6)
𝑉𝑦𝑓 = 0.48 𝑚/𝑠
{IM-App-2.gif}
Horizontal Force and Horizontal Impulse in Walking
The horizontal force and horizontal velocity for walking are shown in the figure below. Horizontal force
in walking typically begins with a brief propulsion phase, following by braking and propulsion.
Example: Use the average force and formula to compute the horizontal velocity (Vx) for each phase of
the horizontal force curve for the walking curve below.
Average Force times Time = Change in Linear Momentum
First Propulsion Phase
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
̅̅̅̅ (∆𝑡) = 𝑚𝑉𝑥𝑓 − 𝑚𝑉𝑥𝑖
𝑅𝑥
26.1 (. 04 𝑠) = 61𝑘𝑔(𝑉𝑥𝑓 ) − 61𝑘𝑔(1.8
𝑚
)
𝑠
𝑉𝑥𝑓 = 1.82 𝑚/𝑠
Braking Phase
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
̅̅̅̅ (∆𝑡) = 𝑚𝑉𝑥𝑓 − 𝑚𝑉𝑥𝑖
𝑅𝑥
−65.3 𝑁 (. 4 𝑠) = 61𝑘𝑔(𝑉𝑥𝑓 ) − 61𝑘𝑔(1.82
𝑚
)
𝑠
𝑉𝑥𝑓 = 1.39 𝑚/𝑠
Second Propulsion Phase
𝐹̅ (∆𝑡) = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
̅̅̅̅
𝑅𝑥 (∆𝑡) = 𝑚𝑉𝑥𝑓 − 𝑚𝑉𝑥𝑖
73.1 (. 3 𝑠) = 61𝑘𝑔(𝑉𝑥𝑓 ) − 61𝑘𝑔(1.39
𝑉𝑥𝑓 = 1.75 𝑚/𝑠
𝑚
)
𝑠
{IM-App-3.gif}
Vertical Force and Vertical Impulse in Walking
The figure below shows vertical force and vertical velocity for a walker during ground contact. In
addition, the free-body diagram is shown for three phases of the motion: heel-strike, mid-stance and
toe-off. The vertical force (Ry) is positive throughout ground contact. At heel-strike the vertical velocity
of the center of mass (Vy) is negative. Since the force Ry is opposing the walker’s motion it will cause
the vertical velocity of the walker’s center of mass to change to change from negative (Vyi = −.18 m/s) at
heel-strike to positive (Vyf = .13 m/s) at toe-off.
Example: Use the area under the curve impulse-momentum formula to compute the Vyf for the walker
below.
The Area under the Force – Time Curve = Change in Linear Momentum
𝑡1
∫ 𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝑡0
.76
∫ 𝑅𝑦 + 𝑚𝑔 𝑑𝑡 = 61𝑘𝑔(𝑉𝑦𝑓 ) − 61𝑘𝑔(−.18)
0
18.8 𝑁 · 𝑠 = 61𝑘𝑔(𝑉𝑦𝑓 ) − 61𝑘𝑔(−.18)
𝑉𝑦𝑓 = 0.13 𝑚/𝑠
{IM-App-4.gif}
Impulse – Momentum in the Vertical Jump
The vertical force and vertical velocity for a vertical jump are shown in the figure below. At the
start of the jump the vertical velocity Vy = 0 m/s. At approximately 0.26 s the vertical velocity
attains the greatest downward value of −0.8 m/s. The jumper stops moving downward at t = .42
s and finally at takeoff the vertical velocity is 2.06 m/s. Use the area under the force – time
formula to compute the vertical velocity at takeoff as follows.
𝑡1
∫ 𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
.678
∫
𝑡0
(𝑅𝑦 + 𝑚𝑔) 𝑑𝑡 = 69𝑘𝑔(𝑉𝑦𝑓 ) − 69𝑘𝑔(0)
0
142 𝑁 · 𝑠 = 69𝑘𝑔(𝑉𝑦𝑓 ) − 69𝑘𝑔(0)
𝑉𝑦𝑓 = 2.06 𝑚/𝑠
The red dashed line and red circles on the graph below shows the point in the motion when the jumper
is moving downward the fastest, at this point the jumper’s center of mass has a vertical velocity of −0.8
m/s and the vertical force is equal to the jumper’s body weight. The blue dashed line and blue circles on
the velocity and force curves indicate the point in the motion when the jumper has stopped moving
down and is about to begin moving upward. From the perspective of muscle mechanics this point is
then end of the eccentric loading of the knee extensors and the hip extensors.
{IM-App-5.gif}
Relationship between Force and Acceleration
Remember from Newton’s second law of motion that acceleration is directly related to force and that
acceleration occurs in the direction of the net force. The shape of the acceleration curve is ALWAYS the
same as the shape of the force curve. If you divide the force curve by mass then you have an
acceleration curve. For a vertical force you must subtract the weight of the object/person before
dividing by mass. The graph below shows the direct relationship between force and acceleration for a
vertical jump.
{IM-App-5a.gif}
Speed Up – Slowing Down and Constant Velocity Run/Walk Curves
The graph below shows typical horizontal force curves with braking and propulsion impulses and the
change in horizontal velocity for: speeding up, constant velocity and slowing down. When running or
walking at a constant velocity the area of braking is equal to the area of propulsion. According to the
impulse – momentum relationship the change in horizontal velocity will be zero or close to zero. When
speeding up (accelerating) during walking or running the area of propulsion will be greater than the area
of braking and the change in velocity will be a positive number. Finally when slowing down
(decelerating) the area of braking will be greater than the area of propulsion and the change in
horizontal velocity will be negative.
{IM-App-6.gif}
Comparison between Horizontal and Vertical Force for Walking and Running
The figure below shows both horizontal and horizontal forces for walking and running. The magnitude
of the forces for running is greater than the magnitude for walking. The time of ground contact is much
longer for walking than running.
{IM-App-7.gif}
Computing Acceleration and Velocity in Excel from a Force Curve
Download the Excel File: Lesson 9 walk Practice.xls
When you first open up the file it should look like the figure below. Follow the instructions below to
compute ax, ay, vx, vy and impulses.
{IM-Excel-0-a.png}
Horizontal Acceleration
To compute ax enter the following equation in cell D2 and copy it down the rest of the column to row
757, where the data end.
=B2/$L$2
The equation above is derived as follows.
{IM-App-8.gif}
Vertical Acceleration
To compute ax enter the following equation in cell E2 and copy it down the rest of the column to row
757, where the data end.
=(C2+($L$2*-9.8))/$L$2
The equation above is derived as follows.
{IM-App-9.gif}
Horizontal Velocity
To compute ax enter the following equation in cell F3 and copy it down the rest of the column to row
757, where the data end.
=((B3*0.001)+($L$2*F2))/$L$2
The equation above is derived as follows.
{IM-App-10.gif}
Vertical Velocity
To compute ax enter the following equation in cell G3 and copy it down the rest of the column to row
757, where the data end.
= (((C3 + ($L$2*-9.8))*0.001) + ($L$2*G2)) / $L$2
The equation above is derived as follows.
{IM-App-11.gif}
Computing Impulses in Excel
The figure below shows the horizontal force curve and the Horizontal Impulse-Momentum section of the
Excel spreadsheet. After making a graph of horizontal force, you must enter the start and end row
numbers where the force curve CROSSES the Y = 0 AXIS. These row numbers should be placed in
columns M and N. The graph below show the corresponding rows where the force curve crosses the Y =
0 axis.
{IM-Excel-0.png}
To find these row numbers look in Column B. As shown in the figure below the force is negative from
rows 2 – 8 and row 9 is the start of the positive phase. Enter 2 in cell M6 and 8 in cell N6 to indicate the
start and end of the first negative phase. Then scroll down to find the end of the positive phase that
begins in row 9.
{IM-Excel-1.png}
Enter 9 in cell M7 and 51 in cell N7 to indicate the start and end of the first positive phase. Then scroll
down to find the end of the negative phase that begins in row 52.
{IM-Excel-2.png}
Enter 52 in cell M8 and 455 in cell N8 to indicate the start and end of the first negative phase. Then
scroll down to find the end of the positive phase that begins in row 456.
{IM-Excel-3.png}
Finally, enter 456 in cell M9 and 757 in cell N9 to indicate the start and end of the first negative phase.
Then scroll down to find the end of the positive phase that begins in row 9.
{IM-Excel-4.png}
When you have completed your work you can compare the results to the completed file.
Lesson 9 walk.xls
The horizontal and vertical impulse-momentum variable sections are designed to automatically compute
impulses using both the change in momentum and the average force times dt. Once you enter the start
and end rows, it computes the velocity for the start and endpoints, the time between the two points,
the average force and the impulse using both methods. The horizontal and vertical impulse sections are
programmed to compute the impulse variables for up to 8 different regions on a curve.
{IM-Excel-5.png}