Kinetics of Particle

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Impulse and Momentum

Prof. Dr. Mehmet Zor

2 . 3

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2.3.1 General Information: The Impulse-Momentum method is a different interpretation of the equation F = ma. It is especially useful in cases where the force is defined as a function of time or in collision problems where it has a very short-term effect and allows us to reach the result more easily. Our aim is to determine the position, velocity, acceleration or unknown external forces of an object at a time t through kinetic analysis.

2.3.2 Concepts of Impulse and Momentum:

We can write Newton's 2nd law, F=ma, in vector form and put it into a different format with the following steps:

(Linear) impulse :

(Linear) momentum :

At a time t

2.3 Kinetics of Particle / Impulse and Momentum

If we rearrange equation (I) a little more,..>>

(I)

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• This last 2.20 vector equation can be reduced scalarly to 3 independent equations on each cartesian axis:..>>

2.3.3 Equations of Impulse-Momentum:

(2.20)

(2.21a)

2.3 Kinetics of Particle / Impulse and Momentum

To put it more clearly:

• Initial momentum + External force impulse = Final momentum.
• In other words, the impulse of external forces causes the momentum of the object to change.

(2.21b)

(2.21c)

a-) For single particle:

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b-) For a System of Particles:

Now let's try to understand the subject better by solving examples..>>

(2.22.a)

2.3 Kinetics of Particle / Impulse and Momentum

(for x direction):

Vector, total 1 equation:

The explicit and general expressions of the impulse-momentum equations in a system formed by the combination of more than one particle are summarized below, both vectorial and scalar.

(2.22.b,c,d)

Scaler total 3 equations:

Figure 2.40

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A

P (t)

μ = 0.4

Solution:

1

P (t)

2

x

P (t)

In region A2 the force P rotates in the « –x » direction.

Constant force

Example 2.3.1 A 1 kg object moving to the right at a velocity of 2 m/s is affected by a time-varying force P as shown in the diagram. Since the friction coefficient is 0.4, a-) Calculate the velocity of the object when P=-4N,

b-) Calculate the maximum velocity that the force P will impart to the object. in the +x direction.

2.3 Kinetics of Particle / Impulse and Momentum

y

a-)

The impulse-momentum equation in the x-direction:

Figure 2.41

Figure 2.42

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b-)

(the maximum velocity occurring in the +x direction.)

t(s)

force

16N

-4N

8

4

0

5

Homework: Calculate the maximum speed that the object can gain in the «-x» direction. (While calculating this, pay attention to the fact that the friction force will also turn to the +x direction from the moment the object's speed is in the -x direction. Compare the results you find with those of your friends and discuss your solutions among yourselves until you find the correct result.)

2.3 Maddesel Nokta Kinetiği / İmpuls ve Momentum

P equation between 0-5 seconds :

As long as the net force remains positive in the +x direction, the speed increases continuously in that direction. According to the equation (I), the net force remains positive until the 3rd second. It becomes zero in the 3rd second and then negative. This situation can also be seen from the graph. Therefore, the maximum speed in the +x direction will occur at the 3rd second. To calculate this speed, let's write the impulse-momentum equation between 0-3 seconds:

(I)

3

12N

We will calculate the maximum velocity that the P force will impart to the object in the +x direction.

Result:

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2kg

2.3.4 Comparison of kinetic solution methods with an example

F=10N

t =8th second

A constant and horizontal force of F=10N acts on a 2kg box at rest on a frictionless horizontal surface.

Calculate the velocity of the box at 8 seconds,

a-) using D’Alembert’s Principle (F=ma),

b-) using Work-Energy Method,

c-) using Impulse-Momentum Method.

a-)

with D’alembert’s Principle

: constant

with Work-Energy Method

c-) with İmpuls – Momentum Method

External Forces Sys.

Efektive Force Sys.

F=10N

(DDDH) :

b-)

(x direction)

Example 2.3.2

Solution

2.3 Kinetics of Particle / Impulse and Momentum

Figure 2.43

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The time-varying force F is applied to the 1 kg block in the horizontal direction along the AB path for 4 seconds as shown in the graph. The initial velocity of the block is 2 m/s to the right. When the F force is removed, the block reaches point B. The AB path is frictionless. Starting from point B, the block enters the path with a kinetic friction coefficient of μ = 0.6 and stops at point C. Find the distance of point C from B.

Example 2.3.3

Solution:

C

2.3 Kinetics of Particle / Impulse and Momentum

Figure 2.44

Figure 2.45.a

Figure 2.45.b

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2.3 Kinetics of Particle / Impulse and Momentum

Question 2.3.1(*): When an object weighing 50 N is moving down a frictionless road with a 30° slope at a velocity of 1 m/s, a force F that changes with time acts on the object as in the graph. Accordingly; calculate the maximum value of the object’s velocity. (𝑔=10m/s²)

Answer: 21m/s

Figure 2.46

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2.3 Kinetics of Particle / Impulse and Momentum

Figure 2.47

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2.3.5 Impulsive Forces :

High-intensity forces that act on an object in a very short time interval but change the momentum of the object at a level that cannot be neglected are called impulsive forces. (The force exerted by a baseball bat on the ball is an example of impulsive forces.)

=

2.3 Kinetics of Particle / Impulse and Momentum

The impulses of gravity forces are very small compared to impulsive forces and are ignored unless stated otherwise.

Figure 2.48

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2.3.6 Conservation of Momentum :

In a particle or system of particles, if the sum of the impulses of external forces is zero, the total momentum of the system does not change, that is, momentum is conserved.

Note: If momentum is conserved, we cannot say that energy is also conserved.

Conservation of Momentum Equation:

total momentum does not change.

(2.23)

2.3 Kinetics of Particle / Impulse and Momentum

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Example : The man in boat A is pulling the empty boat B with a rope (frictions are ignored)

2.3.7 Two important cases in Conservation of Momentum

Case 1: If external forces constantly balance each other during motion, the momentum of the system is conserved.

Boat A:

Boat B:

Entire System:

(the total momentum of the system is conserved)

+

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

Figure 2.49

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Case 2: The time interval (Δt) is very short, and since the external forces are not of the impulsive type, their total impulses are at a level that can be neglected.

(the total momentum of the system is conserved.)

Example: A bullet thrown from below and stuck into a ball attached to a rope. (The rope becomes loose and no force is created in the rope.)

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.50

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If the bullet is sent from above:

An impulsive force occurs in the rope, in this case momentum is not conserved in the direction of the force.

Momentum is conserved in the x-direction.

Impulse-Momentum Equations of the System:

Vectorial:

Scaler:

Momentum is not conserved in the y-direction.

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.51

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2.3.8 Impact (or Collision)

Type 1: Central Impact:

Line of Impact : It is the direction that connects the centers of gravity of colliding objects.

The velocity lines of the objects before and after the impact coincide with line of impact.

We can divide impact into 2 different types:

Line of impact

Type 2: Oblique Impact

The velocity lines of the objects before and after the impact do not coincide with line of impact.

Line of impact

Our goal now is to find the velocities after impact for both impact types, while velocities before impact are known.. >>

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.52

Figure 2.53

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2.3.8.1) Type 1: Central Impact

Momentum is conserved for the entire system:

A

B

A

B

A

B

Before Impact

(At the moment of impact, the objects change shape for a very short time depending on the type of material. At the maximum deformation position, they have a common velocity (u). Then the objects return to their initial state and gain their final velocity. )

Moment of impact

After Impact

Impulse-momentum equations in the x-direction:

For ball A:

For ball B:

Before Impact

Impulse during

Max Deformation

Impulse during restitution (return period)

After Impact

(2.24)

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.54

Figure 2.55

Figure 2.56

�

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2.3.8.1.a) Coefficient of restitution (e):

If we consider the motion of balls A and B in two stages:

Calculation of Coefficient of restitution (e):

Moment of impact

Before Impact

Impulse during

max deformation

Impulse during restitution

After Impact

Moment of impact

Before Impact

Impulse during

max deformation

Impulse during restitution

After Impact

Moment of impact

It is the coefficient related to the materials of the colliding objects.

(2.25)

From the movement of Ball A

From the movement of Ball B

( II )

( I )

From equ. (I) and (II)

A

A

A

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.57

Figure 2.58

Figure 2.59

Figure 2.60

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Train cars (wagons) A and B, moving in opposite directions on the same track, slowly collide with each other and merge.

a-) Calculate their common velocity after impact,

b-) Calculate the average force they exert on each other.

Masses: m_A = 15 tons, m_B = 12 tons,

Velocities before impact:

v_A = 1.5m/s and v_B = 0.75m/s,

Impact time: 0.8 seconds

Solution:

During the impact, the momentum of the entire system is conserved.

If we write for the A wagon:

a-)

b-)

Common velocity after merging (or after impact): v₂

Example 2.3.4

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Momentum is not conserved for a single wagon. Impulse-Momentum equation can be written.

Figure 2.61

Figure 2.62.a

Figure 2.62.b

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Example 2.3.5

While a 2kg box A is moving to the right at a velocity of 18m/s, a force varying in time in the form of F = 3t² –t +3 in the same direction starting from point C acts on box A for 2 seconds (up to point D). After the force is removed at point D, box A collides with the 1kg box B coming to the left at a constant velocity of 12m/s at exactly point E. Calculate the velocities of boxes A and B after the collision by taking the coefficient of restitution e=0.5. Neglect friction.

A

B

F=3t²-t+3

D

E

C

Solution..>>

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.63

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A

B

F=3t²-t+3

A

C

D

A

B

E

Solution:

(1)

• From (1) and (2)

Impulse - Momentum Equation for Box A between C-D:

(Since there is no friction between D and E, the velocity of box A does not change.)

e=0.5

:constant

Impact at point E

• The momentum of the system in the x-direction is conserved.

• coefficient of restitution :

(2)

2.3 Maddesel Nokta Kinetiği / / İmpuls ve Momentum /Momentumun Korunumu

Figure 2.64

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A

B

k

k

The rings A and B, whose weights are W_A = 2N and WB = 3N respectively, can slide frictionlessly on a horizontal bar as shown in the figure. After the spring at C is compressed 18cm, ring A is released and the ring slides and hits ring B, and then ring B compresses spring D. Accordingly, calculate the maximum compression amount of spring D. The coefficient of restitution is e = 0.8 and the spring coefficient for both springs is k = 4N/cm.

(Answer: Spring D is compressed 15.9cm).

C

D

Question 2.3.3 (*)

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

Figure 2.65

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2.3.8.2) Type 2- Oblique Impact

4 independent linear equations containing unknowns are found with the following operations:

: The velocity directions of the objects and the line of impact are different.

4-) coefficient of restitution is written in the x direction:

In this case, since no impulse will act in the y direction, the momentum of the objects in this direction is preserved separately.

(2.26.a)

(2.26.b)

(2.26.c)

(2.26.d)

Line of impact

Therefore, the momentum of the entire system in the x-direction is conserved:

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

Figure 2.66

Figure 2.67

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Example 2.3.6

A billiard player wants to hit ball B with ball A, and put both balls into the holes. To do this, he hits ball A with a billiards stick at an angle of 30° to the vertical and gives it a speed of 6 m/s. Determine the conditions for the billiard player to achieve his goal. (Initially, both balls are at rest.Neglect friction and the dimensions of the balls. The coefficient of restitution is e = 0.5, the masses of the balls are the same, and their radii are 30 mm. At the moment of collision, the common tangent of the contact points of the balls is in the vertical direction.)

A

B

0,142m

1.4m

r_A=r_B=30mm

v_A =6m/s, v_B=0

e=0.5

Do balls go into holes?

Solution..>>

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

Figure 2.68

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Solution:

If it is in the AD direction, it enters hole D.

This type of impact is the oblique impact.

1 and2) The y-components of the momentums of balls A and B are conserved.

If it is in the BC direction, it enters hole C.

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

Figure 2.69

4 unknowns are found with the following 4 equation:

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3-) The momentum of the system in the x direction is conserved:

(I)

4-) The coefficient of restitution in the x direction

(II)

From (I) and (II)

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

is found.

Condition 2: The coefficient of restitution is related to the material type of the billiard balls. The material of the balls must be such that they provide a coefficient of restitution of e = 0.5.

Figure 2.70

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Question 2.3.4 (*)

2.3 Kinetics of Particle / Impulse and Momentum / Conservation of Momentum

Figure 2.71

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Vectorial Calculation:

point at which moment is taken

The point where the object is located (or any point on the tangent)

The linear momentum at that instant

The 1st derivative of the Angular Momentum with respect to time is equal to the moment of the external force about that point.

(2.27)

(2.28)

2.3.9 Angular Momentum and Angular Impulse : These are concepts that enable us to reach solutions to

some special problems more easily.

(The vector product of vectors in the same direction is zero.)

2.3.9.2 Angular Impuls: It is the impulse of the moment of the external force. It is calculated for a time interval:

(2.29)

2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.72

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2.3.9.3 Angular Impulse – Angular Momentum Principle:

For particles system:

Angular momentum-angular impulse equation between positions 1-2:

(Angular Impuls)

(2.30a)

(2.31)

Recalling Equation 2.28 again::

(2.30b)

More clearly :

In Planar Motion, 1 scalar equation can be written. (d₁ and d₂ are perpendicular distances):

(2.32)

3D Motion

angular momentum is conserved..>>

(2.33)

2.3.9.4 Conservation of Angular Momentum : We pay attention to equations 2.30a and 2.30b,

2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.73

Figure 2.74

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A

B

O

A rope is passed through a hole in the middle of a table, and one end is attached to ball B on the table, and the other end is attached to cart A. While cart A is stationary, ball B makes a circular motion at a constant speed of 2m/s. Cart A starts moving downward at a constant speed of 3m/s. Accordingly, calculate the speed of ball B 0.1 seconds after cart A starts moving. (𝑂𝐵=0.6𝑚)

Example 2.3.7

2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.75

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At a time t,

Solution:

- Since N and W balance each other at every moment of the movement, they have no effect on the movement.

- While ball B makes a circular motion with an initial velocity of 𝑣₁ on the table plane, it gains an additional motion towards the center O (in the radial direction) due to the movement of the car A. Since it is attached to the car with a rope, the radial velocity of the ball (𝑣_𝑅) is equal to the velocity of the car A.

We think the car has not started moving yet:

Position 2:

Position 1:

The radial velocity of the ball is equal to the velocity of the car:

(given in the question)

(will be calculated)

Length of rope at position 2 :

2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.76

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2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.77

Figure 2.78

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Example 2.3.8 : A mass of 10kg C is attached to the end of the ⊢ shaped element that rotates around the z axis. While the initial speed of the C mass is 2m/s in the + x direction, a constant force of F = 60N and a variable moment of M = 8t² + 5 are applied as shown in the figure. Calculate the speed of the C mass after 2 seconds. Due to the connections, there is only rotation in the z axis.

Solution:

In order to avoid making sign errors, we will make a vector solution.

According to the right-hand rule,

the moment vector M is in the –z direction:

The total moment of the singular force of 60 N:

The total moment of the singular force of 60 N:

Angular momenta of mass C with respect to point O :

2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.79

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Question 2.3.5 (*)

2.3 Kinetics of Particle / Impulse and Momentum / Angular Impuls and Angular Momentum

Figure 2.80