1. Chapter 7
Linear Momentum

2. Chap07- Linear Momentum: Revised 3/3/2010
Definition of Momentum
Impulse
Conservation of Momentum
Center of Mass
Motion of the Center of Mass
Collisions (1d, 2d; elastic, inelastic)
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

3. Momentum From Newton’s Laws
Consider two interacting bodies with m2 > m1:
F21
F12 m1 m2
If we know the net force on each body then
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 7, Questions 11 and 18.
The velocity change for each mass will be different if the masses are different.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

4. Chap07- Linear Momentum: Revised 3/3/2010
Rewrite the previous result for each body as:
From the 3rd Law
Combine the two results:
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

5. Define Momentum The quantity (mv) is called momentum (p).
p = mv and is a vector.
The unit of momentum is kg m/s; there is no derived unit for momentum.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

6. Chap07- Linear Momentum: Revised 3/3/2010
From slide (4):
The change in momentum of the two bodies is “equal and opposite”. Total momentum is conserved during the interaction; the momentum lost by one body is gained by the other.
Δp1 +Δp2 = 0
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

7. Impulse Definition of impulse:
We use impulses to change the momentum of the system.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 7, Questions 2, 3, 4, 7, 8, 9, 12, 14, and 16.
One can also define an average impulse when the force is variable.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

8. Chap07- Linear Momentum: Revised 3/3/2010
Impulse Example
Example (text conceptual question 7.2): A force of 30 N is applied for 5 sec to each of two bodies of different masses.
30 N
m1 or m2
Take m1 < m2
(a) Which mass has the greater momentum change?
Since the same force is applied to each mass for the same interval, p is the same for both masses.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

9. Chap07- Linear Momentum: Revised 3/3/2010
Example continued:
(b) Which mass has the greatest velocity change?
Since both masses have the same p, the smaller mass (mass 1) will have the larger change in velocity.
(c) Which mass has the greatest acceleration?
Since av the mass with the greater velocity change will have the greatest acceleration (mass 1).
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

10. Change in Momentum Example
An object of mass 3.0 kg is allowed to fall from rest under the force of gravity for 3.4 seconds. Ignore air resistance.
What is the change in momentum?
Want p = mv.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

11. Change in Momentum Example
What average force is necessary to bring a 50.0-kg sled from rest to 3.0 m/s in a period of 20.0 seconds? Assume frictionless ice.
The force will be in the direction of motion.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

12. Conservation of Momentum
v1i
v2i m1 m2
m1>m2
A short time later the masses collide. m1 m2
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 7, Questions 1 and 15
What happens?
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

13. Chap07- Linear Momentum: Revised 3/3/2010
During the interaction: N1 w1
F21 N2 w2
F12 x y
There is no net external force on either mass.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

14. Chap07- Linear Momentum: Revised 3/3/2010
The forces F12 and F21 are internal forces. This means that:
In other words, pi = pf. That is, momentum is conserved. This statement is valid during the interaction (collision) only.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

15. Chap07- Linear Momentum: Revised 3/3/2010
Rifle Example
A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s.
What is the recoil speed of the rifle as the bullet leaves the barrel?
As long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

16. Center of Mass
The center of mass (CM) is the point representing the mean (average) position of the matter in a body.
This point need not be located within the body.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 7, Questions 6 and 20.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

17. Chap07- Linear Momentum: Revised 3/3/2010
Center of Mass
The center of mass (of a two body system) is found from:
This is a “weighted” average of the positions of the particles that compose a body. (A larger mass is more important.)
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

18. Chap07- Linear Momentum: Revised 3/3/2010
Center of Mass
In 3-dimensions write
where
The components of rcm are:
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

19. Chap07- Linear Momentum: Revised 3/3/2010
Center of Mass Example
Particle A is at the origin and has a mass of 30.0 grams. Particle B has a mass of 10.0 grams.
Where must particle B be located so that the center of mass (marked with a red x) is located at the point (2.0 cm, 5.0 cm)? x y A
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

20. Chap07- Linear Momentum: Revised 3/3/2010
Center of Mass Example
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

21. Chap07- Linear Momentum: Revised 3/3/2010
Find the Center of Mass
The positions of three particles are (4.0 m, 0.0 m), (2.0 m, 4.0 m), and (1.0 m, 2.0 m). The masses are 4.0 kg, 6.0 kg, and 3.0 kg respectively.
What is the location of the center of mass? x y 3 2 1
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

22. Chap07- Linear Momentum: Revised 3/3/2010
Example continued:
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

23. Motion of the Center of Mass
For an extended body, it can be shown that p = mvcm.
From this it follows that Fext = macm.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 7, Question 13.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

24. Center of Mass Velocity
Body A has a mass of 3.0 kg and vx = m/s. Body B has a mass of 4.0 kg and has vy = 7.0 m/s.
What is the velocity of the center of mass of the two bodies?
Consider a body made up of many different masses each with a mass mi.
The position of each mass is ri and the displacement of each mass is ri = vit.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

25. Center of Mass Velocity
For the center of mass:
Solving for the velocity of the center of mass:
Or in component form:
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

26. Center of Mass Velocity
Applying the previous formulas to the example,
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

27. Collisions in One Dimension
When there are no external forces present, the momentum of a system will remain unchanged. (pi = pf)
If the kinetic energy before and after an interaction is the same, the “collision” is said to be perfectly elastic. If the kinetic energy changes, the collision is inelastic.
Could incorporate personal response system questions from the College Physics by G/R/R 2E ARIS site ( Instructor Resources: CPS by eInstruction, Chapter 7, Questions 5, 10, 17 and 19.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

28. Collisions in One Dimension
If, after a collision, the bodies remain stuck together, the loss of kinetic energy is a maximum, but not necessarily a 100% loss of kinetic energy. This type of collision is called perfectly inelastic.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

29. Inelastic Collisions in One Dimension
In a railroad freight yard, an empty freight car of mass m rolls along a straight level track at 1.0 m/s and collides with an initially stationary, fully loaded, boxcar of mass 4.0m. The two cars couple together upon collision.
(a) What is the speed of the two cars after the collision?
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

30. Inelastic Collisions in One Dimension
(b) Suppose instead that both cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one had v1i = 1.0 m/s.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

31. One Dimension Projectile
A projectile of 1.0 kg mass approaches a stationary body of 5.0 kg mass at 10.0 m/s and, after colliding, rebounds in the reverse direction along the same line with a speed of 5.0 m/s.
What is the speed of the 5.0 kg mass after the collision?
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

32. Collision in Two Dimensions
Body A of mass M has an original velocity of 6.0 m/s in the +x-direction toward a stationary body (B) of the same mass. After the collision, body A has vx = +1.0 m/s and vy = +2.0 m/s.
What is the magnitude of body B’s velocity after the collision? B A
Final A B
vAi
Initial
Angle is 90o
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

33. Collision in Two Dimensions
x momentum:
y momentum:
Solve for v2fx:
Solve for v2fy:
The mag. of v2 is
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

34. Chap07- Linear Momentum: Revised 3/3/2010
Summary
Definition of Momentum
Impulse
Center of Mass
Conservation of Momentum
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

35. Chap07- Linear Momentum: Revised 3/3/2010
Extra Slides
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

36. Conservation of Momentum
Two objects of identical mass have a collision. Initially object 1 is traveling to the right with velocity v1 = v0. Initially object 2 is at rest v2 = 0.
After the collision:
Case 1: v1’ = 0 , v2’ = v0 (Elastic)
Case 2: v1’ = ½ v0 , v2’ = ½ v0 (Inelastic)
In both cases momentum is conserved.
Case 1 Case 2
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

37. Conservation of Kinetic Energy?
Case 1 Case 2
KE After = KE Before
KE After = ½ KE Before
(Conserved)
(Not Conserved)
KE = Kinetic Energy = ½mvo2
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

38. Where Does the KE Go? Case 1 Case 2 KE After = KE Before
(Conserved)
(Not Conserved)
In Case 2 each object shares the Total KE equally.
Therefore each object has KE = 25% of the original KE Before
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

39. Chap07- Linear Momentum: Revised 3/3/2010
50% of the KE is Missing
Each rectangle represents KE = ¼ KE Before
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010

40. Chap07- Linear Momentum: Revised 3/3/2010
Turn Case 1 into Case 2
Each rectangle represents KE = ¼ KEBefore
Take away 50% of KE. Now the total system KE is correct
But object 2 has all the KE and object 1 has none
Use one half of the 50% taken to speed up object 1. Now it has 25% of the initial KE
Use the other half of the 50% taken to slow down object 2. Now it has only 25% of the the initial KE
Now they share the KE equally and we see where the missing 50% was spent.
MFMcGraw
Chap07- Linear Momentum: Revised 3/3/2010