# Chapter 7 Linear Momentum.

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Chapter 7 Linear Momentum

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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