Chapter 7 Linear Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Linear Momentum Definition of Momentum Impulse Conservation of Momentum Center of Mass Motion of the Center of Mass Collisions (1d, 2d; elastic, inelastic)

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Momentum From Newton’s Laws Consider two interacting bodies with m 2 > m 1 : If we know the net force on each body then The velocity change for each mass will be different if the masses are different. F 21 F 12 m1m1 m2m2

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Rewrite the previous result for each body as: Combine the two results: Momentum From the 3rd Law

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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. Define Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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. Δp 1 +Δp 2 = 0 Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Impulse Definition of impulse: One can also define an average impulse when the force is variable. We use impulses to change the momentum of the system.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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 m 1 or m 2 Take m 1 < m 2 (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. Impulse Example

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ (b) Which mass has the greatest velocity change? Example continued: 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-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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. Change in Momentum Example

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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. Change in Momentum Example

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Impulse Since we are applying the change of momentum over a time interval consisting of the instant before the impact to the instant after the impact, we ignore the effect of gravity over that time period

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Impulse and Momentum The force in impulse problems changes rapidly so we usually deal with the average force

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Work and Impulse Comparison Work changes E T Work = 0 E T is conserved Work = Force x Distance Impulse changes Impulse = 0 is conserved Impulse = Force x Time

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Conservation of Momentum v 1i v 2i m1m1 m2m2 m 1 >m 2 m1m1 m2m2 A short time later the masses collide. What happens?

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ During the interaction: N1N1 w1w1 F 21 x y There is no net external force on either mass. N2N2 w2w2 F 12

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ The forces F 12 and F 21 are internal forces. This means that: In other words, p i = p f. That is, momentum is conserved. This statement is valid during the interaction (collision) only.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Momentum Examples

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Generic Zero Initial Momentum Problem These problems occur whenever all the objects in the system are initially at rest and then separate.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Changing the Mass While Conserving the Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Changing the Mass While Conserving the Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Changing the Mass While Conserving the Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Astronaut Tossing How Long Will the Game Last?

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ How Long Will the Game Last?

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ The Ballistic Pendulum Conservation of Momentum Conservation of Energy

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ The High Velocity Ballistic Pendulum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Relative Speed Relationship The relative speed relationship applies to elastic collisions. In an elastic collision the total kinetic energy is also conserved. If a second equation is needed to solve a momentum problem, the relative speed relationship is easier to deal with than the KE equation.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Relative Speed Relationship

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Relative Speed in Problem Solving Need a second equation when solving for two variables in an elastic collision problem? The relative speed relationship is preferred over the conservation ofkinetic energy relationship

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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. Rifle Example

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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.) Center of Mass

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ In 3-dimensions write where The components of r cm are: Center of Mass

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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 x Center of Mass Example

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Example

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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 Find the Center of Mass

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Example continued:

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Motion of the Center of Mass For an extended body, it can be shown that p = mv cm. From this it follows that  F ext = ma cm.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ For the center of mass: Solving for the velocity of the center of mass: Or in component form: Center of Mass Velocity

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Reference System

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Treatment

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Treatment

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Treatment

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Treatment

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Center of Mass Reference System Center of Mass System Laboratory System

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Collisions in One Dimension When there are no external forces present, the momentum of a system will remain unchanged. (p i = p f ) 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.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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. Collisions in One Dimension

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Elastic Collision - Unequal Masses In the later slides, the initially moving mass, m n, will called be m 1 while the initially stationary mass, m c, will be m 2.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Elastic Collision - KE Transfer

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Elastic Collision - Fractional KE “Loss” For equal masses all of the KE is transferred from m 1 to m 2 The curve is like an energy transfer function from m 1 to m 2. For small m 2 the mass m 1 just rolls on through. For large m 2 the smaller m 1 just bounces off without transferring much energy. Efficient energy transfer requires equal masses.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Elastic Collision - KE Transfer For m 2 > m 1 the velocity of the 2nd mass gets a smaller velocity and even the larger mass can’t overcome the shrinking v 2 factor in KE 2

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Inelastic Collision - Equal Masses v1v1 v 2 = 0 u m1m1 m2m2 m 1 + m 2 Before After Momentum Kinetic Energy

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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? Inelastic Collisions in One Dimension

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ (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 v 1i = 1.0 m/s. Inelastic Collisions in One Dimension

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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? One Dimension Projectile

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ 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 v x = +1.0 m/s and v y = +2.0 m/s. What is the magnitude of body B’s velocity after the collision? A B v Ai Initial B A Final Collision in Two Dimensions Angle is 90 o

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ x momentum: Solve for v 2fx : Solve for v 2fy : y momentum: The mag. of v 2 is Collision in Two Dimensions

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Collisions in Two and Three Dimensions

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Totally Inelastic Collision

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Totally Inelastic Collision If the objects stick together after the collision, then the collision is INELASTIC. Just because the objects DON’T stick together after the collision, doesn’t mean the collision is ELASTIC.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ General Glancing Collision Initially only one object is moving

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ General Glancing Collision Initially only one object is moving

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Another 3-Vector Problem Momentum problems where one of the objects is initially at rest yield 3- vector problems that can be solved using triangles, instead of simultaneous equations.

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Combining Conservation of Energy, Momentum, and KE Transfer in Equal Mass Inelastic Collisions

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Conservation of Energy Conservation of Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Summary Definition of Momentum Impulse Center of Mass Conservation of Momentum

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Extra Slides

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Conservation of Momentum Two objects of identical mass have a collision. Initially object 1 is traveling to the right with velocity v 1 = v 0. Initially object 2 is at rest v 2 = 0. After the collision: Case 1: v 1 ’ = 0, v 2 ’ = v 0 (Elastic) Case 2: v 1 ’ = ½ v 0, v 2 ’ = ½ v 0 (Inelastic) In both cases momentum is conserved. Case 1Case 2

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Conservation of Kinetic Energy? KE = Kinetic Energy = ½ mv o 2 Case 1Case 2 KE After = KE Before KE After = ½ KE Before (Conserved) (Not Conserved)

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Where Does the KE Go? Case 1Case 2 KE After = KE Before KE After = ½ 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-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ % of the KE is Missing Each rectangle represents KE = ¼ KE Before

MFMcGraw-PHY 1401Chap07b- Linear Momentum: Revised 6/28/ Turn Case 1 into Case 2 Each rectangle represents KE = ¼ KE Before 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.