Physics 111. Lecture 20 (Walker: 9.4-6) Momentum Conservation Collisions Center of Mass March 16, Quiz Wednesday - Chaps. 7 & 8, plus 9.

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1 Physics 111 Lecture 20 (Walker: 9.4-6) Momentum Conservation Collisions Center of Mass March 16, 2009 Quiz Wednesday - Chaps. 7 & 8, plus Lecture 20 1/30 Conservation of Linear Momentum The net force acting on an object is the rate of change of its momentum: If the net force is zero, the momentum does not change: Can extend this to systems of 2 or more particles. Lecture 20 2/30

Conservation of Linear Momentum Internal versus External Forces: Internal forces act between objects within the system. As with all forces, they occur in action-reaction pairs.

2 Conservation of Linear Momentum Internal versus External Forces: Internal forces act between objects within the system. As with all forces, they occur in action-reaction pairs. As all pairs act between objects in the system, the internal forces always sum to zero: Therefore, the net force acting on a system is the sum of the external forces acting on it. Lecture 20 3/30 Conservation of Linear Momentum Furthermore, internal forces cannot change the momentum of a system. However, the momenta of components of the system may change. Lecture 20 4/30

3 Conservation of (System) Momentum When no external forces do work on a system consisting of objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision Lecture 20 5/30 Conservation of (System) Momentum Mathematically: m1 v1i + m2v2i = m1 v1 f + m2v2 f Momentum is conserved for an isolated system of objects The system includes all the objects interacting with each other Assumes only internal forces are acting during the collision Can be generalized to any number of objects Lecture 20 6/30

Conservation of Linear Momentum An example of internal forces moving components of a system: Lecture 20 7/30 ConcepTest Suppose a person jumps on the surface of Earth. The Earth 1.

4 Conservation of Linear Momentum An example of internal forces moving components of a system: Lecture 20 7/30 ConcepTest Suppose a person jumps on the surface of Earth. The Earth 1. Will not move at all 2. Will recoil in the opposite direction with tiny velocity 3. Might recoil, but there is not enough information provided to see if that could happened Lecture 20 8/30

Problem Solving Strategy Picture: Determine that net external force ΣF ext (or ΣF ext x ) on system is negligible (or does no work). (If net external force NOT negligible, do not proceed.) Solve: 1.

5 Problem Solving Strategy Picture: Determine that net external force ΣF ext (or ΣF ext x ) on system is negligible (or does no work). (If net external force NOT negligible, do not proceed.) Solve: 1. Draw a sketch showing system before and after the collision. Include coordinate axes and label the initial and final velocity vectors. 2. Equate total initial momentum to total final momentum and express as a vector equation (or one or more scalar equations involving x and y components.) 3. Substitute the given information into the equation(s) and solve for the quantity or quantities of interest. Check: Make sure you include any minus signs that accompany velocity components, because momentum can have either sign. Lecture 20 9/30 Example: A Space Repair Astronaut pushes a solar panel, giving it a velocity of m/s in the x-direction. The astronaut s mass is 60 kg, and the panel s mass is 80 kg. Both astronaut and panel are initially at rest. What is the astronaut s velocity after the push? External forces do no work, so system momentum conserved. r r r r mv + mv = mv + mv = 0 r mv r = mv P Pf A Af P Pi A Ai P Pf A Af r mp r (80 kg) v ( 0.30 m/s) ˆ (0.40 m/s) ˆ Af = vpf = x= x m (60 kg) A Lecture 20 10/30

Example: Runaway Railroad Car Runaway 10,000 kg railroad car is rolling horizontally at 4.00 m/s toward a switchyard. As it passes a grain elevator, 2,000 kg of grain suddenly drops into the car.

6 Example: Runaway Railroad Car Runaway 10,000 kg railroad car is rolling horizontally at 4.00 m/s toward a switchyard. As it passes a grain elevator, 2,000 kg of grain suddenly drops into the car. Assume that the grain drops vertically and that rolling friction and air drag are negligible. How fast is car going after grain drops in? Lecture 20 11/30 Example: Runaway Railroad Car External forces do no work, so momentum of system (car & grain) is conserved. P = P sys fx sys ix ( m + m ) v = m v + m (0) c g f x c ix g m v c fx = v ix m m c + g v fx = (4.0 m/s)(10,000 kg)/(12,000 kg) = 3.3 m/s Lecture 20 12/30

7 Collisions Collision: two objects striking one another. Time of collision is short enough that external forces may be ignored; system momentum conserved. What about system mechanical energy? Elastic collision: momentum and kinetic energy conserved; p f = p i & K f = K i Inelastic collision: momentum is conserved but kinetic energy is not: p f = p i but K f K i. Completely inelastic collision: objects stick together afterwards: p f = p i1 + p i2 Lecture 20 13/30 Sketches for Collision Problems Draw before and after sketches Label each object include the direction of velocity keep track of subscripts Lecture 20 14/30

8 Sketch: Perfectly Inelastic Collisions The objects stick together Include all the velocity directions The after collision combines the masses Lecture 20 15/30 Inelastic Collisions A completely inelastic collision: mv + mv = ( m + m ) v 1 1, i 2 2, i 1 2 f Lecture 20 16/30

Inelastic Collisions Solving for the final momentum in terms of the initial momenta and masses: Lecture 20 17/30 Example: Goal-Line Stand 95.0 kg running back runs toward end zone at 3.75 m/s.

9 Inelastic Collisions Solving for the final momentum in terms of the initial momenta and masses: Lecture 20 17/30 Example: Goal-Line Stand 95.0 kg running back runs toward end zone at 3.75 m/s. A 111 kg line backer moving at 4.10 m/s meets the runner in a head-on collision and locks his arms around the runner. (a) Find their velocity immediately after the collisions. (b) Find initial & final kinetic energies. v mv + mv f = 1 1, i 2 2, i = m1+ m i 2 1 1, i 2 2 2, i K = mv + m v = f ( ) K = m + m v f Lecture 20 18/30

10 Explosions An explosion in which particles of a system move apart from each other after a brief, intense interaction, is the opposite of a collision. The explosive forces are internal forces. If the system is isolated, its total momentum will be conserved during the explosion, so the net momentum of the fragments equals the initial momentum. Lecture 20 19/30 Example: Recoil of a Rifle 10 g bullet is fired from 3.0 kg rifle with speed of 500 m/s. What is the recoil speed of rifle? p ix = 0 P = m ( v ) + m ( v ) = P = 0 fx B fx bullet R fx rifle ix mb ( vfx) rifle = ( vfx) bullet mr (0.010 kg) = (500 m/s) = m/s (3.0 kg) Lecture 20 20/30

11 Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. One-dimensional elastic collision: Lecture 20 21/30 Momentum Conservation Elastic Collisions in 1D mv + mv = mv + mv 1 1f 2 2 f 1 1i 2 2i Energy Conservation mv + mv = mv + mv f f 2 1 1i 2 2 2i Lecture 20 22/30

12 Elastic Collisions We have two equations (conservation of momentum and conservation of kinetic energy) and two unknowns (the final speeds). Solving for the final speeds: Lecture 20 23/30 Elastic Collisions: 3 Cases-Equal Mass 2m ( v ) = ( v ) ; fx 1 2 ix 1 m1+ m2 m m ( v ) ( ) 1 2 fx 1 = vix 1 m1+ m2 m = m : ( v ) = ( v ) and ( v ) = 0 (knock-on) 1 2 fx 2 ix 1 fx 1 Lecture 20 24/30

13 m >> m : ( v ) = 2( v ) and ( v ) = ( v ) (boost-ahead) 1 2 fx 2 ix 1 fx 1 ix 1 m << m : ( v ) = 0 and ( v ) = ( v ) (bounce-off) 1 2 fx 2 fx 1 ix 1 Lecture 20 25/30 Center of Mass The center of mass of a system is the point where the system can be balanced in a uniform gravitational field. Lecture 20 26/30

14 For two objects: Center of Mass The center of mass is closer to the more massive object. Lecture 20 27/30 Center of Mass The center of mass need not be within the object: Lecture 20 28/30

Center of Mass The total mass multiplied by the acceleration of the center of mass is equal to the net external force: The center of mass accelerates just as though it were a point particle of mass M

15 Center of Mass The total mass multiplied by the acceleration of the center of mass is equal to the net external force: The center of mass accelerates just as though it were a point particle of mass M acted on by Lecture 20 29/30 End of Lecture 20 Before Wednesday, read Walker Homework Assignment #9b is due at 11:00 PM on Wednesday, March 18. Quiz Wednesday - Chaps. 7 & 8, plus Lecture 20 30/30

Physics 111. ConcepTest. Lecture 19 (Walker: 9.4-7) Momentum Conservation Collisions Center of Mass Oct. 16, r (80 kg)

Physics 111. ConcepTest. Lecture 19 (Walker: 9.4-7) Momentum Conservation Collisions Center of Mass Oct. 16, r (80 kg) Physics 111 Lecture 19 (Walker: 9.4-7) Momentum Conservation Collisions Oct. 16, 2009 Conservation of (System) Momentum When no external forces do work on a system consisting of objects that interact with