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 each other, the total momentum of the system before the interaction is equal to the total momentum of the system after the interaction Lecture 19 1/24 Lecture 19 2/24 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 19 3/24 Example: A Space Repair Astronaut pushes solar panel, giving it velocity of -0.30 m/s in x- direction. Astronaut s mass is 60 kg; panel s mass 80 kg. Both astronaut and panel initially at rest. What is astronaut s velocity after push? External forces do no work, so system momentum conserved. r r r r mv P Pf + mv A Af = mv P Pi + mv A Ai = 0 r r mv P Pf = mv 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 19 4/24
Example: Runaway Railroad Car Runaway 10,000 kg railroad car is rolling horizontally at 4.00 m/s toward a switchyard. As it passes grain elevator, 2,000 kg of grain suddenly drops into car. Assume that grain drops vertically and that rolling friction and air drag are negligible. How fast is car going after grain drops in? Lecture 19 5/24 Example: Runaway Railroad Car External forces do no work, so momentum of system (car & grain) is conserved. Psys fx = Psys ix ( mc + mg) vf x = mcvix + mg(0) v mc fx = v ix m + m c g v fx = (4.0 m/s)(10,000 kg)/(12,000 kg) = 3.3 m/s Lecture 19 6/24 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 19 7/24 Sketches for Collision Problems Draw before and after sketches Label each object include the direction of velocity keep track of subscripts Lecture 19 8/24
Sketch: Perfectly Inelastic Collisions The objects stick together Include all the velocity directions The after collision combines the masses Inelastic Collisions A completely inelastic collision: mv + mv = ( m + m ) v 1 1, i 2 2, i 1 2 f Lecture 19 9/24 Lecture 19 10/24 Inelastic Collisions Solving for the final momentum in terms of the initial momenta and masses: Lecture 19 11/24 Example: Goal-Line Stand 95.0 kg running back runs forward at 3.75 m/s. 111 kg line backer moving at 4.10 m/s meets runner in head-on collision and locks arms around runner. (a) Find their velocity immediately after collision. (b) Find initial & final kinetic energies. mv 1 1, i + mv 2 2, i v f 1 2 1 2 i 2 1 1, i 2 2 2, i = = m + m 1 2 (95.0 kg)(3.75 m/s) + (111.0 kg)( 4.10 m/s) = = 0.480 m/s (95.0 kg) + (111.0 kg) K = mv + mv = 1600J ( ) K = m+ m v = 24J 1 2 f 2 1 2 f Lecture 19 12/24
Explosions An explosion in which particles of system move apart from each other after 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 explosion, so net momentum of fragments equals initial momentum. Lecture 19 13/24 Example: Recoil of Rifle 10 g bullet fired from 3.0 kg rifle with speed of 500 m/s. What is 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) = -1.67 m/s (3.0 kg) Lecture 19 14/24 Elastic Collisions In elastic collisions, both kinetic energy and momentum are conserved. One-dimensional elastic collision: 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 1 2 1 2 1 2 1 2 2 1 1f 2 2 2 f 2 1 1i 2 2 2i Lecture 19 15/24 Lecture 19 16/24
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: Elastic Collisions: 3 Cases* 2m ( v ) = ( v ) ; 1 fx 2 m1+ m2 ix 1 m m ( v ) ( ) 1 2 fx 2 ix 1 fx 1 1 2 fx 1 = vix 1 m1+ m2 m = m : ( v ) = ( v ) and ( v ) = 0 (knock-on) Lecture 19 17/24 Lecture 19 18/24 m >> m : ( v ) = 2( v ) and ( v ) = ( v ) (boost-ahead) 1 2 fx 2 ix 1 fx 1 ix 1 The center of mass of a system is the point where the system can be balanced in a uniform gravitational field. m << m : ( v ) = 0 and ( v ) = ( v ) (bounce-off) 1 2 fx 2 fx 1 ix 1 Lecture 19 19/24 Lecture 19 20/24
For two objects: The center of mass need not be within the object: The center of mass is closer to the more massive object. Lecture 19 21/24 Lecture 19 22/24 Motion Total mass multiplied by the acceleration of the center of mass is equal to the net external force: End of Lecture 19 Before Monday, read Walker 10.1-2. Homework Assignment #9b is due at 11:00 PM on Tuesday, Oct. 20. Center of mass accelerates as though it were a point particle of mass M acted on by Lecture 19 23/24 Lecture 19 24/24