P H Y S I C S Chapter 9: Collisions, CM, RP Since impulse = change in momentum, If no impulse is exerted on an object, the momentum of the object will not change. If no external forces act on a system, the total momentum of the system will not change. Such a system is called an isolated system. is conserved in every isolated system. This leads into our next discussion. Another way to think about it is: Internal forces can never change the total momentum of a system. In practice, for any event in an isolated system: after = before What are internal forces? Forces that act between the objects 1
Types of Collisions Equations Type Kinetic Energy Conserved Conserved Elastic P P Inelastic Some KE converts to thermal energy P Stick Together Perfectly Inelastic P P See your Best Friend for all equations: Elastic collision problem Elastic collision problem A 5.00-kg ball, moving to the right at a velocity of +2.00 m/s on a frictionless table, collides head-on with a stationary 7.50- kg ball. Find the final velocities of the balls if the collision is elastic. A 5.00-kg ball, moving to the right at a velocity of +2.00 m/s on a frictionless table, collides head-on with a stationary 7.50-kg ball. Find the final velocities of the balls if the collision is elastic. 2
Identify the number and types of collisions in the animation below. Identify the number and types of collisions in the animation below. in Two Dimensions in Two Dimensions Before After 3
12/21/16 P in 2D P in 2D So what happens when two objects are not acting along the same plane? Consider two objects moving as shown. How could the problem be solved if they move in two directions? How can momentum be shown to be conserved? P in 2D Keep them separated Collision Example m1v1i + m2 v2i = m1v1 f + m2 v2 f m1v1ix + m2 v2ix = m1v1 fx+ m2 v2 fx Ø Two cars approach an intersection with a malfunctioning stop light. Ø The red car (m = 2000.00 kg) approaches the intersection from the North at 54.0 km/hr. m1v1iy + m2 v2iy = m1v1 fy+ m2 v2 fy If initial conditions are known, this gives 2 equations, with 4 unknowns, so more information is needed. v1,iy = 11 m/s m2 = 2250.00 kg v1,ix = 15 m/s Ø What will be the resulting velocity of the two cars if they stick together? The Center of Mass Collision Example m1 = 2000.00 kg Ø The orange car (m = 2250.00 kg) approaches the intersection from the West at 39.6 km/hr. Vf =? How should we de+ine the position of the moving body? What is h for GPE = mgh? Take the average position of mass, called Center of Mass (CM) Ø What will be the resulting velocity of the two cars if they stick together? 4
The Center of Mass There is a special point in a system or object, called the center of mass, that moves as if all of the mass of the system is concentrated at that point The CM of an object or a system is the point, where the object or the system can be balanced in the uniform gravitational field The Center of Mass The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry If the object has uniform density The CM may reside inside the body, or outside the body Center of Mass for a System of Particles Two bodies and one dimension x cm = m 1 x 1 x 2 m 1 General case: 3 or more bodies and 2D Where is the Center of Mass? Assume m 1 = 1 kg, m 2 = 3 kg, and x 1 = 1 m, x 2 = 5 m, where is the center of mass of these two objects? m1x1 + x CM = + m 1 m x m 2 2 2 x cm = m 1 x 1 x 2 + m 3 x 3 +... M y cm = m 1y 1 y 2 + m 3 y 3 +... M where M = m 1 + m 3 + Sample Problem : Three particles of masses m 1 = 1.2 kg, m 2 = 2.5 kg, and m 3 = 3.4 kg form an equilateral triangle of edge length a = 140 cm. Where is the center of mass of this system? x CM = m 1 x 1 x 2 + m 3 x 3 m 1 + m 3 y CM = m 1 y 1 y 2 + m 3 y 3 m 1 + m 3 Systems With Changing Mass: Rocket Propulsion In a system involving a rocket, a change in mass is present, which affects the momentum and acceleration of the system. Up to now, we have focused on systems keeping mass as a constant with other changing variables. 5
9-8 Systems With Changing Mass: Rocket Propulsion When a rocket begins to take off, it expels a certain amount of fuel, which changes its mass ( m) at a speed v. If we have a conservation of momentum and treat this as a collision, the momentum of the fuel being expelled ( m)v must be equal to that momentum propelling the rocket. 9-8 Systems With Changing Mass: Rocket Propulsion So the momentum increase of the rocket is: Δp = (Δm) v If we think of the mass being ejected in a certain time, t, then we have the equation: F net = Δp ( ) v = Δm Where the force is referred to as the thrust. Thrust = ( Δm ) v 9-8 Systems With Changing Mass: Rocket Propulsion By grouping the changing mass and changing time, we can refer to this quantity as the fuel ejection rate. F = Δp = Δm ( ) v Thrust Example In order to lift itself at a constant speed, a 450 kg helicopter must force air through the blades of the propeller. What is the fuel ejection rate if the air is pushed through at 65 m/s? In order for the rocket to lift off, the thrust needs to be than the of the rocket. 6