Chapter 9. Linear momentum and collisions. PHY 1124 Fundaments of Physics for Engineers. Michael Wong PHY1124 Winter uottawa.

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1 Chapter 9 Linear momentum and collisions Michael Wong PHY1124 Winter 2019 PHY 1124 Fundaments of Physics for Engineers uottawa.ca

2 Goals 2 Chapter 9 Momentum and collisions Linear momentum Focus on isolated system Collisions in 1D (elastic and inelastic) Collisions in 2D Center of mass Particle model only

3 Introduction 3 Beyond Newton s laws and energy When two cars crash, they go from high v to zero v in a short time. F and a vary with time. Using Newton s laws to analyze the situation is difficult. We can analyze the collision using a momentum picture which depends on mass and velocity. Also useful to analyze rocket propulsion.*

4 Section Ex. 9.1 The archer Consider a 60 kg archer on a frictionless ice who fires a 0.03 kg arrow at 85 m/s. With what velocity does the archer move? We have no info about the forces, the acceleration of the arrow, the work or the energy/energies involved. A new quantity is needed linear momentum.

5 Section Linear momentum The linear momentum of a particle of mass m and velocity v is the product: p = mv A vector quantity in same direction as velocity. Can be expressed as components: p x = mv x p y = mv y p z = mv z SI unit is kg m s. Newton called this the quantity of motion (term in French).

6 Section Momentum and kinetic energy Both values involved mass and velocity. Differences? K is scalar, p is vector. K can be transformed into other types of energy, only one type of linear momentum. Analysis models are separate. Momentum is an independent tool to use in problem solving.

7 Section Momentum Newton s 2 nd law Newton s 2 nd law is used to relate momentum of a particle to the resultant force acting on it: ԦF = ma = m dv dt = d mv dt = dp dt The rate of change of momentum of a particle is equal to the net force acting on it. This was how Newton originally presented 2 nd law. This form also allows for change in mass (more general).

8 Section Quick quiz 9.2 Same v, p, or K? Your phys-ed teacher throws a baseball to you at a certain speed and you catch it. The teacher is next going to throw you a medicine ball whose mass is 10x as large. You are given the following choices: (a) Throw with same speed (b) Throw with same momentum (c) Throw with same kinetic energy. Rank the choices from easiest to hardest to catch.

9 Section Conservation of linear momentum When two or more particles in an isolated system interact, the total momentum of the system remains constant. Individual particle s momentum may change. Ie. Initial momentum equals final momentum. For system of two particles : d dt p 1 + p 2 = 0 p tot = constant p 1,i + p 2,i = p 1,f + p 2,f Can be separated into component form as well.

10 Section Forces and conservation of p There is no statement about what types of forces act on the particles. Can be conservative or nonconservative. Can be constant or not constant. Only requirement: forces are internal to system. Back to the archer (Ex. 9.1). Apply cons. of momentum in isolated system in x. Initial momentum is zero, final momentum is: p 1,f + p 2,f = 0 m 1 v 1,f + mv 2,f = 0

11 Section Characteristics of a collision A collision represents an event where two particles come close then interact through forces. Can have contact or be contact-less. Assume the interaction forces are much greater than other external forces.

12 Section Types of collisions In an elastic collision, both momentum and kinetic energy are conserved. True elastic collisions occur on microscopic scale. Approximate elastic collisions for anything else. Use isolated system model for p and K. In an inelastic collision, momentum conserved but not kinetic energy (transferred to sound/heat). If object stick together after the collision then it is perfectly inelastic collision.

13 Section Perfectly inelastic collisions The total momentum before the collision is equal to the total momentum of the composite system after collision. The final velocity of both objects is the same (they are stuck together). m 1 v 1,i + m 2 v 2,i = m 1 + m 2 v f

14 Section Elastic collisions Both momentum and kinetic energy are conserved. m 1 v 1,i + m 2 v 2,i = m 1 v 1,f + m 2 v 2,f 1 2 m 1v 2 1,i m 2v 2 2,i = 1 2 m 1v 2 1,f m 2 2v 2,f Typically there are two unknowns to solve for and so we need two equations.

15 Section Elastic collisions (cont.) Combining the conservation of p and K equations gives us a simpler equation: v 1,i v 2,i = v 1,f + v 2,f Can be used for one-dimensional, elastic collision between two objects. We can drop the vector notation for 1-D collisions and use negative and positive for velocities to left and right. Some special cases exist: m 1 = m 2, m 1 m 2, m 1 m 2, etc

16 Section Ex. 9.4 Newton s cradle Five identical hard balls lined up. Demonstrates elastic collisions, conservation of momentum and conservation of energy. Is it ever possible for one ball to come in from the left and two balls exit from the right???

17 Section Two-dimensional collisions Extending from 1D to 2D isn t difficult. The momentum is conserved in both x and y directions. Balls on a pool table is the best example. Now we use three subscripts for our velocities: Object (1 or 2) Initial or final value Direction (x or y) For elastic collisions, we use conservation of energy equation too.

18 Section Two-dimensional collisions, example Particle 1 moving at Ԧv 1,i and particle 2 at rest. In x, initial momentum is m 1 v 1,i. In y, initial momentum is 0. After the collision: Momentum in x : m 1 v 1,i = m 1 v 1,f cos θ + m 2 v 2,f cos φ Momentum in y : 0 = m 1 v 1,f sin θ m 2 v 2,f sin φ (negative because v 2,f,y is down)

19 Section Ex. 9.8 Collision at an intersection Blue car has mass 1500 kg and speed 25 m/s. Grey truck has mass 2500 kg and speed 20 m/s. What is final velocity and direction of the wreckage after the collision? Assume cars stick together.

20 Section The center of mass The center of mass is a point in a system or object that moves as if the whole mass is concentrated at that point. The system moves as if an external force was applied to a single particle of mass M located at the CM. We ll keep it simple: particle model only (no weird shapes).

21 Section Center of mass in 3D We can have objects in three dimensions. Each dimension will have a CM coordinate. (figure shows only 1D). x CM = σ i m i x i M, y CM = σ i m i y i M, z CM = σ i m i z i M Treat each object as a particle. M is the total mass of the system.