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PHYS 1443 – Section 001 Lecture #12 Tuesday June 21, 2005 Dr. Andrew Brandt • • • Momentum Impulse Collisions Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 1 Announcements • Homework: – HW7 on ch 8 due Tuesday 6/21 at 8pm Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 2 Linear Momentum The principle of energy conservation can be used to solve problems that are harder to solve using Newton’s laws. It is used to describe motion of an object or a system of objects. A new concept of linear momentum can also be used to solve physical problems, especially problems involving collisions of objects. Linear momentum of an object whose mass is m and is moving at a velocity of v is defined as What can you tell from this definition about momentum? What else can use see from the definition? Do you see force? Tuesday June 21, 2005 1. 2. 3. 4. p mv Momentum is a vector quantity. The heavier the object the higher the momentum The higher the velocity the higher the momentum Its unit is kg.m/s The change of momentum in a given time interval dp d dv ma F mv m dt dt dt PHYS 1443-001, Summer I 2005 3 Dr. Andrew Brandt Linear Momentum and Forces dp d F mv dt dt • • • What can we learn from this Force-momentum relationship? The rate of the change of particle’s momentum is the same as the net force exerted on it. When the net force is 0, the particle’s linear momentum is constant as a function of time. If a particle is isolated, the particle experiences no net force, therefore its momentum does not change and is conserved. What else can this relationship be used for? Can you think of a few cases like this? Tuesday June 21, 2005 The relationship can be used to study the case where the mass changes as a function of time. Motion of a meteorite PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt Motion of a rocket 4 Linear Momentum Conservation r r ur ur p1i p 2i m1 v1 m2 v 2 ur ur r' r' p1 f p 2 f m1 v1 m2 v 2 Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 5 Conservation of Linear Momentum in a Two Particle System Consider a system with two particles that does not have any external forces exerting on it. What is the impact of Newton’s 3rd Law? If particle#1 exerts a force on particle #2, there must be another force that particle #2 exerts on #1 as the reaction force. Both the forces are internal forces and the net force in the SYSTEM is still 0. Let’s say that the particle #1 has momentum First define the momenta p1 and #2 has p2 at a given time. of the two particles We use the momentum-force relationship to define the force F21 (exerted on object 1 by object 2) Since net force is 0 F F 21 F 12 F 21 d p1 dt this implies F 12 d p2 dt d p 2 d p1 d p p 0 2 1 dt dt dt Therefore p 2 p1 const The total linear momentum of the system is conserved!!! Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 6 More on Conservation of Linear Momentum in a Two Particle System From the previous slide we’ve learned that the total momentum of the system is conserved if no external forces are acting on the system. p p ur ur ur ur p 2i p1i p 2 f p1 f Mathematically this statement can be written as xi system P xf system P yi system This can be generalized into conservation of linear momentum in many particle systems. Tuesday June 21, 2005 p1 const As in the case of energy conservation, this means that the total vector sum of all momenta in the system is the same before and after any interaction What does this mean? P 2 P yf system P zi system P zf system Whenever two or more particles in an isolated system interact, the total momentum of the system remains constant. PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 7 Example for Linear Momentum Conservation Estimate an astronaut’s resulting velocity after he throws a book. vA vB From momentum conservation, we can write p i 0 p f mA v A mB v B Assuming the astronaut’s mass is 70kg, and the book’s mass is 1kg and using linear momentum conservation vA 1 mB v B vB 70 mA Now if the book gained a velocity of 20 m/s in +x-direction, the Astronaut’s velocity is Tuesday June 21, 2005 1 v A 70 20i 0.3 i m / s PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 8 Example for Linear Momentum Conservation A type of particle, a neutral kaon (K0), decays (breaks up) into a pair of particles called pions (p+ and p-) that are oppositely charged but have equal mass. Assuming the K0 is initially produced at rest, prove that the two pions must have momenta that are equal in magnitude and opposite in direction. This reaction can be written as K0 pp p p K p p 0 pp Since this system consists of a K0 in the initial state which results in two pions in the final state, the momentum must be conserved. So we can write Since the K0 is produced at rest its momentum is 0. p K 0 pp pp p K 0 pp pp 0 pp pp Therefore, the two pions from this kaon decay have momenta with same magnitude but opposite in direction. Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 9 Impulse and Linear Momentum Net force causes change of momentum Newton’s second law By integrating the above equation in a time interval ti to tf, one can obtain impulse I. So what do you think an impulse is? tf ti F dp dt tf d p p f p i D p F dt t i d p Fdt tf I F dt D p ti Impulse of the force F acting on a particle over the time interval Dt=tf-ti is equal to the change of the momentum of the particle caused by that force. Impulse is the degree to which an external force changes momentum. The above statement is called the impulse-momentum theorem and is equivalent to Newton’s second law. What are the units of Impulse? Defining a time-averaged force 1 tf F F dt t i Dt Tuesday June 21, 2005 Impulse can be rewritten I FDt If the force is constant I F Dt It is generally assumed that the impulse force acts for a short PHYS 1443-001, Summer I 2005 10 time but is much greater than any other forces present. Dr. Andrew Brandt Example 9-6 (a) Calculate the impulse experienced when a 70 kg person lands on firm ground after jumping from a height of 3.0 m. Then estimate the average force exerted on the person’s feet by the ground, if the landing is (b) stiff-legged and (c) with bent legs. In the former case, assume the body moves 1.0cm during the impact, and in the second case, when the legs are bent, about 50 cm. We don’t know the force. How do we do this? Obtain velocity of the person before striking the ground. KE DPE 1 2 mv mg y yi mgyi 2 Solving the above for velocity v, we obtain v 2 gyi 2 9.8 3 7.7m / s Then as the person strikes the ground, the momentum becomes 0 quickly giving the impulse I FDt Dp p f pi 0 mv 70kg 7.7m / s 540 N s Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 11 Example 9 – 6 cont’d In coming to rest, the body decelerates from 7.7m/s to 0m/s in a distance d=1.0cm=0.01m. 0 vi 7.7 3.8m / s v 2 2 0.01m d 3 The time duration of the landing is 2.6 10 s Dt v 3.8m / s I FDt 540N s Since the magnitude of impulse is I 540 5 The average force on the feet during 2.1 10 N F 3 this landing is Dt 2.6 10 2 2 How large is this average force? Weight 70kg 9.8m / s 6.9 10 N The average speed during this period is F 2.1105 N 304 6.9 102 N 304 Weight If landing is stiff-legged, the feet must sustain 300 times the body weight. The person will likely break his leg(s). d 0.50m 0.13s Dt 3.8 m / s For bent-legged landing: v 540 F 4.1103 N 5.9Weight 0.13 Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 12 Example for Impulse A golf ball of mass 50g is struck by a club. The force exerted on the ball by the club varies from 0, at the instant before contact, up to some maximum value at which the ball is deformed and then back to 0 when the ball leaves the club. Assuming the ball travels 200m, estimate the magnitude of the impulse caused by the collision. vB qB m The range R of a projectile is vB2 sin 2q B 200m R g Let’s assume that launch angle qi=45o. vB 200 g 1960 44m / s Then the speed becomes: (immediately before the collision) Considering the time interval for the vi 0 collision, ti and tf , the initial and v f 44m / s (immediately after the collision) the final speed are Therefore the magnitude of the impulse I D p mvBf mvBi on the ball due to the force of the club is 0.05 44 2.2kg m / s Tuesday June 21,can 2005solve impulsePHYS 1443-001,through Summer Iknowing 2005 Remember, you problems Dr. Andrew Brandt the force or knowing the change in momentum 13 Another Example for Impluse In a crash test, an automobile of mass 1500kg collides with a wall. The initial and final velocities of the automobile are vi=-15.0i m/s and vf=2.60i m/s. If the collision lasts for 0.150 seconds, what would be the impulse caused by the collision and the average force exerted on the automobile? Let’s assume that the force involved in the collision is a lot larger than any other forces in the system during the collision. From the problem, the initial and final momentum of the automobile before and after the collision is p i mvi 1500 15.0i 22500i kg m / s p f mv f 1500 2.60i 3900i kg m / s Therefore the impulse on the I D p p pi 3900 22500 i kg m / s automobile due to the collision is 26400i kg m / s 2.64 104 i kg m / s f The average force exerted on the automobile during the collision is Tuesday June 21, 2005 4 D p 2.64 10 F Dt 0.150 1.76 105 i kg m / s 2 1.76 105 i N PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 14 Collisions Generalized collisions must cover not only the case of physical contact but also collisions without physical contact such as particles on a microscopic scale. The collisions of these ions never involves Consider the case of a collision physical contact because the electromagnetic between a proton and a helium ion. repulsive force between these two becomes great as they get closer causing a collision. F F12 t F21 Using Newton’s 3rd Assuming no external forces, the force exerted on particle 1 by particle 2, F21, changes the momentum of particle 1 by ur ur d p 2 F 12 dt Likewise for particle 2 on particle 1 law we obtain ur ur d p 2 F 12 dt So the momentum change of the system in the collision is 0 and momentum is conserved Tuesday June 21, 2005 ur ur d p1 F 21dt ur ur F 21dt d p1 ur ur D p d p1 d p 2 0 p system PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt p1 p 2 constant 15 Elastic and Inelastic Collisions Momentum is conserved in any collision as long as external forces are negligible. A collisions is classified as elastic or inelastic depending on whether kinetic energy is conserved during the collision. Elastic Collision A collision in which the total kinetic energy and momentum are the same before and after the collision. Inelastic Collision A collision in which the total kinetic energy is not the same before and after the collision, but the momentum is. Two types of inelastic collisions: Perfectly inelastic and inelastic Perfectly Inelastic: Two objects stick together after the collision moving at a certain velocity together. Inelastic: Colliding objects do not stick together after the collision but some kinetic energy is lost. Note: Momentum is constant in all collisions but kinetic energy is constant only in elastic ones. Tuesday June 21, 2005 PHYS 1443-001, Summer I 2005 Dr. Andrew Brandt 16 Elastic and Perfectly Inelastic Collisions In perfectly Inelastic collisions, the objects stick together after the collision, moving together. Momentum is conserved in this collision, so the final velocity of the stuck system is How about elastic collisions? m1 v1i m2 v 2i (m1 m2 )v f vf m1 v1i m2 v 2i (m1 m2 ) m1 v1i m2 v 2i m1 v1 f m2 v2 f 1 1 1 1 In elastic collisions, both the m1v12i m2 v22i m1v12f m2 v22 f 2 2 2 2 momentum and the kinetic energy m1 v12i v12f m2 v22 f v22i are conserved. Therefore, the final speeds in an elastic collision m1 v1i v1 f v1i v1 f m2 v2 f v2i v2i v2 f can be obtained in terms of initial From momentum m1 v1i v1 f m2 v2 f v2i speeds as conservation above dividing gives: m1 m2 2m2 general solution: v v v 1i 1f 2i Tuesday June 21, 2005 v2 f v1i v2i v1 f m1 m2 m1 m2 2 m m m Summer PHYS 1443-001, v1i 1I 20052 v2i v2 f Dr. 1Andrew Brandt m1 m2 m1 m2 Special cases 17