Chapter 6 Momentum and Collisions

Transcription

1 Chapter 6 Momentum and Collisions Momentum and Its Relation to Force Impulse Conservation of Momentum Collisions Classification Inelastic Collisions in One Dimension Elastic Collisions in One Dimension

2 1. Quantity of Motion A historical preamble Query: How can one quantify motion? Is the mere velocity enough? Historically, philosophers of nature quantified motion using the rather intuitive kinematic concept of velocity: however, simple experiments of removing movement show that the amount of moving substance, that is mass, must play a role in defining the amount of motion Ex: Consider two balls with the same size but different mass. When launched with the same speed, the heavier ball will compress a spring more Wood Steel v 0 v 0 v 0 This became obvious by the XVII-th century, with the novel tendency to look for the causal emergence of mechanical phenomena based on universal principles, such as conservation laws Descartes was the first to notice that, if a quantity of motion comprises both velocity v and mass m, the product mv is preserved in collisions, while Huygens suggested that to solve any collisions the quantity can be also negative, that is, it depends on direction Leibniz had an alternative approach associating motion with the product mv 2 related to the modern concept of kinetic energy Newton adopted the concept of quantity of motion, and built his mechanics as a mathematical compilation of how it can be changed in proportion to the applied forces

3 Momentum Its Relation to Force Def: The translational momentum of a particle of mass m moving with a velocity v is a vector given by the product of mass and velocity: p mv s In modern formalism, Newton stated his 2 nd law in terms of momentum: the net force applied to an object is equal to the rate of change of its momentum This is equivalent with the popular form F = ma only when the mass of the moving particle is constant: m m p kg m or Ns SI F p p t Ex: 1D uniform motion: Consider a particle of mass m moving in a straight displacement Δx, acted for a time Δt by a constant force F parallel with the motion. F F p0 mv0 Δx = x 1 x 0 p mv Δt = t x t 0, x 1 t 0 0 t 1, x 1 p F Ft p p0 Ft mv1 mv0 t By Newton s 2 nd Law, the momentum will increased as given by p mv v m m F m v ma v t t t t t

4 Newton s 2 nd law in terms of momentum can be used to demonstrate that the net momentum is conserved within isolated systems In order to understand how momentum is conserved, recall the concept of internal forces: forces between the parts of a system to be contrasted with the forces acted by objects outside the system, called external forces A system acted by no external forces is called an isolated system Therefore, in general, for an isolated system of n objects (indexed 1,2 n) the net momentum, equal to the vector sum of individual momenta is constant: Momentum Conservation If we consider an isolated system of particles, the only forces involved are pairs of action-reaction which cancel each other out two by two. Hence, the net momentum is conserved: rd 3 Law nd by 2 Law net i internal 0 net pnet i t i i F F F p 0 p cons ant p p p... p m v m v... m v const. 1 2 n n n Notice that individual momenta can change, but only such that the sum stays constant p 5 p 4 p 1 p 3 p p7 6

5 Quiz 1: Momentum and Newton s 2 nd Law: A kid catches a tennis ball and then a basketball, both moving with the same momentum. The kid applies the same force on each ball until it stops. Which of the two balls travels a longer time until it stops? a) Both balls require the same stopping time b) The tennis ball c) The basketball Exercise 1: Conservation of momentum: A man of mass m = 80 kg is initially at rest on a raft of mass M = 150 kg immobile with respect to the still water. Suddenly, the man starts to move to the right with a speed v = 0.50 m/s with respect to the water. Neglect water resistance. a) Write the momenta of the system before and after the man starts walking. b) Calculate the velocity of the raft relative to the water after the man starts walking. m M pinitial pm pm 0 p final p m p M mv MV v p final p initial mv MV m V v M 0

6 Impulse-Momentum Theorem How can we use momentum to calculate the force? The details of the force that determines a change in the state of motion of an object can be often obscured by a complex dependency of time However, the overall effect of applying a force F for a certain time interval Δt can be integrated into a vector physical quantity called impulse. Let s look at two situations: 1. If the force F is constant during Δt, the impulse is simply given by J Ft 2. If the force F is not constant during Δt, the impulse can be written in terms of the average force F av : we see that the impulse is the area under the F vs t curve J F t Impulse-Momentum Theorem: The impulse of the average net force acting on a particle for an interval of time is equal to the change in particle s momentum: av Force Combining the impulse with Newton s 2 nd Law, we find a way to estimate the average force during motion changing events using: J p Favt t t p F av 0 Δt actual force impulse = area Time

7 Impulse Examples Notice that the Impulse-Momentum Theorem tells us that the same change in momentum can be obtained either by applying a large force a short time interval, or a small force a long time Ex: 1. Instinctual knee protection: when we land after a jump, the change in momentum and consequently the impulse is the same, no matter how long it takes to stop. However, by bending out knees we increase the stopping time which results in a decreased average force onto the knees 2. Car crash air bag protection: The average force suffered by the body during a car crash can be decreased by increasing the time the body changes its momentum from its value before the impact to zero. This is the job of air bags which first inflate extremely fast and then deflate in a controlled time. 3. The table cloth trick: The force (friction) between the objects on a table and the table cloth is constant. However, pulling the cloth away really fast minimizes the impulse, so the change of momentum is small and the objects barely move

8 Exercise 2: How is Superman s chest different from a sidewalk? In the first Superman movie, the man of steel who furtively loves Lois saves her several times from certain death. In one such scene, Lois falls from almost the top of Empire State Building at least 30 meters before Superman catches her, so she can engages him in peppy dialog. Well, while we know that Superman can take a hit, let s see if Lois s body would make any difference between hitting the sidewalk versus Superman s vinyl clad chest. Let s estimate Lois s mass to be m = 50 kg and check out the Physics of the situation. a) Neglecting air resistance, what is Lois s speed after she falls a distance h = 30 meters? b) If it takes 0.03 s for Superman to stop her fall, what is the average force experienced by Lois?

9 Exercise 3: How Neo should ve listened to the Architect Bad Physics has extenuating circumstances in the movie Matrix, since its world is mostly virtual. However, the movie still perpetuates some misconceptions contradicting elementary Physics even within the logic of its computer controlled reality where people die if their matrix persona dies. For instance, Neo (the local Messiah) saves his lover (vinyl clad Trinity) from death as she falls from a tall building, by catching her hastily right before she hits the sidewalk. Let s compare the average forces acting on Trinity with and without Neo s grab. Assume that Trinity s mass m = 120 kg is about to hit the ground with a speed v i = 50 m/s when Neo arrives and imparts her a very underestimated speed v f = 100 m/s. a) If Neo needs 0.05 s to deflect her fall, what is the average force experienced by Trinity? b) If otherwise Trinity needs 0.05 s to stop when she hits the ground, what is the average force she experiences?

10 Collisions - Classification The particles in an isolated system are allowed to interact with each other so the individual momenta can change. At all times the net momentum must stay the same This property is instrumental in handling collisions since the momentum is always conserved during a collision. However, depending on the character of the energy conservation, collisions can be: 1. Inelastic, if the kinetic energy is not conserved during the collision. Some energy is converted irreversibly into heat. If the objects stick together, the collision is called perfectly inelastic and before after p p p p p p p p before after 1 2 n before 1 2 p p m v m v m m v velocity of the composite object n after 2. Elastic, if the kinetic energy is conserved during the collision. Hence, for one dimensional elastic collisions we can write two conservation laws: p p m v m v m v m v before before after after KE KE m v m v m v m v

11 Exercise 4: Mike s Perfectly Inelastic Collision In the film Back to the future, Michael J. Fox plays the role of a teenager who travels in time, back into the wild-west past of his hometown. Among other adventures, he is challenged to a shootout by wicked Mad Dog Tannen. Mike gets shot, but he s smartly bulletproofed with a stove door. However, the impact with the bullet throws him violently flat on his back. Let s take a look at the scene, and then check out if time traveling is the only dereliction from known Physics in the movie. Estimate realistically the mass of the bullet to be m = 5 g, moving with a speed v bullet = 500 m/s at impact. Also, assume Mike s mass M = 60 kg. a) Considering the impact perfectly inelastic, find Mike s speed right after the impact: b) If the bullet needs 10 ms to stop in the stove door, what is the average force exerted on Mike?

12 Collisions One dimensional inelastic collisions In inelastic collisions some of the initial kinetic energy is lost to thermal or potential energy. The energy may also be gained as during explosions, as there is the addition of chemical or nuclear energy The conservation of momentum still works: Problem: m v m v m v m v before after 1. Inelastic collision: A bullet with mass m hits with a speed v a wood block of mass M suspended from the ceiling by cables of length l. a) Knowing that the inelastic collision took a time Δt, find the force exerted by the bullet on the wood in terms of given quantities. b) Find the maximum height h reached by the system with respect to the initial position.

13 Collisions One dimensional elastic collisions Since in the case of elastic collisions both momentum and kinetic energy are conserved, we can write two equations which allow us to solve for the two unknown final speeds before m 1 v1 v 2 m 2 For head-on collisions, the conservation of kinetic energy can be reduced to a simpler form, so we get: collision m v v m v m v m v v v v after v 1 v 2 Problems: 2. Head on elastic collision: Demonstrate the second relationship above. 3. Elastic collision with target at rest: Two objects of masses m 1 and m 2 collide head-on. Mass m 1 has an initial speed v 1, and mass m 2 is initially at rest. a) Calculate the speeds of the masses after collision in terms of the given quantities. b) Comment on what is going to happen if m 1 = m 2, and if one the objects is much more massive than the other c) What is the force exerted by mass m 1 on m 2?