Chapters 9-10 Impulse and Momentum Momentum Concept Newton s 2 nd Law restated Conservation of momentum Impulse. Impulse-Momentum Theorem Collisions 1D inelastic and elastic collisions 2D collisions
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 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 century, with the 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
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 m p kg m s or Ns SI p 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 dp d mv dv dm dm F m v ma v dt dt dt dt dt 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 F dp dt
Momentum Conservation of Net Momentum 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 If we consider an isolated system of particles, the only forces involved are pairs of action-reaction forces which cancel each other out two by two. Hence, the net momentum is conserved: p 7 p 1 p 6 p 5 p 2 p 3 p 4 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: rd by 3 Law nd by 2 Law Fnet Fi Finternal 0 dpnet 0 pnet pi constant i p p p... p m v m v... m v const. 1 2 n 1 1 2 2 n n Note that individual momenta can change, but only such that the sum stays constant i
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 v M
Impulse Definition. Impulse-Momentum Theorem The overall effect of applying a force F for a certain time interval Δt can be integrated into a vector physical quantity called impulse a vector denoted J given by the product between a force and the interval of time it acts: 1. If the force F is constant during Δt, the impulse is simply given by 2. If the force is not constant and depend on time as given by a function F(t), the impulse can be obtained analytically by integration: 3. If the force F is not constant during Δt, the impulse can also be found graphically 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 Force Combining the definition of impulse with Newton s 2 nd Law, we find a way to estimate the average force during motion changing events using: 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: J av J p Favt t t t 0 F t dt p F av 0 J Ft Δt actual force impulse = area Time
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
Exercise 2: Average Force on a Tennis Ball A tennis ball of mass m = 57 g hits a vertical wall with a horizontal speed v = 60 m/s. The collision is filmed with a high speed camera and a profile of the force exerted by the ball onto the wall is plotted versus time. Use the graph to calculate the average force exerted by the wall onto the ball during the collision. Assume the collision elastic. a) What is the collision time interval? Force b) What is the change of momentum during the collision? (Assume the right direction positive.) 0 Δt 0 1 2 3 4 5 t (ms) c) Use the Impulse-Momentum Theorem to estimate the average force.
Problem: 1. Using the impulse: A rubber ball of mass m = 2.5 kg is dropped from a height y 0 on the floor and bounces back. The graph shows the time evolution of the force of the floor on the ball. a) What is impulse of the ball as it hits the floor? b) What is average force exerted by the floor on the ball? c) How high does the ball bounce?
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 = 60 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?
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 be conserved:...... p p p p p p p p before after 1 2 n before 1 2 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 objects deform irreversibly during the collision, kinetic energy is not conserved. Yet the momentum is conserved In 1D: p p m v m v m v m v before after 1 1 2 2 1 1 2 2 If the objects stick together, the collision is called perfectly inelastic and p p m v m v m m v before after 1 1 2 2 1 2 velocity of the composite object 2. Elastic, if the objects regain shape after the deformation during the collision the kinetic energy is conserved, as well as the momentum In 1D: p p m v m v m v m v before after 1 1 2 2 1 1 2 2 During collisions, objects exert forces upon each other. The average force exerted on each colliding object is F av n after mv mv t
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?
Collisions Elastic collisions An application of the concept of kinetic energy is in elastic collisions where both momentum and kinetic energy are conserved since the net work done by the forces exerted by the colliding objects upon each other is overall zero Only for head-on elastic collisions, the conservation of kinetic energy can be reduced to a simpler form: before m 1 collision W net = 0 v1 v 2 m 2 m v v m v m v m v 1 1 2 2 1 1 2 2 v v v 1 2 1 2 after v v 2 1 Problems: 3. Equations for head-on elastic collision: Demonstrate that the conservation of momentum and kinetic energy in elastic collisions leads to the equations above. 4. Head-on elastic collision with stationary target: 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
Problem: 5. Inelastic collision: A bullet of mass m 1 = 5.00 g is fired with a horizontal speed v 1 into a wooden block of mass m 2 = 1.0 kg moving on a horizontal surface with speed v 2 = 2.0 m/s. The coefficient of kinetic friction between the block and surface is μ k = 0.20. The bullet remains embedded in the block, which is observed to slide a distance d = 0.25 m in the opposite direction than its initial motion before stopping. a) What was the initial speed v 1 of the bullet? b) What is the force exerted by the bullet on the block during the collision, if it took a time Δt = 10 ms to change its direction of motion? m 1 v 1 m 2 v 2 v = 0 μ k d
Collisions Multidimensional conservation of momentum The conservation of momentum during collisions is a vector relationship...... p p p p p p p p before after 1 2 n before 1 2 So, the relationship can be applied to the components of the net momentum, resulting in a number of independent equations corresponding to the dimensionality of the collision: p p p, p, p i x y z p p p... p m v m v... m v const. x 1x 2x nx 1 1x 2 2 x n nx p p p... p m v m v... m v const. y 1y 2 y ny 1 1y 2 2 y n ny p p p... p m v m v... m v const. z 1z 2z nz 1 1z 2 2z n nz n 1D after 2D 3D Ex: If a billiard ball collides with another ball initially at rest, the 2D conservation of momentum can be written: p p before p 1 2 pafter p p p p 1 p 2 p p p 1x 1x 2x 1y 1y 2 y p 1 p 1 p 2
Problem: 6. Two dimensional collision: Ball (2) rests on a flat surface when it is struck by a second identical ball (1), which was originally traveling at v 1 = 40.0 m/s. Ball 1 is deflected an angle α = 30 from its original direction. Also, as a result of the collision, ball 2 starts moving at an angle θ = 45 with respect to the original direction of ball 1. a) Draw a vector diagram with the momenta of the balls before and after the collision b) Write the conservation of momentum in a suitable system of coordinates c) Calculate the speeds of each ball after collision 1 1 v 1 v 1 α θ 2 v 2