Conservation of Momentum -1

Transcription

1 Impulse, Action-Reaction and Change in Momentum: Prolem 1: A pitcher throws a 150g aseall y applying a 50N force for 0.1 second. Assuming that the ase all starts from rest, otain a. the initial velocity the initial momentum and KE d. the final momentum and KE e. the final velocity of the all f. the acceleration of the all g. the work done on the all h. Are the momentum and the KE conserved? Prolem : The velocity of a 150g aseall is changed from 0m/s to 10m/s y applying a 100N force. Otain a. the initial momentum and KE the final momentum and KE d. how long the 100N force is applied on the all. e. the acceleration of the all f. the work done on the all g Impulse: FΔ t Change in Momentum: Δ p= Impulse = Δ p = F Δt FReaction= FAction Are the momentum and the KE conserved? Conservation of Momentum -1 Prolem 4: A 150g aseall traveling 0m/s is hit y a aseall at. The all is in contact with the at for 0.1 second. After the hit the aseall moves 0m/s in the opposite direction.. Otain a. the initial momentum and KE the final momentum and KE d. the force applied. e. the acceleration of the all f. the force exerted on the at y the all g. the force exerted on the all y the at h. the work done on the all i. Are the momentum and the KE conserved? Prolem 5 A 1kg pistol fires a ullet of mass 8g with a velocity of 300m/s. Otain a. the change in the momentum and KE of the ullet during firing the momentum and KE of the ullet after firing c. the force applied on the ullet if the firing of the ullet took 0.1s long d. the force applied on the pistol e. the work done on the ullet if the pistol is 0 cm Prolem 3: The velocity of a 150g aseall is changed from 0m/s to 10m/s y applying a force for seconds.. Otain a. the initial momentum and KE the final momentum and KE d. the force applied e. the acceleration of the all f. the distance the all travels during this s time interval g. Are the momentum and the KE conserved?

2 Conservation of Momentum: The total momentum of a system is conserved when the net impuls on the system is zero or approximately zero. r F net Δt 0, r where P T =p1+p+p3+ is the sum of the momenta of all the ojects in the system and stand for efore and after event, e.g. collision Conservation of Momentum - = constant, i.e. r = r a the suscripts and a Prolem 1: Two ice-skaters of masses 50kg and 80kg are standing across from each other.. a. What is the initial momentum each skater?. What is the total momentum of oth skaters? The skaters push each other away. As a result, 80kg skater acquires a velocity of 5 m/s. c. What is the total momentum of the skaters after the push off? d. What is the momentum of the 80kg skater? e. What is the momentum of the 50kg skater? f. What is the velocity of the 50kg skater? g. Otain the KE of each skater efore and after the push off. Is total energy conserved? Is KET conserved? Prolem : How do rockets travel in space? Prolem 3: A 5000kg rocket needs to reach a velocity of 000m/s. How much exhaust does it need to expel at a velocity of 10000km/s in order to reach this speed? Prolem 4: In figure skating a pair standing push each other away. Otain the ratio of their speeds in terms of their masses. Now assume that one the skaters is 30 kg and the other one is 80 kg. a. If the 80 kg skater is moving 5 m/s, otain the speed of the other skater.. If the two skater pushed each other for a duration of 0.5 seconds, how much force did the skaters apply on each other? c. Otain the KE of each skater efore and after the push off. d. How much work did they do?

3 Collisions Conservation of Momentum -3 Prolem 1: A illiard all of mass M moving with a velocity v1 hits a second illiard all of identical mass at rest. a. Otain the velocity of the second illiard all in terms of the velocity of the first illiard all efore and after the collision.. Assume that the velocity of the first all after the collision is zero. Otain the velocity of the second all after the collision. c. Otain the KE of each all efore and after the collision. Is total energy conserved? Is KET conserved? Prolem : A collision car of mass M1 moving with a velocity of v1 and collide with a second car of mass M at rest. After the collision, two cars stick to each other and move together. a. Otain the velocity of the two car system after the collision in terms of M1, M, and v1.. Assume that the cars are identical and have the same mass. Otain the velocity of the two car system after the collision in terms of v1. c. Assume that M1=10 kg, M=30 kg and v1= m/s. Otain the velocity of the two car system after the collision. d. Otain the KE of each all efore and after the collision. Is total energy conserved? Is KET conserved? Prolem 3: Two illiard alls of mass M1 and M moving towards each other with velocity v1 and v collide head on. a. Otain the momentum and velocity of the second illiard all after the collision in terms of the velocity of the second illiard all after the collision and M1, M,v1, and v.. Assume that M1= kg, M = 5 kg, v1= 4 m/s, v=7 m/s, and v1a= 3 m/s. Otain pa, va. c. Otain the KE of each all efore and after the collision. Is total energy conserved? Is KET conserved? Prolem 4: Two illiard alls of mass M1 and M moving ack to ack with velocity v1 and v. The first all catches up with the second and they collide. a. Otain the momentum and velocity of the second illiard all after the collision in terms of the velocity of the first illiard all after the collision and M1, M,v1, and v.. Assume that M1= kg, M = 5 kg, v1= 4 m/s, v=7 m/s, and v1a= 3 m/s. Otain pa, va. c Calculate KE of each all efore and after the collision. Is KET conserved? Conservation of Energy and Momentum in Elastic Collisions If a collision is elastic, the total KE of the system does not change. As a result, the conservation of energy expression can e reduced to the one where we have to deal with only the kinetic energy part.

4 Conservation of Momentum -4 Conservation of Momentum in Collisions in Dimensions Consider an oject of mass M1 and velocity v1 at angle θ1 with respect to the x-axis colliding with a second oject of mass M and velocity v at angle θ with respect to the x-axis. After the collision oject-1 goes in the direction which makes angle φ1 with respect to the x-axis and the oject- goes in the direction which makes angle φ with respect to the x- axis. We would like to otain the velocities, momenta and kinetic energies of the ojects after the collision. Through out the collision the net force on the two-oject system is zero. Since Fnet=0, the total mometnum is conserved. In order to avoid cluttering the expressions with too many suscripts, we will not use the suscripts and a to show quantities efore and after the collision. I. Momenta efore the collision Since Fnet=0, the total mometnum is conserved. = M 1 v 1 x = cos θ 1 y = sinθ 1 Therefore, a = = M v x = cos θ y = sinθ After the collision = + x = x + x y = y + y = M 1 v 1a x = cos φ 1 y = sinφ 1 = M v a x = cos φ y = sinφ = + x = x + x y = y + y a a θ 1 θ φ φ 1 a x a a II. Kinetic Energies efore the collision KE 1 = 1 M 1v 1 KE = 1 M v II. Kinetic Energies after the collision KE 1 = 1 M 1v 1a KE = 1 M v a KE T = KE 1 + KE KE T = KE 1 + KE = M 1 v 1 The collision is elastic if KE T = KE Ta. Otherwise it is inelasti.c.

5 Conservation of Momentum -5 All the angles are given with respect to the positive x-axis. See figures elow. Before the collision After the collision M 1 = kg, v 1 =5 m =10Ns@ The Law of Sines M =4kg, KE1 = 5J KE = 00J KET = 5J v =10 m =40Ns@ After the collision, the angle etween the two momenta is 75 and the angle across from the total momentum is 105. The total momentum is conserved. Therefore, a =. The angle etween the x-axis and the total momentum is - 31, the angle etween the x-axis and is 30 and the The Law of Sines The angle etween the two momenta is 90. As a result, the total momentum is given y the Pythagorean theorem. = + = 41.Ns. We can use the law of sines (or the Pythagorean theorem in this case) to figure out the angle etween and (or ). sin 90 = sin α = sin β. sin α = sin β = pt =0.46, α=14 pt =0.9704, β=76 The angle etween the total momentum and the x-axis is 14 = 31 elow the x-axis. As a result, =41.Ns@ 31 x =10Ns cos =7.07Ns y =10Ns sin =7.07Ns The Component Method x =40Ns cos( ) =8.8Ns y =40Ns sin( ) =8.8Ns x = y = x + y + x =35.35Ns y =1.1Ns angle etween the x-axis and is. As a result, the angle etween and is 14, and and is 61. sin 105 = sin 61 = sin 14 where =41.Ns. =37.31Ns@ =10.3Ns@30 Check = + cos 105 x = x Ns = 3 y = y + 1.1Ns = 1 The Component Method x = + cos30+ y = sin30+ cos(- 45) Ι sin(- 45) Using expressions I and II together gives =37.3Ns and =10.35Ns. =37.3Ns@, v 1 =18.66 m =10.35s@30, KE1a = 348J KEa = 13.J KETa = 361J KE T KE Ta KE T = 60% ΙΙ v =.59 m 30 = Tx + Ty =41.Ns The collision is inelastic. KE Ta > KE T, therefore, there tanθ = y x = =41.Ns@ 31 and θ = 31 must have een some kind of explosion during the collision

6 Conservation of Momentum -6 Before the Collision y x y After the Collision x