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General Physics (PHY 2130) Lecture 17 • Momentum Momentum and impulse Conservation of momentum http://www.physics.wayne.edu/~apetrov/PHY2130/ Lightning Review Last lecture: 1. Work and energy: elastic potential energy momentum Review Problem: An 8.0 kg crate, initially at rest, slides down a slope as in the 15 m picture. The work of the force of friction is 300 J. What is the speed of the crate at the bottom of the slope? Review problem An 8.0 kg crate, initially at rest, slides down a slope as in the picture. The work of the force of friction is 300 J. What is the speed of the crate at the bottom of the slope? Idea: use energy considerations with nonconservative forces present 15 m Wnc = ΔE = E f − Ei At the top: Ei = mgh + 0 mv 2 At the bottom: E f = 0 + 2 Thus: mv 2 Wnc = − mgh 2 or 2 v= ( mgh + Wnc ) = 19.2 m / s m Recall: Momentum • From Newton’s laws: force must be present to change an object’s velocity (speed and/or direction) Wish to consider effects of collisions and corresponding change in velocity Golf ball initially at rest, so some of the KE of club transferred to provide motion of golf ball and its change in velocity Method to describe is to use concept of linear momentum Linear momentum = product of mass scalar × velocity vector Momentum p = mv • Vector quantity, the direction of the momentum is the same as the velocity’s • Applies to two-dimensional motion as well p x = mv x and p y = mv y Size of momentum: depends upon mass depends upon velocity Impulse • In order to change the momentum of an object (say, golf ball), a force must be applied • The time rate of change of momentum of an object is equal to the net force acting on it • m(v f − vi ) Δ p F net = ma = = Δt Δt or : Δ p = F net Δt • Gives an alternative statement of Newton’s second law • (F Δt) is called the impulse • Impulse is a vector quantity, the direction is the same as the direction of the force Graphical Interpretation of Impulse • Usually force is not constant, but time-dependent impulse = ∑ Fi Δti = area under F (t ) curve Δti • If the force is not constant, use the average force applied • The average force can be thought of as the constant force that would give the same impulse to the object in the time interval as the actual time-varying force gives in the interval If force is constant: impulse = F Δt Example: Impulse Applied to Auto Collisions • The most important factor is the collision time or the time it takes the person to come to a rest • This will reduce the chance of dying in a car crash • Ways to increase the time • Seat belts • Air bags The air bag increases the time of the collision and absorbs some of the energy from the body 9 Example: A car of mass 1500 kg collides with a wall and rebounds as shown. If the collision lasts for 0.150 s, Find (a) the impulse delivered to the car due to the collision and (b) the magnitude and direction of the average force exerted on the car. – Assume force exerted by wall is large compared with other forces – Gravitational and normal forces are perpendicular and so do not effect the horizontal momentum – Can apply impulse approximation (a) Impulse delivered to the car = Δp = p f − pi = (1500kg × 2.60m / s) − (1500kg × (−15.0m / s)) = 2.64 × 10 4 kg .m / s (b) The average force exerted on the car Δp 2.64 × 10 4 kg .m / s Fav = = = +1.76 × 10 5 N Δt 0.150s ConcepTest Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare? 1. It takes less time to stop the ping-pong ball. 2. Both take the same time. 3. It takes more time to stop the ping-pong ball. ConcepTest Suppose a ping-pong ball and a bowling ball are rolling toward you. Both have the same momentum, and you exert the same force to stop each. How do the time intervals to stop them compare? 1. It takes less time to stop the ping-pong ball. 2. Both take the same time. 3. It takes more time to stop the ping-pong ball. Note: Because force equals the time rate of change of momentum, the two balls loose momentum at the same rate. If both balls initially had the same momenta, it takes the same amount of time to stop them. Problem: Teeing Off A 50-g golf ball at rest is hit by “Big Bertha” club with 500-g mass. After the collision, golf leaves with velocity of 50 m/s. a) Find impulse imparted to ball b) Assuming club in contact with ball for 0.5 ms, find average force acting on golf ball Problem: teeing off 1. Use impulse-momentum relation: Given: impulse = Δp = mv f − mvi = (0.050 kg )(50 m s ) − 0 mass: m=50 g = 0.050 kg velocity: v=50 m/s Find: impulse=? Faverage=? = 2.50 kg ⋅ m s 2. Having found impulse, find the average force from the definition of impulse: Δp 2.50 kg ⋅ m s = Δt 0.5 ×10 −3 s = 5.00 ×10 3 N Δp = F ⋅ Δt , thus F = Note: according to Newton’s 3rd law, that is also a reaction force to club hitting the ball: of club F ⋅ Δt = − F R ⋅ Δt , or ( ) mv f − mv i = − M V f − M V i , or mv f + M V f = mv i + M V i CONSERVATION OF MOMENTUM Conservation of Momentum • Definition: an isolated system is the one that has no external forces acting on it Momentum in an isolated system in which a collision occurs is conserved (regardless of the nature of the forces between the objects) • A collision may be the result of physical contact between two objects • “Contact” may also arise from the electrostatic interactions of the electrons in the surface atoms of the bodies Conservation of Momentum The principle of conservation of momentum states when no external forces act on a system consisting of two objects that collide with each other, the total momentum of the system before the collision is equal to the total momentum of the system after the collision Conservation of Momentum • Mathematically: m1 v1i + m2 v 2i = m1 v1 f + m2 v 2 f • Momentum is conserved for the system of objects • The system includes all the objects interacting with each other • Assumes only internal forces are acting during the collision • Can be generalized to any number of objects Problem: Teeing Off (cont.) Let’s go back to our golf ball and club problem: Ball : Δp = 2.50 kg ⋅ m s , m = 50 gramm Δv = 50 m s ( ( ) ) Club : m v f − v i = −2.50 kg ⋅ m s , so − 2.50 kg ⋅ m s v f − vi = = −5m s 0.5 kg factor of 10 times smaller 18 More on Conservation of Momentum v1i m1>m2 v2i m1 m2 A short time later the masses collide. m1 m2 What happens? 19 During the interaction: N1 N2 y F21 F12 w1 ∑F ∑F y = N1 − w1 = 0 x = − F21 = m1a1 w2 ∑F ∑F y = N 2 − w2 = 0 x = F12 = m2 a2 There is no net external force on either mass. x 20 Conservation of Momentum F21 = − F12 F21Δt = − F12 Δt Δp1 = −Δp 2 p1 f − p1i = p 2 f − p 2i p1i + p 2i = p1 f + p 2 f If net external force acting on a system is zero, then the momentum of the system is conserved If ∑ Fext = 0, pi = p f 21 Example: A rifle has a mass of 4.5 kg and it fires a bullet of 10.0 grams at a muzzle speed of 820 m/s. What is the recoil speed of the rifle as the bullet leaves the barrel? Given: mr = 4.5 kg mb = 10.0 g =0.01 kg vb = 820 m/s vir = vib = 0 m/s Find: vr = ? Idea: as long as the rifle is horizontal, there will be no net external force acting on the rifle-bullet system and momentum will be conserved. pi = p f 0 = mb vb + mr vr " 0.01 kg % mb vr = − vb = − $ ' 820 m/s mr # 4.5 kg & = −1.82 m/s ConcepTest Suppose a person jumps on the surface of Earth. The Earth 1. will not move at all 2. will recoil in the opposite direction with tiny velocity 3. might recoil, but there is not enough information provided to see if that could happened ConcepTest Suppose a person jumps on the surface of Earth. The Earth 1. will not move at all 2. will recoil in the opposite direction with tiny velocity 3. might recoil, but there is not enough information provided to see if that could happened Note: momentum is conserved. Let’s estimate Earth’s velocity after a jump by a 80-kg person. Suppose that initial speed of the jump is 4 m/s, then: Person : Δp = 320 kg ⋅ m s Earth : Δp = M EarthVEarth = −320 kg ⋅ m s , so VEarth = − 320 kg ⋅ m s − 23 = − 5 . 3 × 10 m s 6 ×10 24 kg tiny negligible velocity, in opposite direction 24 Do you know why police use skid marks to find velocities of vehicles before collision??? See page 246 Rocket Propulsion • The basic equation for rocket propulsion is: ⎛ Mi v f − v i = v e ln⎜⎜ ⎝ Mf ⎞ ⎟⎟ ⎠ • Mi is the initial mass of the rocket plus fuel • Mf is the final mass of the rocket plus any remaining fuel • The speed of the rocket is proportional to the exhaust speed Thrust of a Rocket • The thrust is the force exerted on the rocket by the ejected exhaust gases • The instantaneous thrust is given by Δv ΔM Ma = M = ve Δt Δt • The thrust increases as the exhaust speed increases and as the burn rate (ΔM/Δt) increases