Physics 2514 Lecture 26 P. Gutierrez Department of Physics & Astronomy University of Oklahoma Physics 2514 p. 1/12
Review We have defined the following using Newton s second law of motion ( F net = d p dt ): Impulse J = t f t Fnet i dt; Momentum p = m v; Showed the impulse and momentum related by J = p. Using Newton s third law we arrived at momentum conservation ( F 12 = F 21 d P net dt = 0): For two objects p 1i + p 2i = p 1f + p 2f ; Holds if no external forces acting on system; Holds for any number of isolated objects. Components independently conserved. Physics 2514 p. 2/12
Cases to Consider Three basic problem types Use impulse-momentum theorem to calculate changes in a system Collisions Perfectly inelastic objects stick together Perfectly elastic objects bounce apart (requires energy to be conserved) Inelastic part way between the above two cases. Explosions Objects initially at rest relative to each other Collisions in 2 dimensions Includes cases given above. Physics 2514 p. 3/12
Example Impulse A sled slides along on a horizontal surface on which the coefficient of kinetic friction is 0.25. Its velocity at point A is 8.0 m/s and at point B is 5.0 m/s. Use the impulse-momentum relation to find how long the sled takes to travel from A to B. An object is being slowed by friction (µ k = 0.25). How long ( t) does it take the object to reduce its speed from v i = 8.0 m/s to v f = 5.0 m/s? replacements Momentum Forces p i p f PSfrag replacements f k n A B mg Physics 2514 p. 4/12
Example Impulse An object is being slowed by friction (µ k = 0.25). How long ( t) does it take the object to reduce its speed from v i = 8.0 m/s to v f = 5.0 m/s? replacements Momentum Forces p i p f PSfrag replacements f k n A B mg J = F t = p f k t = mv f mv i f k = nµ k = mgµ k mgµ k t = m(v f v i ) t = v i v f gµ k = 1.2 s Physics 2514 p. 5/12
Clicker A block of mass m is placed on a frictionless inclined plane that makes a 30 angle relative to the horizontal. The block is released from rest, travels a distance of 20 cm along the incline. At the bottom of the incline is a spring that applies a force given by F = 100 s, where s is the amount the spring is compressed. After striking the spring, the block starts moving upward and eventually comes to rest 10 cm below where it started. What is the total impulse from the moment the block is released until it comes to rest again. A) What? B) 0 N-s C) 30 N-s D) 100 N-s E) 200 N-s Physics 2514 p. 6/12
Example Momentum Conservation Two ice skaters, with masses of 50 kg and 70 kg, are at the center of a 60 m diameter circular rink. The skaters push off each other and glide to the opposite edges of the rink. If the heavier skater reaches the edge in 20 s, how long does it take the lighter skater to reach the edge? Objects A (m = 50 kg) and B (m = 70 kg) have a combined zero net momentum. If object B takes 20 s to travel 30 m, how long does it take object A? Physics 2514 p. 7/12
Example Momentum Conservation cements Objects A (m = 50 kg) and B (m = 70 kg) have a combined zero net momentum. If object B takes 20 s to travel 30 m, how long does it take object A? p = 0 Before x Velocity object B: x = v B t v B = x t Velocity object A: m B v B m A v A = 0 = 1.5 m/s p A p B After v A = m B m A v B = 2.1 m/s Time for A to reach edge of rink: x = v A t t = x v A = 14.3 s Physics 2514 p. 8/12
Clicker Consider a perfectly inelastic collision between two objects (a collision where the objects stick together after they interact). Object A has a mass m and an initial speed v, while object B has a mass 2m and is initially at rest. After they collide what is the speed of the system A) v B) v/2 C) 2v D) v/3 E) 3v Physics 2514 p. 9/12
Example 2-d Collision The figure shows a collision between 3 clay objects. What is the speed and direction of the resulting object? Before After m = 90 g PSfrag replacements p = 90 v f Physics 2514 p. 10/12
Example 2-d Collision The figure shows a collision between 3 clay objects. What is the speed and direction of the resulting object? Momentum conservation, vertical m 20 v 2 m 40 v 4 sin(45 ) = m 90 v yf v yf = 0.81 m/s Momentum conservation, horizontal m 40 v 4 cos(45 ) m 30 v 3 = m 90 v xf v xf = 0.26 m/s Result v = q v 2 xf + v2 yf = 0.85 m/s θ = tan 1 v yf v xf = 72.4 Physics 2514 p. 11/12
Assignment Start reading Chapter 10 Physics 2514 p. 12/12