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11/6/2014 Momentum Momentum and Collisions • Momentum: p = mv • Units are kg(m/s): no derived units • A vector quantity: same direction as velocity Phy 114 v=2m/s Linear momentum p= 3 kg (2m/s) Important points about linear momentum • Linear momentum is a vector quantity; it is important to consider the direction in which the colliding objects are moving before and after the collision. • Momentum depends on the velocity of the object, and the velocity depends on the choice of the reference frame. Different observers will measure different momenta for the same object. • To establish that momentum is a conserved quantity, we need to ensure that the momentum of a system changes in a predictable way for systems that are not isolated. © 2014 Pearson Education, Inc. Momentum constancy of an isolated system • For a system with more than two objects, we simply include a term on each side of the equation for each object in the system. © 2014 Pearson Education, Inc. © 2014 Pearson Education, Inc. Example 5.1: Two rollerbladers • Jen (50 kg) and David (75 kg), both on rollerblades, push off each other abruptly. Each person coasts backward at approximately constant speed. During a certain time interval, Jen travels 3.0 m. • How far does David travel during that same time interval? © 2014 Pearson Education, Inc. 1 11/6/2014 From Newton’s 2nd ΣF = ma ∆v ΣF = m ∆t ΣF ( ∆t ) = m(∆v) ΣF ( ∆t ) = mv2 − mv1 • Impulse: J=ΣF Δt • Units are same as momentum • Impulse causes a change in momentum • What happens to ΣF with Impulse due to a force exerted on a single object • We need a way to account for change in momentum when the net external force on a system is not zero • A relationship can be derived from Newton's laws and kinematics: – Small Δt? – Large Δt? ΣF ( ∆t ) = ∆p © 2014 Pearson Education, Inc. • Elastic – Momentum is Conserved – Kinetic Energy is Conserved • Inelastic – Momentum is Conserved – Kinetic Energy is NOT Conserved Impulse: The product of the external force exerted on an object and the time interval Graphs A force is acting in the x-direction on a 4.00kg particle. The force varies in time and is shown on the graph. Find: • the impulse • the final velocity if initially at rest • the final velocity if initial velocity is –2.00 m/s. Force vs Time for a Moving Cart 4.5 4 3.5 3 Force (N) Collisions 2.5 2 1.5 1 0.5 0 0 5 10 Time (sec) 15 20 Impulse-momentum equation for a single object • If the magnitude of the force changes during the time interval considered in the process, we use the average force. © 2014 Pearson Education, Inc. © 2014 Pearson Education, Inc. 2 11/6/2014 Example 5.2: Abrupt stop in a car • A 60-kg person is traveling in a car that is moving at 16 m/s with respect to the ground when the car hits a barrier. The person is not wearing a seat belt, but is stopped by an air bag in a time interval of 0.20 s. • Determine the average force that the air bag exerts on the person while stopping him. Problem Solving Multiple Plans • Draw a MD – What new motion equation can you add? • Draw a FBD – What new equation do we have that relates force and motion? • A 0.500-kg football is thrown toward the east with a speed of 15.0 m/s. A stationary receiver catches the ball and brings it to rest in 0.020 0 s. • What is the average force exerted on the receiver? Ans: F avg = 375 N © 2014 Pearson Education, Inc. Problem Solving: …..this way Problem Solving: solve this way or… Multiple Plans • Draw a MD Multiple Plans • Draw a MD – What new motion equation can you add? • Draw a FBD x=0 t=0 v=15 m/s x= t= 0 a= – What new equation do we have that relates force and motion? – What new motion equation can you add? • Draw a FBD – What new equation do we have that relates force and motion? Impulse --- Momentum equation − nhand on football ∆t = m∆v − nhand on football = ma x Ans: F avg = 375 N Ans: F avg = 375 N Problem in 2-D Collision • Momentum always conserved – Do this first – Remember it is a vector • Kinetic Energy conserved in elastic collision – Check this next A billiard ball moving at 5.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.2 m/s at an angle of 30° with respect to the original line of motion. • Find the velocity (magnitude and direction) of the second ball after collision. • Was the collision inelastic or elastic? Conservation of Momentum Motion Momentum Momentum (x) Momentum (y) Mass 1: Initial 5 m/s Mass 2: Initial 0 m/s Mass 1: Final 4.2 m/s at 30° Mass 2: Final Conserve Momentum 3 11/6/2014 Conservation of Momentum The generalized impulse-momentum principle Assume 0.5 kg balls Motion Momentum Momentum (x) Momentum (y) Mass 1: Initial 5 m/s 2.5 kgm/s to the right +2.5 kgm/s 0 Mass 2: Initial 0 m/s 0 0 0 Mass 1: Final 4.2 m/s at 30° 2.1 kgm/s at 30° +2.1 cos30° = 1.82 kgm/s +2.1 sin30° = 1.05 kgm/s Mass 2: Final unknown Conserve Momentum px py 2.5 + 0 = 1.82 + px Px = +0.68 0 = 1.05 + py Py=-1.05 Use pythagorean equation to get p=1.25 kgm/s @57 degrees below +x axis. Use p=mv to get final vel of Mass 2: v=2.5 m/s @57 degrees below +x axis © 2014 Pearson Education, Inc. Impulse-momentum bar charts Example 5.5: Stopping the fall of a movie stunt diver • The record for the highest movie stunt fall without a parachute is 71 m, held by 80-kg A. J. Bakunas. His fall was stopped by a large air cushion, into which he sank about 4.0 m. His speed was approximately 36 m/s when he reached the top of the air cushion. • Estimate the average force that the cushion exerted on this stunt diver's body while stopping him. © 2014 Pearson Education, Inc. © 2014 Pearson Education, Inc. Determining the stopping time interval from the stopping distance © 2014 Pearson Education, Inc. 4