Solution: Electricity: 1 kw-hr = (1 x10 3 J/s)(3.6 x10 3 s) = 3.6 x10 6 J 1 kw-hr = (1 x10 3 J/s)(3.6 x10 3 s) = 3.6 x10 6 J E/$ = 3.6 x10 6 /$0.1 = 3.6 x10 7 J/$ Gasoline: E/$ = 3.6 x10 6 /$0.1 = 3.6 x10 7 J/$ Gasoline: 1 gal = 1.3 x10 8 J 1 gal = 1.3 x10 8 J E/$ = 1.3 x10 8 /$2 = 6.5 x10 7 J/$ E/$ = 1.3 x10 8 /$2 = 6.5 x10 7 J/$ Cost (gas/elec) ~ 180% for same efficiency.Cost (gas/elec) ~ 180% for same efficiency. Physics 1710—Chapter 9 Momentum

Consider this: How high will the tennis ball bounce relative to its starting height when they are dropped as shown below? How high will the tennis ball bounce relative to its starting height when they are dropped as shown below? Physics 1710—Chapter 9 Momentum

1’ Lecture Linear momentum is the product of the velocity and the mass and is, therefore, a vector quantity. Linear momentum is the product of the velocity and the mass and is, therefore, a vector quantity. p = m v p = m v Momentum is always conserved. Momentum is always conserved. Impulse is the time integrated force. Impulse is the time integrated force. Physics 1710—Chapter 9 Momentum

Linear Momentum (Latin: “movement”) p ≡ m v p x = m v x p y = m v y p z = m v z Physics 1710—Chapter 9 Momentum

Newton’s Second Law of Motion (What Newton actually said:) ∑F = d p/dt The net external force is equal to the time rate of change in the linear momentum. First law: In the absence of any net external force (∑F = 0)the linear momentum p of a system is conserved (d p/dt = 0). Physics 1710—Chapter 9 Momentum

The Collision of Two Bodies ①⇒ ⇐② F 12 = - F 21 F 21 + F 12 = 0 d p 1 /dt +d p 2 /dt = 0 d( p 1 + p 2 )/dt = 0 Thus, the total momentum is conserved in a collision. Physics 1710—Chapter 9 Momentum

Conservation of Linear Momentu m p 1i + p 2i = p 1f + p 2f ∑p ix =∑p fx ∑p iy =∑p fy ∑p iz = ∑p fz ∑p iz = ∑p fz Physics 1710—Chapter 9 Momentum

Big Ball – Little Ball Collision Demonstration!!! ??? ⃘ ⃗ ⃖ ⃝ ??? Physics 1710—Chapter 9 Momentum

Elastic Collision P conserved p 1i + p 2i = p 1f + p 2f m 1 v 1i + m 2 v 2i = m 1 v 1f + m 2 v 2f (Assume) E conserved K 1i + K 2i = K 1f + K 2f (Assume) E conserved K 1i + K 2i = K 1f + K 2f ½ m 1 v 1i 2 +½ m 2 v 2i 2 = ½ m 1 v 1f 2 + ½ m 2 v 2f 2 Solve equations: v 1f = [(m 1 -m 2 )/ (m 1 +m 2 )]v 1i +[(2 m 2 )/ (m 1 +m 2 )]v 2i v 2f = [(m 2 –m 1 )/ (m 1 +m 2 )]v 2i +[(2 m 1 )/ (m 1 +m 2 )]v 1i Physics 1710—Chapter 9 Momentum

v 1f = [(m 1 -m 2 )/ (m 1 +m 2 )]v 1i +[(2 m 2 )/ (m 1 +m 2 )]v 2i v 2f = [(m 2 –m 1 )/ (m 1 +m 2 )]v 2i +[(2 m 1 )/ (m 1 +m 2 )]v 1i Elastic Collision v 1f = [(m 1 -m 2 )/ (m 1 +m 2 )]v 1i +[(2 m 2 )/ (m 1 +m 2 )]v 2i v 2f = [(m 2 –m 1 )/ (m 1 +m 2 )]v 2i +[(2 m 1 )/ (m 1 +m 2 )]v 1i Physics 1710—Chapter 9 Momentum Let m 2 >> m 1 ; v 1i = -v 2i v 1f = [(0-m 2 )/ (0+m 2 )]v 1i -[(2 m 2 )/ (0 +m 2 )](-v 1i ) v 1f = (-v 1i -2 v 1i ) = - 3 v 1i !! K 1f /K 1i = v 1f 2 /v 1i 2 = 9 = h 1i /h 2f

What about a ball that bounces off of the earth? What is the change in the momentum of the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up? What is the change in the momentum of the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up? What is the change in velocity of the earth? (mass of earth = 6.0x10 24 kg) What is the change in velocity of the earth? (mass of earth = 6.0x10 24 kg) Physics 1710—Chapter 9 Momentum

Peer Instruction Time What is the change in the momentum of the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up?What is the change in the momentum of the earth when a 0.60 kg ball falls from a height of 1.0 meter and bounces up? What is the change in velocity of the earth? (mass of earth = 6.0x10 24 kg) What is the change in velocity of the earth? (mass of earth = 6.0x10 24 kg) Physics 1710—Chapter 9 Momentum Σpi = P i +p i = 0 = M V i + mv i 6.0x10 24 kg V i = - (m/M)v i v i = √2gh = √2(9.8 m/s 2 )(1.0 m h ) = 4.4 m/s V i = - (0.6 kg/ 6.0x10 24 kg )v i = x 10 – 25 m/s If ball rebounds to same height v 1f = v 1i ; V 1f = V 1i ; ΔV = 8.8 x 10 – 25 m/s (~1proton/36yr)

Elastic Collision Special Case m 1 = m 2 v 1f = [(2 m 2 )/ (m 1 +m 2 )]v 2i v 2f = [(2 m 1 )/ (m 1 +m 2 )]v 1i K 1f = [(4 m 1 m 2 )/ (m 1 +m 2 ) 2 ] K 2i K 2f = [(4 m 1 m 2 )/ (m 1 +m 2 ) 2 ] K 1i v 1f = v 2i K 1f = K 2i v 2f = v 1i K 2f = K 1i Physics 1710—Chapter 9 Momentum

Elastic Collision Special Case m 1 = m 2 Newton’s Cradle Physics 1710—Chapter 9 Momentum

Totally Inelastic Collisions (no bounce) [The objects stick together after collision.] v 1f = v f v 2f = v f m 1 v 1i + m 2 v 2i = (m 1 + m 2 ) v f v f = (m 1 v 1i + m 2 v 2i )/ (m 1 + m 2 ) K i = ½ m 1 v 1i 2 +½ m 2 v 2i 2 K f = ½ (m 1 + m 2 )v f 2 K f = ½ (m 1 + m 2 )v f 2 Physics 1710—Chapter 9 Momentum

Impulse and Momentum d p = F dt ∆p = ∫d p = ∫F dt = Impulse The impulse on a body equals the change in momentum. Physics 1710—Chapter 9 Momentum

Impulse and “Follow Through” Demonstration Physics 1710—Chapter 9 Momentum

∆p = ∫F dt ∆p =F ave ∆t For a given force the longer it is applied the greater will be the impulse and the change in momentum. applied the greater will be the impulse and the change in momentum. Physics 1710—Chapter 9 Momentum

Impulse and Seat Belts Seat Belts ( and air bagsSeat Belts ( and air bags and crumple zones) increase the stopping time ∆t. If ∆p is the same in two instants the impulse will be the same. The case with the longer ∆t will exhibit the smaller average force.If ∆p is the same in two instants the impulse will be the same. The case with the longer ∆t will exhibit the smaller average force. Physics 1710—Chapter 9 Momentum

Stopping Force ∆ p = mv ∆t = s/v ave = s/(v/2) F ave = ∆ p/ ∆t = mv 2 /(2s) Speed kills? : v 2 What about the sudden stop? :1/s Physics 1710—Chapter 9 Momentum

Physics 1710 Chapter 9 Linear Momentum and Collisions Center of Mass R CM = ∑m i r/ M Or R CM = ∫r dm / M

Physics 1710 Chapter 9 Linear Momentum and Collisions Summary: Linear momentum is the product of the mass and velocity. Linear momentum is the product of the mass and velocity. Linear momentum is conserved.Linear momentum is conserved. Impulse is the time integral of the momentum.Impulse is the time integral of the momentum. The impulse is equal to the change in momentum of a system.The impulse is equal to the change in momentum of a system.