1. PHY 101: Lecture 7 7.1 The Impulse-Momentum Theorem 7.2 The Principle of Conservation of Linear Momentum 7.3 Collision in One Dimension 7.4 Collisions in Two Dimensions 7.5 Center of Mass

2. PHY 101: Lecture 7 Impulse and Momentum 7.1 The Impulse-Momentum Theorem

3. Force and Collision Time

4. Definition of Impulse Vector quantity Impulse J of a force is product of average force F avg and time interval  t during which the force acts  J = F avg  t Direction of Impulse is same as direction of average force SI Units of Impulse: Newton second (Ns)

5. Definition of Linear Momentum Linear momentum p of an object is the product of the object’s mass m and velocity v  p = mv Linear momentum is a vector that has the same direction as velocity Symbol: p SI Units of Linear Momentum: kilogram meter/second (kg m/s)

6. Linear Momentum – Example 1 What is magnitude of linear momentum of a 7.1-kg bowling ball traveling at 12 m/s?  p = mv = 7.1(12) = 85.2 kg(m/s)

7. Linear Momentum – Example 2 Linear momentum of a runner in a 100-m dash is 7.5 x 10 2 kg(m/s) The runner’s speed is 10 m/s What is his mass?  p = mv  750 = m(10)  m = 750/10 = 75 kg

8. Momentum vs. Kinetic Energy Momentum and Kinetic Energy are two different properties of matter Massless radiation has Momentum but not Kinetic Energy, but that is PHY 102 Formula relationship is  KE = p 2 / 2m

9. Impulse and Momentum Relationship  p = mv  p =  (mv) = m(  v)  p/  t = m(  v/  t) = ma = F (Newton’s 2 nd Law)  p = F(  t) = J Note:  p = p f – p i,  t = t f – t i Note: We assumed that m is constant, but m could also change. We will see examples where this is the case

10. Impulse – Momentum Theorem When a net force acts on an object, the impulse of this force is equal to the change in momentum of the object  F)  t = mv f - mv i Impulse = Change in Momentum

11. Impulse-Momentum Theorem Example 1 A 0.20-kg softball is tossed upward and hit horizontally by a batter Softball receives an impulse of 3.0 Ns With what horizontal speed does the ball move away from the bat?  J = p f – p i = mv f - mv i  Initial momentum is 0  J = mv f  v f = J/m = 3/0.20 = 15 m/s

12. Impulse-Momentum Theorem Example 2 Automobile with linear momentum of 3.0x10 4 kg(m/s) in +x-direction is brought to a stop in 5.0 s What is the average braking force?  J = Ft = p f – p i  Final momentum is zero  F = – p i /t = -(+3.0 x 10 4 ) / 5 = - 6000 N

13. Impulse-Momentum Theorem Example 3 A pool player imparts an impulse of 3.2 Ns to a stationary 0.25-kg cue ball with a cue stick What is the speed of the ball just after impact?  J = F  t = p f – p i = mv f - mv i  Initial momentum is zero  J = mv f  v f = J/m = 3.2/0.25 = 12.8 m/s

14. Impulse-Momentum Theorem Example 4 A car of 1500kg is traveling in +x-direction at 20m/s Force acts on car in the +x-direction for 30 seconds Final velocity of car is 40m/sec in the +x-direction What is the force?  J = F  t = p f – p i = mv f - mv i  J = 1500(+40) – 1500(+20)  J = 60000 – 30000 = 30000  F = 30000/  t = 30000/30 = 1000 N

15. Impulse-Momentum Theorem Example 5 Machine gun fires 15 bullets (mass 0.020 kg) per second Velocity of bullets is 800 m/sec in the +x-direction What is the force on the machine gun?  Change in momentum of 1 bullet = mv f – mv i = (0.020)800 = 16  Change in momentum for all bullets in second = 16(15) = 240 N  This is average force on the bullets  Force on gun is equal and opposite, -240 N

16. PHY 101: Lecture 7 Impulse and Momentum 7.2 The Principle of Conservation of Linear Momentum

17. Conservation of Momentum Internal / External Forces Two types of forces act on the system  Internal forces – Forces that the objects within the system exert on each other  External forces – Forces exerted on the objects by agents external to the system

18. Conservation of Momentum Collision between Two Balls 1 System is two balls that are colliding F 12 is the force exerted on ball 1 by ball 2 F 21 is the force exerted on ball 2 by ball 1 Forces are internal forces Forces are equal in magnitude but opposite in direction  F 12 = - F 21 Force of gravity also acts on the balls The weights are external forces The weights are W 1 and W 2 Impulse-Momentum Theorem is applied to the two balls

19. Conservation of Momentum Collision between Two Balls 2 (W 1 + F 12 )  t = m 1 v f1 – m 1 v i1 (W 2 + F 21 )  t = m 2 v f2 – m 2 v i2 Add these equations (W 1 +W 2 +F 12 +F 21 )  t = (m 1 v f1 +m 2 v f2 ) – (m 1 v i1 +m 2 v i2 ) (sum external forces+sum internal forces)  t = p f - p i Sum of internal forces is zero (Newton’s Third Law) (sum of external forces)  t = p f – p i Suppose that the sum of external forces is zero p f = p i This is conservation of momentum

20. Conservation of Momentum Total linear momentum of an isolated system remains constant An isolated system is one for which the vector sum of the external forces acting on the system is zero Note: It is important to realize that the total linear momentum may be conserved even when the kinetic energy is not conserved

21. Conservation of Momentum Example 1 A 60-kg astronaut floats at rest in space He throws his 0.50-kg hammer such that it moves with a speed of 10 m/s in the -x-direction What happens to the astronaut?  Astronaut is object 1 and hammer is object 2  Momentum is conserved, p f = p i  Initial momentum is zero  m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i = 0  60v 1f + 0.5(-10) = 0  v 1f = 5/60 = +0.0833 m/s

22. Conservation of Momentum Example 2 An 800-kg car travelling in the +x-direction with a velocity of 30 m/sec It hits a stationary truck of 3200 kg The car and truck stick together What is speed of combined masses after the collision?  The car is object 1 and the truck object 2  p f = p i  (m 1 + m 2 )v f = m 1 v 1i + m 2 v 2i  4000v f = 800(+30) + 3200(0) = 24000  v f = 24000 / 4000 = +6 m/s

23. Conservation of Momentum Example 3 A car of mass 1200 kg is traveling 30 m/s in the +x- direction The car collides with a truck of 3200 kg traveling in the –x-direction at 20 m/s After the collision, the truck comes to a stop What is the velocity of the car?  The car is object 1 and the truck object 2  p f = p i  m 1 v 1f + m 2 v 2f = m 1 v 1i + m 2 v 2i  (1200)v 1f = 1200(30) + (3200)(-20) = -28000  v 1f = -28000/1200 = -23.33 m/s

24. PHY 101: Lecture 7 Impulse and Momentum 7.3 Collisions in One Dimension

25. 1-Dimension Collisions Total linear momentum is conserved when two objects collide If they are an isolated system (no external forces) Collisions are classified according to whether the total kinetic energy changes during the collision:  Elastic collision – Total kinetic energy of system after collision is equal to total kinetic energy before collision  Inelastic collision – Total kinetic energy of system is not the same before and after collision  If the objects stick together after colliding, the collision is said to be completely inelastic When a collision is completely inelastic, the greatest amount of kinetic energy is lost

26. Elastic Collision 1 Ball 1 initially is moving in the +x-direction Ball 2 initially is not moving The two balls collide and bounce off each other with no loss of total kinetic energy Each ball moves off with a separate velocity What are the velocities of the two balls after the collision?

27. Elastic Collision 2 m 1 = m 2 v 1 = 0v 2 = u 1 m 1 >>> m 2 v 1 = u 1 v 2 = 2u 1 m 1 <<< m 2 v 1 = -u 1 v 2 = 0

28. PHY 101: Lecture 7 Impulse and Momentum 7.4 Collisions in Two Dimension Skipped

29. PHY 101: Lecture 7 Impulse and Momentum 7.5 Center of Mass Skipped