Presentation is loading. Please wait.

Presentation is loading. Please wait.

Oct. 5, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 7 Linear Momentum Conservation of Momentum.

Similar presentations


Presentation on theme: "Oct. 5, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 7 Linear Momentum Conservation of Momentum."— Presentation transcript:

1

2 Oct. 5, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 7 Linear Momentum Conservation of Momentum

3 2 Recall: Generalized Work-Energy Theorem K 1 + U 1 + W nc = K 2 + U 2 Conservation of Energy means the total energy doesn’t change. Conservation of Momentummeans the total momentum doesn’t change. What’s Momentum????

4 3 Momentum p = mv 1.Momentum = mass * velocity 2.Momentum is a vector. 3.It is parallel to the velocity.

5 4 Equivalent Formulation of Newton’s Second Law The rate of change of momentum of a body is equal to the net force applied to it.

6 5 Equivalent Formulation of Newton’s Second Law The rate of change of momentum of a body is equal to the net force applied to it.

7 6 Newton’s Second Law & Momentum When there is a net force: Momentum changes.  p =  F  t = Impulse When there is no net force: Momentum remains constant.  p =  F  t = 0

8 7 CAPA 1 & 2 A golf ball of mass 0.05 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 5.0x10 -3 s. 1.Find the impulse imparted to the ball. Impulse =  F  t =  p = mv f – mv o Impulse = m(v f – v o ) = 0.05kg * (45 – 0)m/s Impulse = 2.25 kg-m/s = 2.25 N-s

9 8 CAPA 1 & 2 A golf ball of mass 0.05 kg is hit off the tee at a speed of 45 m/s. The golf club was in contact with the ball for 5.0x10 -3 s. 2.Find the average force imparted to the golf ball by the club. Impulse =  F  t ( = 2.25 N-s ) Impulse/  t =  F  F = 2.25 N-s/5.0x10 -3 s = 450 N

10 9 Application to Collisions When  p = 0 then: Momentum Before Collision = Momentum After Collision Mathematically this means: M 1 V 1b + M 2 V 2b = M 1 V 1a + M 2 V 2a

11 10 Types of Collisions An Elastic Collision – Kinetic Energy does not change. An Inelastic Collision – Kinetic Energy changes

12 11 Inelastic Collision: CAPA 4 Momentum Before Collision = Momentum After Collision M t V tb + M s V sb = M t V ta + M s V sa V tb V ta = V sa = 0V sb M t V tb + M s V sb = (M t +M s ) 0 M t V tb = - M s V sb  V sb = -V tb *M t /M s

13 12 Inelastic Collision Momentum Before Collision = Momentum After Collision M 1 V 1b + M 2 V 2b = M 1 V 1a + M 2 V 2a V 1b = V b V 1a = V 2a = V a V 2b = 0 M 1 V b + M 2 0 = (M 1 +M 2 )V a

14 13 Inelastic Collisions (cont) VbVb V a M 1 V b + M 2 0 = (M 1 +M 2 )V a M 1 V b = (M 1 +M 2 )V a M 1 /(M 1 +M 2 ) V b = V a

15 14 Center of Mass Velocity M 1 V 1b + M 2 V 2b  (M 1 + M 2 )V cm V 1b V 2b U 1b U 2b U = V - V cm and V = U + V cm

16 15 Elastic Collision in the CM U 1b U 2b U 1a U 2a M 1 U 1b + M 2 U 2b = 0 M 1 U 1a + M 2 U 2a = 0 ½M 1 U 2 1b + ½M 2 U 2 2b = ½M 1 U 2 1a + ½M 2 U 2 2a

17 16 Elastic Collision in the CM M 1 U 1b + M 2 U 2b = 0M 1 U 1a + M 2 U 2a = 0 ½M 1 U 2 1b + ½M 2 U 2 2b = ½M 1 U 2 1a + ½M 2 U 2 2a Solution – Mirror Image: U 1a = - U 1b U 2a = - U 2b

18 17 Elastic Collision in the CM: CAPA #6-9 A pair of bumper cars in an amusement park ride collide elastically as one approaches the other directly from the rear. One has mass m 1 = 467 kg and the other mass m 2 = 567 kg. If the lighter one approaches at v 1 = 4.23 m/s and the other one is moving at v 2 = 3.88 m/s, calculate: the velocity of the lighter car after the collision. the velocity of the heavier car after the collision. the change in momentum of the lighter car. the change in momentum of the heavier car. 4.23 m/s3.88 m/s

19 18 Elastic Collision in the CM: CAPA #6-9 1.Calcuate V cm = (M 1 V 1b +M 2 V 2b )/(M 1 +M 2 ) 2.Calcuate U 1b and U 2b 1.U 1b = V 1b – V cm and U 2b = V 2b – V cm 3.Set U 1a = -U 1b and U 2a = -U 2b 4. Calcuate V 1a and V 2a 1.V 1a = U 1a + V cm and V 2a = U 2a + V cm 3.88 m/s

20 19 Elastic Collision in the CM: CAPA #6-9 1.Calcuate V cm = (M 1 V 1b +M 2 V 2b )/(M 1 +M 2 ) 2.Calcuate U 1b and U 2b A.U 1b = V 1b – V cm and U 2b = V 2b – V cm V cm = (M 1 V 1b +M 2 V 2b )/(M 1 +M 2 ) V cm = (467*4.23 + 567*3.88)/(467+567) = 4.038 m/s U 1b = V 1a - V cm = 4.23 m/s - 4.038 m/s = 0.192 m/s U 2b = V 2b - V cm = 3.88 m/s - 4.038 m/s = -0.158 m/s U 1b U 2b U 1a U 2a

21 20 Elastic Collision in the CM: CAPA #6-9 3.Set U 1a = -U 1b and U 2a = -U 2b 4. Calcuate V 1a and V 2a A.V 1a = U 1a + V cm and V 2a = U 2a + V cm U 1a = - U 1b = -0.192 m/s U 2a = -U 2b = 0.158 m/s V 1a = U 1a + V cm = -0.192 m/s + 4.038 m/s = 3.846 m/s V 2a = U 2a + V cm = 0.158 m/s + 4.038 m/s = 4.196 m/s U 1b U 2b U 1a U 2a 3.85 m/s4.20 m/s

22 21 Elastic Collision in the CM: CAPA #6-9 Impulse = mV a – mV b (for each of cars 1 and 2) Note: Impulse on car 1 = - Impulse on car 2.

23 22 Next Time Chapter 7 – Conservation of Momentum. Quiz on Chapter 7. Please see me with any questions or comments. See you on Monday.


Download ppt "Oct. 5, 2001 Dr. Larry Dennis, FSU Department of Physics1 Physics 2053C – Fall 2001 Chapter 7 Linear Momentum Conservation of Momentum."

Similar presentations


Ads by Google