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Impulse, Momentum, Work & Energy

Impulse, Momentum, Work & Energy. Impulse. Force acting for a period of time J = F∆t Unit of measurement: N·s or kg·m/s Describes how hard and how long we push to change a motion. Momentum. Impulse causes a change in momentum. Impulse-Momentum Theorem F∆t = ∆p

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Impulse, Momentum, Work & Energy

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  1. Impulse, Momentum, Work & Energy

  2. Impulse • Force acting for a period of time • J = F∆t • Unit of measurement: N·s or kg·m/s • Describes how hard and how long we push to change a motion Finkster '07

  3. Momentum • Impulse causes a change in momentum. • Impulse-Momentum Theorem • F∆t = ∆p • Linear momentum is defined as the product of mass x velocity • p = m∆v Finkster '07

  4. Unit of Measurement for ‘p’ • MKS N·s or kg·m/s • CGS dyn·s or g·cm/s • FPS lb·s Finkster '07

  5. Vector Quantities Both impulse and momentum are vector quantities. They have both magnitude and direction. Finkster '07

  6. Newton’s Second Law • F∆t = m∆v • F/m = ∆v/t • a = a • F = m ∆v/t • F = ma • Force is directly proportional to the change in velocity if the mass is held constant. Finkster '07

  7. Exerting a given impulse • A big force for a short time (F∆t) • A smaller force for a longer time (F∆t) • A force that changes while it acts (F∆t) Finkster '07

  8. Direction of Impulse • Impulse points in the same direction as the force and the change in momentum. Finkster '07

  9. Impulse and Momentum Sample Problems

  10. 1. A baseball of mass 0.14 kg is moving at +35 m/s. • Find the momentum of the baseball. m = 0.14 kg p = m∆v v = +35 m/s = (0.14 kg)(+35 m/s) p = ? p = +4.9 kg·m/s Finkster '07

  11. b. Find the velocity at which a bowling ball, mass 7.26 kg, would have the same momentum as the baseball. p = +4.9 kg·m/s p = mv m = 7.26 kg ∆v = p/m ∆V = ? = +4.9 kg·m/s 7.26 kg ∆v = +0.67 m/s Finkster '07

  12. c. Find the average acceleration of the ball during its contact with the bat if the average force is -1.4 x 104 N and m = 0.144 kg. F = -1.4 x 104 N F = ma m = 0.144 kg a = F/m a = ? = -1.4 x 104N 0.144 kg a = -9.7 x 104 m/s2 Finkster '07

  13. 2. A 54 N·s impulse if given to a 6.0 kg object. What is the change of momentum for the object? J = 54 N·s p = F∆t m = 6.0 kg p = J p = ? P = 54 N·s Finkster '07

  14. 3. Which quantities do not always occur in equal and opposite pairs when an interaction takes place within a system? • Impulses • Acceleration • Forces • Momenta changes Finkster '07

  15. Answer to 3 • accelerations Finkster '07

  16. Conservation of Momentum Sum of Momenta before = Sum of Momenta after

  17. Any gain in momentum by an object occurs only by the lost of a corresponding amount of momentum by another object. Finkster '07

  18. In a system consisting of objects upon which no external force is acting, the momentum of the system is conserved. Finkster '07

  19. Two objects of equal mass approach each other head-on with the same speed. The total momentum of the system before the collision takes place equal zero. Finkster '07

  20. Scenario 1 Before collision After Collision -----> at rest ----> ------> m1v1 + m2v2 = m1v1’ + m2v2’ p1 + p2 = p1’ + p2’ Finkster '07

  21. Scenario 2 Before collision After Collision -----> < ---- <---- ------> m1v1 - m2v2 = - m1v1’ + m2v2’ p1 - p2 = -p1’ + p2’ Finkster '07

  22. Scenario 3 Before collision After Collision at rest <---- <----- < ---- m1v1 + m2v2 = m1v1’ + m2v2’ p1 + p2 = - p1’ + - p2’ Finkster '07

  23. Scenario 4 Before collision After Collision -----> at rest --- >------> m1v1 + m2v2 = (m1 + m2) v’ p1 + p2 = p’ Finkster '07

  24. Examples

  25. 1. Object A has a momentum of 60 N·s. Object B, which has the same mass, is standing motionless. Object A strikes object B and stops. What is the velocity of object B after the collision if the mass of object B is 6 kg? Finkster '07

  26. Before collision After Collision -----> at rest stops ? m1v1 + m2v2 = m1v1’ + m2v2’ p1 + p2 = p1’ + p2’ 60 N·s + 0 N·s = 0 N·s + (6.0 kg)v2’ v2’ = 10 m/s, to the right Finkster '07

  27. 2. Ball A of mass 0.355 kg moves along a frictionless surface with a velocity of +0.095 m/s. It collides with ball B of mass 0.710 kg moving in the same direction at a speed of +0.045 m/s. After the collision, ball A continues in the same direction with a velocity of +0.035 m/s. What is the velocity and direction of ball B after the collision? Finkster '07

  28. 2. Before collision After Collision -----> ----- > ------ > ? m1v1 + m2v2 = m1v1’ + m2v2’ (0.355 kg)(0.095 m/s) + (0.710 kg)(0.045 m/s) = (0.355 kg)(0.035 m/s) + (0.710 kg)V2’ v2’ = 0.075 m/s, to the right Finkster '07

  29. 3. A 0.105 kg hockey puck moving at +48 m/s is caught by a 75 kg goalie at rest. With what speed does the goalie slide on the ice? Finkster '07

  30. 3. Before collision After Collision -----> at rest ? m1v1 + m2v2 = (m1 + m2)v’ (0.105 kg)(48 m/s) + 0 N·s = (0.105 kg + 75 kg)v’ v’ = 0.067 m/s Finkster '07

  31. 4. A 0.50 kg ball traveling at +6.0 m/s collides head on with a 1.00 kg ball moving in the opposite direction at a velocity of -12.0 m/s. The 0.50 kg ball moves away at -14 m/s in the opposite direction after the collision. Find v2’. Finkster '07

  32. 4. Before collision After Collision -----> < ---- < ---- ? m1v1 + m2v2 = m1v1’ + m2v2’ (0.50 kg)(+6.0 m/s) + (1.00 kg)(-12.0 m/s) = (0.50 kg)(-14 m/s) + (1.00 kg)v2’ v2’ = -2.0 m/s Finkster '07

  33. Momentum at an Angle Right Angle Other Angles

  34. 1. A 1325 kg car is moving due east at 27 m/s. It collides with a 2165 kg car moving due north at 17.0 m/s. • Calculate the momentum of each car before the collision. • The two cars stick together and move off at an angle. Calculate the resultant momentum, speed and angle of the cars. Finkster '07

  35. a. PN = mv =(1325 kg)(27 m/s) PN =3.6 x 104 N·s PE= mv = (2165 kg)(17.0 m/s) PE = 3.7 x 104 N·s Finkster '07

  36. b. p’ 2 = (3.6 x 104 N·s)(3.7 x 104 N·s) p’ = 5.2 x 104 N·s v’ = p’ / mt = 5.2 x 104 N·s / (1325 kg + 2165 kg) v’ = 14.9 m/s tan Ө = PN / PE = 3.6 x 104N·s / 3.7 x 104 N·s Ө = 44o N of E Finkster '07

  37. 2. A 6.0 kg ball (A) is moving due east at 3.0 m/s. It collides with another 6.0 kg ball (B) that is at rest. Ball A moves off at an angle of 40.o north of east. Ball B moves off at an angle of 50.o south of east. • Calculate the momentum of ball A before the collision. • Calculate the x-component of ball A. • Calculate the y-component of ball B. • Calculate the velocity of ball A. • Calculate the velocity of ball B. Finkster '07

  38. PA = mv = (6.0 kg)(3.0 m/s) PA = 18 kg·m/s b) PA’ = PA cos 40.o = 18 kg·m/s (cos 40.o) PA’ = 14 kg·m/s Finkster '07

  39. PB’ = PA sin 40.o = 18 kg·m/s (sin 40.o) PB’ = 12 kg·m/s • VA’ = PA’ / mA = 14 kg·m/s / 6.0 kg vA = 2.3 m/s Finkster '07

  40. e) vb’ = PB / mB = 12 kg·m/s / 6.0 kg vb’ = 2.0 m/s Finkster '07

  41. 3. Two objects of masses M1 = 1 kg and M2= 4 kg are free to slide on a horizontal frictionless surface. M1 is moving due east at 16 m/s. M2 is at rest. The objects collide and the magnitudes and directions of the velocities of the two objects before and after the collision are M2 moves 37o N of E at 5 m/s; M1 movess 90o due S at 12 m/s. Finkster '07

  42. Calculate the x and y components of the momenta of the balls before and after the collision. • Show, using the calculations that momentum is conserved. • Calculate the kinetic energy of the two-object system before and after the collision. • Is kinetic energy conserved in the collision? Finkster '07

  43. Before Collision Px = M1v1 Py = M1v1 = (1 kg)(16 m/s) = (1 kg)(0 m/s) Px = 16 kg·m/s py = 0 kg·m/s Px = M2v2 py = M2v2 =(4 kg)(0 m/s) = (4 kg)(0 m/s) Px = 0 kg·m/s py = 0 kg·m/s Finkster '07

  44. After Collision P’1x = M1v1cos 900 p’1y = M1v1 sin 90o =(1kg)(12m/s)(0) =(1kg)(12 m/s)(1) P’1x = 0 kg·m/s p’1y = 12 kg·m/s P’2x = M2v2 cos 37o P’2x = - M2v2 sin 37o = (4 kg)(5 m/s)(0.80) = (4 kg)(5 m/s)(0.60) P’2x = 16 kg·m/s P’2x = - 12 kg·m/s Finkster '07

  45. b. Sum of Px before: (16 + 0) kg·m/s = 16 kg·m/s Sum of Py before: (0 + 0) kg·m/s = 0 kg·m/s Total = 16 kg·m/s Sum of Px after = (0 + 16 kg·m/s) = 16 kg·m/s Sum of Py after = [(+12) + (-12)] = 0 kg·m/s Total = 16 kg·m/s Px & py are conserved. Finkster '07

  46. c. K.E. = ½ mv2 Before: K.E. = ½ (1 kg)(16 m/s)2 = 128 J After: K.E.= ½ [(1 kg)(12 m/s)2 + (4 kg)(5m/s)2 K.E. = 122 J d. K.E. isn’t conserved. Finkster '07

  47. 4. A fireworks rocket is moving at a speed of 50.0 m/s. The rocket suddenly breaks into two pieces of equal mass, and they fly off with velocities v1 and v2. v1 is acting at an angle of 30.0o with the original direction and v2 is acting at an angle of 60.0o with the original direction. What are the magnitudes of v1 and v2? Finkster '07

  48. X - components 2mv1 = m1v1’+ m2v2’ 2mv1 = m1v1’cos 30.0o + m1v1’cos 60.0o 2(50.0 m/s) = 0.866v1’ + 0.500 v2’ v1 = 1.73 v2 Y-components 2mv1 = m1v1’sin 30.0o + m1v1’sin 60.0o 0 = 0.500 v1 + 0.866v2 =(0.500)(1.73 v2) + 0.866 v2 V2 = 50.0 m/s V1 = (1.73)(50.0 m/s) V1 = 86.5 m/s Finkster '07

  49. Work

  50. Work is done only when a force moves an object. Finkster '07

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