1 / 29

4.1a Further Mechanics Momentum concepts

4.1a Further Mechanics Momentum concepts. Breithaupt pages 4 to 17. September 3 rd , 2010. AQA A2 Specification. Momentum, p. momentum = mass x velocity p = mv m in kilograms (kg) v in metres per second (ms -1 ) p in kilograms metres per second (kg ms -1 )

kendall
Download Presentation

4.1a Further Mechanics Momentum concepts

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. 4.1a Further MechanicsMomentum concepts Breithaupt pages 4 to 17 September 3rd, 2010

  2. AQA A2 Specification

  3. Momentum, p momentum = mass x velocity p = mv m in kilograms (kg) v in metres per second (ms-1) p in kilograms metres per second (kg ms-1) Momentum is a VECTOR quantity – direction the same as the velocity

  4. Newton’s 2nd law (A2 version) The resultant force acting on an object is proportional to the rate of change of momentum of the object and is in the same direction as the resultant force. ΣFαΔ(p) Δ(t)

  5. Inserting a constant of proportionality ‘k’ ΣF = kΔ(p) Δ(t) but: p = mv hence:ΣF = kΔ(mv) Δ(t) If the mass, m remains constant: ΣF = kmΔ(v) Δ(t)

  6. but: Δ(v) = a(acceleration) Δ(t) hence: ΣF = km a A force of one newton is defined as that required to cause an acceleration of 1 ms -2 with a mass of 1 kg. Inserting these values into: ΣF = km a gives: 1 = k x 1 x 1 and so: k = 1 giving: ΣF = m a(the AS version of Newton’s 2nd law) Note: This simplified version only applies for an object of constant mass.

  7. Force and Momentum Force is equal to the rate of change of momentum. F = Δ(mv) / Δt F in newtons (N) Δ(mv) in kilograms metres per second (kg ms-1) Δt in seconds (s)

  8. A car of mass 800 kg moving at a velocity of 30 ms -1 is brought to rest by a braking force of 1200 N. Calculate: (a) its initial momentum (b) the time taken to stop the car. (a) p = mv = 800 kg x 30 ms-1 momentum = 24 000 kg ms-1 (b) F = Δ(mv) / Δt 1200N = 24 000 kg ms-1 / Δt Δt = 24 000 kg ms-1 / 1200N time = 20 seconds Question 1

  9. A car of mass 750kg travelling at a speed of 4.0ms -1 is struck from behind by another vehicle. The impact lasts for 0.30s and causes the speed of the car to increase to 6.0ms -1 . Calculate: (a) the change in momentum of the car due to the impact. (b) the impact force. (a) Δp = Δmv mass is constant, so: Δp = m Δv = 750 kg x (6.0 – 4.0) ms-1 momentum change = 1 500 kg ms-1 (b) F = Δ(mv) / Δt = 1 500 kg ms-1 / 0.30s force = 5000 N Question 2

  10. Impulse, Δp Impulse is equal to the change of momentum produced by a force over a period of time. Impulse, Δp = FΔt = Δ(mv) Δp is measured in newton seconds (Ns)

  11. Impulse caused by a golf club(Breithaupt page 8)

  12. Force – time graphs(Breithaupt page 6) Impulse is equal to the area under a force-time graph.

  13. Calculation Example(Breithaupt page 9)

  14. F / N 3.0 2 5 t / s Graph Question Calculate the impulse and change in velocity caused to mass of 6kg from the graph opposite. Area = impulse = 3N x (5 - 2 )s impulse = 9 Ns = Δ(mv) = 6kg x Δ(v) therefore, Δ(v) = 9 / 6 velocity change = 1.5 ms-1

  15. Conservation of Linear Momentum The total linear momentum of an isolated system of bodies remains constant An isolated system is one where no external forces (e.g. friction or air resistance) acts on the interacting bodies.

  16. Question A trolley of mass 4kg moving at 5ms-1 collides with another initially stationary trolley of mass 3kg. If after the collision the trolleys move off attached together calculate their common final velocity. Initial total linear momentum of the system: = momentum of 4kg trolley + momentum of 3kg trolley = (4kg x 5ms-1) + (3kg x 0ms-1) = 20 kgms-1

  17. Conservation of linear momentum: Final total linear momentum of the system must also = 20 kgms-1 (total mass x final common velocity) = 20 kgms-1 (4kg + 3kg) x v = 20 kgms-1 7v = 20 v = 20 / 7 Final common velocity = 2.86 ms-1

  18. Elastic and inelastic collisions ELASTIC– KINETIC energy is conserved INELASTIC– Some (or all) KINETIC energy is transformed into thermal or other forms of energy. In both types of collision both the total energy and momentum are conserved.

  19. Collision question continued Was the collision in the previous example elastic or inelastic? Kinetic energy = ½ x mass x (speed)2 Total initial KE = KE of 4kg trolley = ½ x 4kg x (5 ms-1)2 = 2 x 25 = 50 J Total final KE = KE of combined 7kg trolley = ½ x 7kg x (2.86 ms-1)2 = 3.5 x 8.18 = 28.6 J Kinetic energy reduced – Collision INELASTIC

  20. Explosions KINETIC energy is increased Both the total energy and momentum are conserved

  21. Question A gun of mass 3kg fires a bullet of mass 15g. If the bullet moves off at a speed of 250ms-1 calculate the recoil speed of the gun. Initial total linear momentum of the system: = momentum of the gun + momentum of the bullet = (3kg x 0ms-1) + (15g x 0ms-1) = 0 kgms-1 Conservation of linear momentum: Final total linear momentum of the system must also = 0 kgms-1

  22. Therefore: (bullet mass x velocity) + (gun mass x velocity) = 0 (0.015kg x 250ms-1) + (3kg x gun velocity) = 0 (3.75) + (3 x gun velocity) = 0 3 x gun velocity = - 3.75 gun velocity = - 3.75 / 3 = - 1.25 ms-1 The MINUS sign indicates that the gun’s velocity is in the opposite direction to that of the bullet Gun recoil speed = 1.25 ms-1

  23. Internet Links • Effect of impulse - NTNU • Collisions along a straight line - NTNU • 1D collision showing momentum and ke - NTNU • 2D collisions - NTNU • 2D Collisions - Explore Science • Two dimensional collisions - Virginia • Elastic & Inelastic Collisions - Fendt • Newton's Cradle - Fendt • Gaussian gun - NTNU • Dropping a load onto a trolley - momentum - netfirms • Ballistic Pendulum - NTNU

  24. Core Notes from Breithaupt pages 4 to 17 • Define what is meant by momentum. State its unit. • Explain how force is related to the rate of change of momentum. • What is meant by ‘impulse’? How can impulse be found graphically? Copy Figure 3 on page 6. • State the principle of conservation of momentum. • Redo the worked example on page 13 this time with the first rail wagon moving at 4ms-1 colliding with another now of mass 2000kg. • Define what is meant by (a) an elastic and (b) an inelastic collision. • Redo the worked example on page 15 this time with the first rail wagon moving at 4ms-1 colliding with another now of mass 6000kg. • Explain how the principle of conservation of momentum applies in an explosion. • Redo summary question 1 on page 17 this time with a shell of mass 3.0kg being fired from a gun of mass 1000kg.

  25. Notes from Breithaupt pages 4 to 7Force & Momentum • Define what is meant by momentum. State its unit. • Explain how force is related to the rate of change of momentum. • What is meant by ‘impulse’? How can impulse be found graphically? Copy Figure 3 on page 6. • Show how the version of Newton’s 2nd law of motion on page 5 can be used to derive the equation: F = ma • Redo the worked example on page 7 this time with a force of 20N on a mass of 200kg. • Try the summary questions on page 7

  26. Notes from Breithaupt pages 8 to 10Impact Forces • Redo the worked example on page 8 this time with a velocity increase of 25ms-1 over a time of 20ms. • Explain how the relationship between force and momentum change is relevant to vehicle safety. • Explain why a greater force is needed to send a ball back along its initial path than to deflect it at an angle. • Redo the worked example on page 10 this time with a ball of mass 0.30kg moving at an initial speed of 15ms-1. • Try the summary questions on page 10

  27. Notes from Breithaupt pages 11 to 13Conservation of momentum • State the principle of conservation of momentum. • Redo the worked example on page 13 this time with the first rail wagon moving at 4ms-1 colliding with another now of mass 2000kg. • Show how the version of Newton’s 3rd law of motion on page 11 can be used to derive the principle of conservation of momentum. • Explain how the principle of conservation of momentum can be verified experimentally. • Explain how the principle of conservation of momentum applies in a head-on collision. • Try the summary questions on page 13

  28. Notes from Breithaupt pages 14 & 15Elastic and inelastic collisions • Define what is meant by (a) an elastic and (b) an inelastic collision. • Redo the worked example on page 15 this time with the first rail wagon moving at 4ms-1 colliding with another now of mass 6000kg. • What is a totally inelastic collision? • Explain how conservation of energy can still apply to an inelastic collision. What else is conserved in this type of collision? • Try the summary questions on page 15

  29. Notes from Breithaupt pages 16 & 17Explosions • Explain how the principle of conservation of momentum applies in an explosion. • Redo summary question 1 on page 17 this time with a shell of mass 3.0kg being fired from a gun of mass 1000kg. • Explain how the principle of conservation of momentum can be verified experimentally with an explosive interaction. • Try the other summary questions on page 17

More Related