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Momentum and changing momentum

Momentum and changing momentum. 1 In a car collision, which causes greater damage?. A A massive car. B A fast car. C A massive and fast-moving car. 2 A car collides with a fixed wall. How can the damage be reduced?. A It collides at high speed.

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Momentum and changing momentum

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  1. Momentum and changing momentum

  2. 1 In a car collision, which causes greater damage? A A massive car. B A fast car. C A massive and fast-moving car.

  3. 2A car collides with a fixed wall. How can the damage be reduced? A It collides at high speed. BIt has a softer bumper at the front so that it stops in a longer time. CIt has a very hard bumper so that it stops in a very short time.

  4. 1 What is momentum? Experience tells us that: a heavy,fast car causes greater damage than a heavy,slow car or a light,fast car. Both mass and velocity are important physical quantities for studying moving objects.

  5. 1 What is momentum? Physicists have defined aquantity called momentum(動量) of amoving object as... Momentum = mass velocity  = mv Unit: kg m s1 A vector quantity

  6. Example 1 Calculating momentum Calculate the momentum of the following objects in kg m s–1. (a)A 20 g bullet moving at 400 m s–1. Momentum of the bullet = 0.02400 = 8 kg m s–1 (towards the right)

  7. 90 = 0.05 3.6 Example 1 (b)A 0.05 kg tennis ball moving at 90 km h–1. Momentum of the tennis ball 0.05 kg 90 km h–1 = 1.25 kg m s –1 (towards theleft) +ve

  8. vu = m ( ) t mvmu = t 2 Momentum and force is also related to theforce on an object. Newton's 2nd law can be expressed in terms of thechange in momentum. F=ma initial  change in momentum final 

  9. mvmu = t 2 Momentum and force change in momentum Force = time  Newton’s 2nd Law can be restated as: The net force acting on an object is equal to the rate of change of momentum of the object.

  10. Q1 Two rams fight... Two rams fight. If ram A wins, Which of the following must be true? A mA>mB BA>B CvA>vB A B mB mA vA vB

  11. Q2 Vincent & Esther... Vincent & Esther are skating.Who is correct? AVincent. BEsther. CNeither. We have the same  as the products of our m & v are equal. 2.0 m s–1 2.5 m s–1 I don't think so, our directions are different. 40 kg 50 kg Esther Vincent

  12. Q3 A 0.4-kg football... A 0.4-kg football is moving at 20 m s–1. What is themomentum of the football? Momentum of the football = mv = __________________ =___________ 0.4 20 8 kg m s–1

  13. = v 2gs = v 2  10  1.8 Q4 A stone of mass 1.5 kg... A stone of mass 1.5 kg is dropped from a height of1.8 m. (a)What is the change in momentum of the stonewhen it reaches the ground? v2u2 = 2as = 6 m s–1 Momentum change of the stone = mv= __________= _________ 9 kg m s–1 1.5  6

  14. Q4 A stone of mass 1.5kg... A stone of mass 1.5 kg is dropped from a height of1.8 m. (b)Why does the momentum of the stone change? It is because a ____________ force acts on thestone for a period of _________. gravitational time

  15. 3 Impact When a tennis ball is hit by a racket, a large force acts on theball in a short time. How can westudy the force duringimpact? Simulation

  16. 30 m s1 20 m s1 0.05 kg Example 2 Average force acting on tennis ball What is the average force acting on the ball? time of impact t = 0.005 s +ve

  17. mvmu F = t 0.05  30  0.05  (20) = 0.005 Example 2 Average force acting on tennis ball massm = 0.05 kg u = 20 m s1 v = 30 m s1 t = 0.005 s +ve = 500 N to the left 500 N

  18. Experiment 8a Investigating the impact of force Set up the following apparatus: Start data-logging. Fix a small spring on the force sensor. slightly pushit from rest so it moves down the runway & collides on the force sensor.

  19. Experiment 8a Investigating the impact of force From the v-t graph, find the velocity of the trolley before and after impact. From the F-t graph, find thearea under the graph. Compare this area withthe change in momentum. Video

  20. a Force-time graph of impact v changes when trolley collides on force sensor velocity Result of experiment 8a: time (s) max. F force (N) F F  v. short time interval 1.65 s 1.60 s

  21. a Force-time graph of impact When a tennis ball is being hit bya racket, the ball is deformed. Why does the force vary during impact? The force of impact tomax. when the ball isdeformed most. As the ball regains its shape,the force .

  22. mvmu F = t a Force-time graph of impact Rearrange terms in  Ft=mv  mu impulse:productof force & time duringwhich the force acts Impulse = change in momentum

  23. a Force-time graph of impact force / N This F-t graph is a straight horizontal line. constant force Area under F-t graph = Ft = impulse time / s area = impulse

  24. a Force-time graph of impact force / N curve line force varies a series of narrow rectangular bars time / s area of each bargives theimpulse during the time interval total area gives the total impulse

  25. a Force-time graph of impact force / N force / N time / s time / s Area under F-t graph = impulse = change in momentum

  26. F F t t b Force of impact For the same change in momentum, shorter time of impact force same area same impulse same  in  larger F, smaller t smaller F, larger t

  27. b Force of impact Impact timedepends on the hardnessof colliding objects. Harder object  shorter impact time e.g. • Golf ball's hardness >> tennis ball's shorter impact time on being struck by golf club • Design of car:  collision time in case of accidence

  28. b Force of impact Video Video Video

  29. Example 3 F-t graph of an impact A trolley hits and rebounds from a force sensor and the F-t graph isobtained. The area under graph is 0.46 N s. 87654321 force (N) 1.75 1.80 1.85 1.90 1.95 2.00 time (s)

  30. Example 3 F-t graph of an impact (a) What is the maximum force acting on the sensor during impact? Maximum force = 7.8 N area under the graph = 0.46 N s 87654321 7.8 N force (N) 1.75 1.80 1.85 1.90 1.95 2.00 time (s)

  31. 87654321 1.75 1.80 1.85 1.90 1.95 2.00 Example 3 F-t graph of an impact (b)Find the change in momentum of the trolley. area under the graph = 0.46 N s force (N) obtained by data-logging program time (s) Areaunder the graph= 0.46 N s Change in momentum = 0.46 kg m s–1

  32. 87654321 1.75 1.80 1.85 1.90 1.95 2.00 in momentum 0.46 0.3 time Example 3 F-t graph of an impact (c)Hence find the average force acting during impact. in momentum = 0.46 m s–1 force (N) time of impact = 2.05 – 1.75 = 0.3 s time (s) Average force = = 1.53 N =

  33. Example 4 Crumple zone of a car A car of mass 1500 kg moving at 20 m s1 (72 km h1) collides with a wall head-on and comes to a stop. Calculate the force of impact on the car in the following cases. (a) A car has a crumple zone in the front section and it stops in 0.5 s. (b) A car has a strong bumper in the front section and it stops in 0.02 s.

  34. mvmu F = t 0  1500  20 = 0.5 Example 4 Crumple zone of a car (a) A car has a crumple zone in the front section and it stops in 0.5 s. mass of car = 1500 kg opposite direction to car’s motion +ve = 6  104 N 20 m s1

  35. mvmu F = t 0  1500  20 = 0.02 Example 4 Crumple zone of a car (b) A car has a strong bumper in the front section and it stops in 0.02 s. mass of car = 1500 kg 25 times greater than that in (a)! +ve = 1.5  106 N 20 m s1

  36. Example 5 Force of impact of a falling can (a) Calculate the speed of impact of the can on the ground. (g = 10 m s2) 0.4 kg +ve 30 m time of impact = 5 ms

  37. Example 5 Force of impact of a falling can u = 0 a = g = 10 m s2 v2u2 = 2as v2 = 2  10  30 30 m v= 24.5 m s1 (or 88.2 km h1) v = ? +ve

  38. mvmu F = t (0  0.40  24.5) = 0.005 s Force of impact 1960 = Weight of can 0.4 10 Example 5 Force of impact of a falling can (b) Find the average force of impact on the ground if the rebound speed is negligible. time of impact = 5 ms speed of impact =24.5 m s1 mass of can = 0.4 kg = 1960N (i.e. upwards) +ve = 490 times!

  39. Example 6 Thrust of a rocket A rocket pushes out 100 kg of hot gas each second. The velocity of the ejected hot gas is 500 m s1. Calculate the forward force (thrust) on the rocket. rate = 100 kg s1 v = 500 m s1

  40. mvmu F = (100  500)  0 = t 1 Example 6 Thrust of a rocket Let F be the force on gas = 50 000 N By Newton’s 3rd law, F = thrust u = 0 = 50 000 N v = 500 m s1 rate = 100 kg s1

  41. Q1 Which of the following situations... Which of the following situations does NOT involveimpact force? ATravelling in a lift moving up at a constant speed. BKicking a ball. CA ping pong ball bouncing up from the ground.

  42. F/N 4 3 2 1 1 2 t / s 12 1 2 3 4 5 0 5 Q2 A boy pushes a ball... A boy pushes a ball and the F-t graph is shown below.Find the average force acting on the ball. Impulse = area under F-t graph = ____________ = _______ Average force acting on the ball = _______=_______  (1+5)  4 12 N s 2.4 N

  43. Q3 A car of mass 1000 kg... A car of mass 1000 kg accelerates from rest to 8 m s–1in 4 s. (a)What is the change inmomentum of the car? Change in momentum = mv mu 1000  (8  0) 8000 kg m s–1 = _____________ = _______________

  44. Q3 A car of mass 1000 kg... A car of mass 1000 kg accelerates from rest to 8 m s–1in 4 s. (b)What is the average force acting on the 70-kgdriver in the car? Change in momentum of the driver = ___________ = _______________ 560 kg m s–1 70  8 Average force on the driver = = ___________ = ____________ change in momentum time taken 560/4 140 N

  45. Q4 Using info. in e.g. 5... Using information in e.g. 5. Find theaverage force of impact on the ground if the canrebounds with 10% of the impact speed. mass of can = 0.4 kg time of impact = 0.005 s impact force (without rebound) = 1960 N +ve Rebound speed v = 10%×_____ = _____ m s–1 24.5 2.45 (downwards as +ve) u = 24.5 m s1

  46. Q4 Using info. in e.g. 5... mass of can = 0.4 kg time of impact = 0.005 s impact force (without rebound) = 1960 N Average force on thecan = = = _________ +ve mv mu t (2.45  24.5) 0.40  v = 2.45 m s1 0.005 2156 N u = 24.5 m s1

  47. Q4 Using info. in e.g. 5... mass of can = 0.4 kg time of impact = 0.005 s impact force (without rebound) = 1960 N impact force (with rebound) = 2160 N The average force of impact on the ground is______________pointing __________________(upwards/downwards). The force is ___________________(greater/smaller) than the case if the can does notrebound. 2156 N +ve downwards greater v = 2.45 m s1 u = 24.5 m s1

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