**1 In a car collision, which causes greater damage?**A A massive car. B A fast car. C A massive and fast-moving car.**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.**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.**1 What is momentum?**Physicists have defined aquantity called momentum(動量) of amoving object as... Momentum = mass velocity = mv Unit: kg m s1 A vector quantity**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 = 0.02400 = 8 kg m s–1 (towards the right)**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**vu**= m ( ) t mvmu = 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 **mvmu**= 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.**Q1 Two rams fight...**Two rams fight. If ram A wins, Which of the following must be true? A mA>mB BA>B CvA>vB A B mB mA vA vB**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**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**=**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? v2u2 = 2as = 6 m s–1 Momentum change of the stone = mv= __________= _________ 9 kg m s–1 1.5 6**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**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**30 m s1**20 m s1 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**mvmu**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 s1 v = 30 m s1 t = 0.005 s +ve = 500 N to the left 500 N**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.**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**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**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 .**mvmu**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**a Force-time graph of impact**force / N This F-t graph is a straight horizontal line. constant force Area under F-t graph = Ft = impulse time / s area = impulse**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**a Force-time graph of impact**force / N force / N time / s time / s Area under F-t graph = impulse = change in momentum**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**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**b Force of impact**Video Video Video**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)**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)**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**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 =**Example 4**Crumple zone of a car A car of mass 1500 kg moving at 20 m s1 (72 km h1) 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.**mvmu**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 s1**mvmu**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 s1**Example 5**Force of impact of a falling can (a) Calculate the speed of impact of the can on the ground. (g = 10 m s2) 0.4 kg +ve 30 m time of impact = 5 ms**Example 5**Force of impact of a falling can u = 0 a = g = 10 m s2 v2u2 = 2as v2 = 2 10 30 30 m v= 24.5 m s1 (or 88.2 km h1) v = ? +ve**mvmu**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 s1 mass of can = 0.4 kg = 1960N (i.e. upwards) +ve = 490 times!**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 s1. Calculate the forward force (thrust) on the rocket. rate = 100 kg s1 v = 500 m s1**mvmu**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 s1 rate = 100 kg s1**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.**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**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 = _____________ = _______________**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**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 s1**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 s1 0.005 2156 N u = 24.5 m s1**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 s1 u = 24.5 m s1