buku physics11

6.1 Introduction Fig.6.1 Change in motionMany people have a vague idea of what momentum is. It is oftenassociated with a change in the flow of support for a candidate in anelection. Or it may be a sports team gaining an advantage by chang-ing the flow of action and pace in a game. The idea of momentumcontains some notion of a change in motion. Newton’s second law from his work, Principia, Lex II, translatesfrom the Latin as: “Change of motion is proportional to the moving force impressed, andtakes place in the direction of the straight line in which that force isimpressed.” Expressed in more modern terms,The momentum of an object is changed when an unbalanced force isapplied over a given period of time. The resulting change occurs in thedirection of the force.The physics definition of momentum is mass times velocity, or p→ ϭ mv→where p→ is the momentum. Notice that momentum is a vector quantity andhas a direction associated with it. The SI units for momentum are kg·m/sand have no special name assigned to them. Perhaps you will become afamous physicist and these units will be named after you.e x a m p l e 1 Calculating momentumCalculate the momentum of a 1.0 kg duck moving at 28 m/s (a new Fig.6.2biotech superperformance duck) and of a sumo wrestler with mass200 kg, moving slowly at 0.5 km/h, as shown in Fig. 6.2.Solution and Connection to TheoryGivenmduck ϭ 1.0 kg, v→duck ϭ 28 m/smsumo ϭ 200 kg, v→sumo ϭ 0.5 km/h ϭ 0.14 m/sp→ ϭ mv→ pduck ϭ 1.0 kg(28 m/s) ϭ 28 kg·m/sand psumo ϭ 200 kg(0.14 m/s) ϭ 28 kg·m/sNotice that the two momenta are similar, even though the objects are not. chapter 6: Momentum 189

Fig.6.3 Mass and velocity When someone asks you whether you would rather be hit by a piece of chalk or tackled by a serious footballdetermine momentum player, at first, the answer seems to be the chalk. But if the chalk were moving at 250 m/s, then you’re basically dealing with a bullet, and the choice becomes obvious. Momentum is an important concept when consid- ering impacts, collisions, and how objects in general interact. It is not just an object’s mass that is important; it is the product of its mass and velocity. gpplyin 1. Calculate the momentum of the following objects:Co a) A 100 kg football player running at 12 km/ha the b) A blue whale of mass 150 tonnes moving at 30 km/h tsncep c) The Saturn V rocket with mass 8.7 ϫ 106 kg and velocity 28 000 km/h 2. How does momentum relate to Newton’s first law?Fig.6.4 A rocket undergoes a large 6.2 Momentum and Newton’s Second Lawchange in momentum From Newton’s second law, F→ ϭ ma→ ϭ ᎏm⌬ᎏv→ . For image ⌬t see student Multiplying both sides by ⌬t, text. F→⌬t ϭ m⌬v→ mv→ is called the momentum, p→, of an object, so m⌬v→ is the change in momentum of an object. F→⌬t, the force multiplied by the time it acts, is called the impulse of the force, and is equal to the change in momentum of the object. The equation for Newton’s second law may be written F→ ϭ ma→ or F→⌬t ϭ ⌬(mv→)190 u n i t a : M ot i o n a n d Fo rc es

The second law equation The case of ⌬m 0 In most momentum problems, we F→⌬t ϭ ⌬p→ expect the mass to be constant. Rockets are an important exception.is a statement illustrating cause and effect. An object just sitting around, Most of the mass of a rocket standingminding its own business (Newton’s first law), has a force applied to it for on its launch pad comprises fuel. Asa period of time, ⌬t. We will call this part the “cause.” Now the object has the rocket lifts off (Fig. 6.4), themoved and its velocity has changed. Because velocity changes, so does burning fuel provides the thrust tomomentum. This part is the “effect.” In general, it is easier to see the move the rocket (following Newton’s“effect” side of an action rather than the “cause” side. The left side of the third law). While the rate of burningequation is called impulse (units N·s and symbol →J ), and the right side is (determining the thrust) does notthe change in momentum (units kg·m/s). change, the mass being pushed is decreasing because fuel is being used up. With a decreasing mass and constant thrust, you get an increas- ing acceleration.Fig.6.5 Cause and Effect from Newton’s Second LawNewton’s Second Law connecti ts F៬⌬t ϭ m⌬៬v the ng ncepCoCause EffectImpulse Change in momentum ៬J ⌬p៬ϭN • s kg • m/s Fig 6.6 Impulse from a club changes the momentum of a golf ball Note that since the vectors → ⌬p→are ϭ in the same direction, J we can perform calculations using their magnitudes only. However, you should always be conscious of the effect direction may have on the solution to a problem. chapter 6: Momentum 191

e x a m p l e 2 Calculating impulse and change in momentum A golfer hits a ball of mass 45 g, causing it to leave the tee at 40 m/s. Find the impulse imparted by the club to the ball and the change in momentum of the ball. Solution and Connection to Theory Given v→1 ϭ 0 m/s v→2 ϭ 40 m/s m ϭ 45 g ϭ 0.045 kg We know that impulse is F→⌬t, but neither F→ nor t is given. However, we also know that F→⌬t ϭ ⌬p→, where ⌬p→ ϭ m(v→2 Ϫ v→1), which we can calculate. ⌬p→ = 0.045 kg(40 m/s Ϫ 0 m/s) ϭ 1.8 kg·m/s Therefore, the impulse is also 1.8 N·s. Notice that the sign of the velocity and the sign of the impulse are the same. The direction of the impulse was acting in the direction of the golf swing. In sports involving something hitting a projectile, we watch the effect part of the equation (change in momentum). If we were to watch the impulse part, our eyes would need to see and interpret events much faster than they can. Times in the tens of milliseconds would be involved. For image For image For imagesee student see student see student text. text. text. Fig 6.7 Impulse in sports Impulse192 The role impulse plays in sports is often subtle. When a goalie catches a puck or an outfielder catches a ball in their gloves, the gloves do more than protect the hand and provide a larger catching surface. The glove is made of a soft material with a webbing stitched into the catching part of the glove. When the puck or ball hits the glove, the puck or ball is cushioned and brought to rest. Why not use a piece of wood instead? It would seem to protect your hand much better. Let’s apply the impulse equation to this situation. unit a: Motion and Forces

Fig.6.8 Calculating Impulse and Momentum m etho s of Kinematics do ces pr m F៬⌬tGivens v៬1 F៬ possible YES Also know v៬2 to ⌬p៬ calculate? ⌬t NO Calculate ⌬p៬ Also know F៬⌬t F→⌬t ϭ ⌬p→ Fig.6.9 How well would a woodenIn both cases, the ball is brought to rest at some point. Assuming that the glove work?ball was travelling at the same initial velocity each time, the change inmomentum is the same for both cases. However, in the case of the woodenglove, the time of interaction between ball and glove is much shorter thanfor the normal glove. This means that the force imparted to the ball caughtwith the wooden glove is greater. This in turn gives the ball a greater recoilvelocity, which creates the tendency for it to fly out of the glove. For the normal glove, the time of interaction between ball and glove islonger. Therefore, the average force imparted to the ball is smaller. Thismeans that the ball’s rebounding ability is decreased, allowing the catchermore time to close the glove. The more time you have in contact with the ball,the more control you gain over it. To help with the complete stopping of apuck, the goalie moves his or her hand back, in the direction of motion of thepuck as it hits the glove. This increases the time of interaction with puck andglove, and ensures that the puck stays in the glove. The reflexes of the playerare now fast enough to close the glove before the puck bounces out of it. Fig.6.10 Moving the glove back increases the time of interaction F៬ F៬ v៬1fϭ0៬v10 chapter 6: Momentum 193

e x a m p l e 3 Force of impact on landing for a parachutistFig.6.11 Increasing impulse time A parachutist lands, flexes her knees, and rolls in order to stop. Calculate the impact force on a 70 kg parachutist falling at 10 m/s if the time to stopdecreases force was 0.8 s. Compare the force of impact if a soldier lands in a rigid position, standing at attention. The time of stopping is 0.05 s.Fig.6.12 A short interaction Solution and Connection to Theorytime means a large force Assume the standard reference system. Given m ϭ 70 kg v→1 ϭ 10 m/s v→2 ϭ 0 m/s ⌬t ϭ 0.8 s for first case and ⌬t ϭ 0.05 s for the second case. F→⌬t ϭ ⌬p→ → ϭ ᎏ⌬⌬ᎏp→t ϭ ᎏ70 kg (0ᎏm/s Ϫ 1ᎏ0 m/s) ϭ Ϫ875 N 0.8 s F For the stiff-legged case, F→ ϭ ᎏ⌬ᎏp→ ϭ ᎏ70 kg (0ᎏm/s Ϫ 1ᎏ0 m/s) ϭ Ϫ14 000 N ⌬t 0.05 s The negative sign indicates that the stopping action is in the opposite direction to the initial velocity. Notice also, that for the small change of 0.75 s, the force has increased by 13 125 N, enough to cause serious injury! In Example 3, where the soldier did not roll and bend his knees, the stopping force required to produce the necessary change in momentum was disastrously high. However, there are many cases, like the actions in Fig. 6.12, where keeping the time of interaction short and the force high is desirable. Fig.6.13 A longer interaction time means a smaller force194 u n i t a : M ot i o n a n d Fo rc es

These events require a violent change in momentum. In events where con- Fig.6.14A Area under the F→-t graphForce ៬F (N)trol is important, like in Fig. 6.13, you increase the time of interaction toyield a smaller force. is equal to the impulse In most cases, the force acting on an object during the collision process 30is a complicated one. It increases as the colliding objects are compressed 25together and decreases as the objects move away from each other. All this 20may be on a microscopic level (for hard objects). We will use average force 15when dealing with such interactions. We can still get a value for the 10impulse if we have a graph of force vs. time available to us. Analysis of thegraph in Fig. 6.14 shows that the area under the graph produces the desired 5units of impulse, N·s. Compression Rebound To find impulse using F→-t graphs, calculate the area under the graph. 0 0.2 0.4 0.6 0.8 1.0 1.2 Time t (s) Fig.6.14BI M P U L S E F RO M A N F→- t G RA P H For image see studentThe graph in Fig. 6.14 shows the impulse involved in a collision. Imagine it to be a ball beingkicked. During the first 0.6 s, the ball compressed as it was kicked. The force of com- text.pression increased from 0 to 25 N. After the ball reached its maximum compression, itbegan to rebound. During the next 0.6 s, the force diminished to zero. The graph showsthe whole interaction over 1.2 s. →The area under the F-t graph is ᎏ21ᎏ(25 N)(1.2 s) ϭ 15 N·s The impulse between the ball and the foot is 15 N·s.Fig.6.15 Solving Problems Using Momentum g uttin it all Object obeying To geth Newton’s first law p erA force is applied Givens and kinematics Creates impulse Solve ៬F⌬t ϭ⌬p៬ ៬F⌬tCauses change in ៬v ⌬p៬ chapter 6: Momentum 195

g pplyin 1. Impulse causes a change in momentum. Explain this change inCo the terms of Newton’s second law.a ncep ts 2. In sports, follow-through is the continuation of the motion (e.g., con-Fig.6.16 tinued swing) after contact with the ball has been made. It is an important factor in performance. Use the fact that impulse involves For image ⌬t to explain why follow-through is important. Which sports are see student most affected by this concept? text. 3. Use the graph in Fig. 6.14 to calculate impulse at 0.4 s and 1.0 s. 4. Find the impulse for the following situations. Each case involves a196 20 kg object. a) The object changes its velocity from 0 to 3.0 m/s. b) The object slows down from 3.0 m/s to 0 m/s. c) The object hits a surface at 3.0 m/s and rebounds with the same speed. Rockets When a rocket blasts off, the burning of fuel causes gases to be expelled from it, so the rocket is pushed upward, away from Earth. The total momentum of the system is zero because the momentum of the gases moving down and the momentum of the rocket moving up add up to zero. As fuel is used up, the mass of the rocket decreases, causing its velocity to increase. But the velocity of the ejected gas with respect to the rocket remains constant. The impulse force driving the rocket upward is called thrust. 5. a) From Newton’s second law, F→ ϭ ᎏ⌬⌬pᎏ→t . Derive the formula for thrust: ΂ ΃→ ᎏ⌬⌬mᎏt Fthrustϭ v→gas , where v→gas is the speed of the gas relative to the rocket and ᎏ⌬⌬mᎏt is the rate at which the rocket burns fuel. b) A rocket of mass 780 kg wishes to attain a speed of 1000 m/s from rest. If the rate at which the rocket consumes fuel is 2.0 kg/s and the speed of the exhaust gas relative to the rocket is 2500 m/s, find the acceleration of the rocket and the time the rocket takes to reach a speed of 1000 m/s. 6.3 Conservation of Momentum Most of us have heard the expression “energy cannot be created or destroyed.” This is called a conservation law. Translated, it means that in a closed system, no matter what the event, the total energy before the event will equal the total energy after the event. It implies that energy can change its form or be distributed, but its total quantity remains the same. This con- cept will be explained further in the energy unit. unit a: Motion and Forces

The conservation of linear momentum is very similar. The totalmomentum of a system before a collision is the same as after a collision.It also implies that the object’s motions will change and the momentumwill be distributed, but the sum total of momentum for the system remainsthe same. In mathematical form, the conservation of momentum is p ϭ p→ → totalinitial totalfinal or p→t0 ϭ p→tf For clarity in calculations involving two or more objects in a collisionprocess, we use “0” to denote “initial” and the “f” to denote “final.” We usethe numbers 1 and 2 to denote the colliding objects. When two objectscollide, the values of their momenta can be related by the equation m1v→10 ϩ m2v→20 ϭ m1v→1f ϩ m2v→2fFig.6.17 Momentum is Conserved in All Situations Collision m etho s of do ces pr 197Bounce off Move in same One object Objectseach other direction stops join together Find initial values of momentum, p៬0 ϭ mv៬0 p៬T0 ϭ ៬p10 ϩ p៬20 Find final values of momentum ៬pf ϭ m៬vf ៬pTf ϭ p៬1f ϩ ៬p2f Solve for unknown in ៬pTf ϭ p៬T0e x a m p l e 4 Calculating the momentum of objects initially at rest (recoil situations) Consider two hockey players, both at rest on the ice. The 90 kg Montreal Canadiens player pushes the 105 kg Ottawa Senators’ player. Find the velocity of the Senator if the Canadien moves back at 10 km/h after the push. chapter 6: Momentum

Fig.6.18 Solution and Connection to Theory For image Given v→1ov→ϭ2f 0 m/s v→2o ϭ 0 m/s see student m1 ϭ 90 kg m2 ϭ 105 kg ϭ? v→1f ϭ Ϫ10 km/h ϭ Ϫ2.8 m/s text. We are using the standard reference system and are assuming that theFig.6.19 Canadien went to the left after the push. For image Conservation of momentum gives see student m1v→10 ϩ m2v→20 ϭ m1v→1f ϩ m2v→2f text. 90 kg(0 m/s) ϩ 105 kg(0 m/s) ϭ 90 kg(Ϫ2.8 m/s) ϩ 105 kg(v→2f)198 Now solve for v→2f : (0 kg и m/s) ϭ Ϫ252 kg и m/s ϩ 105 kg(v→2f) v→2f ϭ ᎏ25210k5gᎏkи gm/s ϭ 2.4 m/s The Senator moves to the right at 2.4 m/s. Notice how the equation pro- duced the expected direction of the Senator. In Example 4, both the initial and final total momenta are zero. Checking the final momentum, you get p→t0 ϭ p→tf ϭ 90 kg(Ϫ2.8 m/s) ϩ 105 kg(2.4 m/s) ϭ Ϫ252 kg·m/s ϩ 252 kg·m/s ϭ 0 This example shows that you must be careful not to automatically assume there is no motion when the total momentum is zero. e x a m p l e 5 Calculating the momentum of objects in motion Calculate the initial velocity of the American rugby player if she tackles a stationary Canadian player, causing both of them to move off with a velocity of 5 km/h. Both rugby players have a mass of 75 kg. Solution and Connection to Theory Given v→v→120f ϭ ? v→20 ϭ 0 m/s m1 ϭ 75 kg m2 ϭ 75 kg ϭ 5 km/h ϭ 1.4 m/s v→1f ϭ 5 km/h ϭ 1.4 m/s m1v→10 ϩ m2v→20 ϭ m1v→1f ϩ m2v→2f 75 kg(v→10) ϩ 0 kg·m/s ϭ 75 kg(1.4 m/s) ϩ 75 kg(1.4 m/s) v→10 ϭ 2.8 m/s Therefore, the American rugby player’s initial velocity is 2.8 m/s. unit a: Motion and Forces

Example 5 shows a collision where two separate bodies join and becomeone body. Car crashes, where the cars do not spring apart, are another exam-ple of this case (see Fig. 6.20). For image For image Fig.6.20 see student see student For image text. text. see student text.e x a m p l e 6 A poor excuse for a pool game, but a valuable insight into solving momentum problemsTwo pool balls are shot towards each other at the same time (Fig. 6.21).They hit and recoil. If the velocity of ball 1 is 2.0 m/s to the right, and thevelocity of ball 2 is 1.2 m/s to the left, find the velocity of ball 2 after thecollision if ball 1 recoils with a velocity of 0.4 m/s.Fig.6.21 0.0 0.1 Time t (s) 0.2 0.3 0.4 Ϫ0.6 Ϫ0.4 Ϫ0.2 0 0.2 0.4 Displacement d៬1(m)Solution and Connection to TheoryGivenFor this example, the masses were not given. However, we can assumethat the masses are equal given the nature of the game. chapter 6: Momentum 199

m1 ϭ m2 ϭ m v→10 ϭ 2.0 m/s v→20 ϭ Ϫ0.4 m/s v→1f ϭ 0 m/s v→2f ϭ ? m(2.0 m/s) m1v→10 ϩ m2v→20 ϭ m1v→1f ϩ m2v→2f ϩ mv→2f ϩ m(–1.2 m/s) ϭ m(Ϫ0.4 m/s) Since the ms are common, they cancel. v→2f ϭ 1.2 m/s The velocity of the second ball after the collision is 1.2 m/s [right]. Another way to solve a problem involving equal masses is to assign your own value for the masses. A useful choice is 1.0 kg, thereby eliminating an extra calculation. The conservation of momentum is applicable to all interactions among objects, and is one of the most important laws in the universe. It deals with all possible situations and any number of participants. For example, if two stars collide, exploding into a million pieces, the total momentum of the system before the collision would still equal the total momentum after the collision. Fig.6.22 Collision Overview Collision utting it all gethTop Kinematics Givens Dynamics pplyiner the ncep Newton’s Conservation Second of200 Law momentum ៬F⌬t ϭ m⌬៬v p៬T0ϭ ៬pTf g 1. For the following cases, assume that object 1 has a mass of 1.5 kgCo and object 2 has a mass of 2.0 kg. Use the conservation of momentuma to find the unknown velocity. ts a) m1 moves toward m2 with a velocity of 3.0 m/s, m2 is at rest, and m1 bounces back at 1.0 m/s. b) m1 moves at 3.0 m/s while m2 moves at 1.0 m/s in the same direction, then moves off at 2.0 m/s. c) m1 moves at 3.0 m/s while m2 moves at 1.0 m/s in the opposite direc- tion. After collision, m1 moves at 0.5 m/s in its original direction. d) The same as part c) except that both objects stick together after collision. 2. In a car accident, conservation of momentum plays a large role in determining what happens during the collision. What factors would affect the final position of a two-vehicle crash? unit a: Motion and Forces

Rubber BulletsTo enforce the law, police occasionally need to use force, especially duringriots. The trick is to control a violent mob without causing fatal injuries byfiring real bullets. The invention of rubber bullets has provided a solution. Consider the case of two bullets with the same mass, fired with thesame velocity. One is a rubber bullet, the other is a metal one. Whichbullet will knock over a wooden test dummy more easily? The answer is the rubber bullet. Because of its elastic material, whena rubber bullet rebounds off a test dummy, it gains twice the impulse it hadwhen it was fired, and hence twice the momentum. The dummy providesthe necessary force to stop the bullet and then to cause it to fly in the oppo-site direction with an equal velocity. According to Newton’s third law, thebullet applies an equal and opposite force (hence impulse) to the dummy,thereby knocking it over (Fig. 6.23(a)). In contrast, the metal bulletmerely penetrates the dummy and remains embedded in it, so it onlytransfers its original momentum to the dummy. The equal and oppositeforce (impulse) applied only stops the bullet, so the dummy remains stand-ing (Fig. 6.23(b)). Fig.6.23(a) 201 (b) When solving problems in the conservation of linear momentum, it is extremely important to designate a system. Consider the case of a falling ball. As the ball falls towards the ground, its velocity increases. If we consider the ball only, conservation of momentum is not true because the initial and final momenta of the ball are not the same. However, if we include Earth (which causes the ball to fall) in the sys- tem, then the total momentum of the system remains constant. We are now implying that Earth is moving up to meet the ball. Because of Earth’s enormous mass, its motion is not observed. In general, if the net force of the system is zero, then momentum is conserved. chapter 6: Momentum