Test, Lesson 4 Energy-Work-Power- Answer Key Page 1

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1 Test, Lesson 4 Energy-Work-Power- Answer Key Page 1 1. What is the axial height for the ond hup on a roller coaster if the roller coaster is traveling at 108 k just before hr clibing the ond hup? The ond hup cannot be higher than 45.9 eters. (A) 39.9 (B) 41.9 (C) 43.9 (D) 45.9 (E) 47.9 This proble asks you to convert pure kinetic energy into potential energy. 2. By accident, a roller coaster car with a ass of 500 kg breaks loose and rolls down an incline fro a height of 49 eters. It strikes another 500 kg roller coaster car sitting at the botto of the incline and, together, they roll along the track. At what speed do they roll along the track? Kinetic energy is ½ v 2 and potential energy is gh. ½ v 2 = gh Mass cancels out, leaving ½ v 2 = gh, and h = v2 2g. (A) 15.5 (B) 17.5 (C) 18.5 (D) 21.5 (E) 23.5 Once 108 k is converted to, v2 = a hr 2g height of 45.9 eters.

2 Test, Lesson 4 Energy-Work-Power- Answer Key Page 2 Does the cobined potential and kinetic energy of the two cars before the collision have to equal the potential and kinetic energy of the two cars after the collision? because this type of collision is an inelastic collision, eaning that soehow energy was lost in this collision in crupling their bupers, or generating heat and noise. Only in elastic collisions, where two objects bounce off each other without losing energy, is the cobined potential and kinetic energy of both objects before ipact equal to the cobined potential and kinetic energy after ipact. Unlike potential and kinetic energy before and after a collision, the cobined oentu of two objects before ipact does equal the cobined oentu of the two objects after ipact, regardless of whether the collision elastic or inelastic. In this case, the total oentu before ipact is the ass of the first car, 1, ties its velocity, v 1 right before ipact, plus the ass of the ond car, 2, ties its velocity, v 2. The total oentu after ipact is 1 plus 2 ties their cobined velocity, v 3. Since v 2 is zero, 1v1 = x v3. In order to figure out v 3, we need to figure out v 1 just before ipact. v 1 results when the roller coaster car s potential energy, fro starting out 49 eters off the ground, is converted into kinetic energy. gh = ½ (v1) 2 Which eans 2gh = v1 2. When the values for the acceleration of gravity and height of 49 eters are plugged in, v 1 turns out to be Plugging this value for v 1 into the equation for total oentu before and after ipact, we get, 500 kg x 31.0 = 1000 kg x v 3 and v 3 = How uch force is needed to stop a 10,000 kilogra truck in 4 onds if the truck is traveling at 108 k hr? (A) 55,000 N (B) 65,000 N (C) 75,000 N (D) 85,000 N (E) 95,000 N

3 Test, Lesson 4 Energy-Work-Power- Answer Key Page 3 In a velocity versus tie graph, the initial speed of the truck is 108 k hr. Multiplying by 1000 an initial velocity of 30. by 1 hr k 3600, we get The truck s ass is 10,000 kilogras. Its deceleration is 30 divided by the tie it takes to stop: 4 onds. The deceleration is Force equals ass ties acceleration, so force equals 10,000 kilogras ties 7.5 kg 2, which is 75,000 2, or 75,000 newtons. 4. What is the inial distance to stop a car traveling at 50 k with a coefficient of hr rubber on concrete of 0.8? (A) 8.3 (B) 9.3 (C) 10.3 (D) 11.3 (E) 12.3 Stopping a oving car takes energy. Energy can be described as force ties distance, or as kinetic energy, or even as potential energy. A oving car is pure kinetic energy, 1 2 v2. To overcoe the car s kinetic energy, we have to exert the exact sae energy but in the opposite direction. That force will be exerted over whatever distance it takes to bring the car to a halt, force ties distance. Since these two opposing energies have to equal each other, 1 2 v2 ust equal force ties distance. For a car traveling 50 k hr, 1 2 v2 is constant. Force and distance, however, can both vary, depending on how uch force is applied to the brakes. If axial force is applied to the brakes so that the car s wheels copletely stop, the distance that the car travels depends on the friction of the car s wheels with the road. The force opposing the car s kinetic energy is the force of friction, which is µ, the coefficient of static friction, ties the weight of the car, which is the car s ass ties the acceleration of gravity.

4 Test, Lesson 4 Energy-Work-Power- Answer Key Page v2 is thus µ ties ass ties g ties distance traveled. ½ (27.8 2) 2 x 0.8 x = a distance of 49.3 eters. Mass cancels out, leaving,½ v 2 equal to µ ties the acceleration of gravity ties distance traveled. The distance traveled equals the velocity squared divided by2µ ties the acceleration of gravity. Since ass is no longer present, it akes no difference whether a heavy truck or a lightweight car is trying to stop. The only thing that atters is the speed of the vehicle and the coefficient of friction. And speed is far ore iportant because v is squared. 5. Shooting a rifle causes the rifle to kick back, or recoil, into your shoulder. What is the velocity of the recoil for a rifle weighing 3.6 kg and shooting a 4 g bullet at 900 /s? In our exaple, we first have to convert 50 k k into. 50 is hr hr Distance = , the coefficient of 2 x 0.8 friction between rubber and concrete, ties d equals The car takes 12.3 eters to coe to a halt. Notice that a car traveling twice as fast, 100 k hr stop., requires four ties the distance to 100 k is 27.8 hr (A) 1.0 (B) 2.0 (C) 3.0 (D) 4.0 (E) 5.0 Before the bullet is shot, nothing is oving, so the total oentu is zero. The bullet s oentu after being shot has to equal its oentu before being shot, naely zero. Therefore, after the trigger is pulled, the bullet s oentu out the barrel has to equal the rifle s recoil oentu so that the two vectors cancel each other out and leave a zero total oentu.

5 Test, Lesson 4 Energy-Work-Power- Answer Key Page 5 The bullet s oentu is kg ties, which equals 3.6 kg ties the rifle s 900 recoil velocity. The rifle recoils with a velocity of 1.0. This is only about 2.2 iles per hour. Moentu is ass ties velocity. Before they were released, the blocks weren t oving, so their total oentu was zero. Their oentu after being released also has to be zero. The ass of block X ties its velocity after release, Vx, plus the ass of block Y ties its velocity after release, Vy, equals zero. That eans that the Mx ties Vx equals negative My ties Vy, and the velocity of block Y after release is therefore inus the ass of block X ties its velocity after release, Vx, divided by the ass of block Y. 6. Here are two blocks connected by a spring. One block has a ass of X and the other, a ass of Y. When pulled apart and then released, the two asses ove toward each other with different velocities. After release, which block has ore kinetic energy? The oentu of each block equals the negative oentu of the other block, because they re oving in opposite directions. Will the kinetic energy of each block also equal the kinetic energy of the other block, or because kinetic energy involves v squared, should we expect the saller block to be oving faster and therefore have ore kinetic energy? Let s see. (A) the saller block (B) the larger block (C) their kinetic energies are equal No energy was added or subtracted after the blocks were released. This eans two things. One is that the total oentu before and after release reains the sae, and two, the total potential and kinetic energy before and after release reains the sae. Iediately before the blocks were released, all the energy was potential and after they were released, all kinetic. So, the total potential energy before release equals the total kinetic energy after release. The total kinetic energy after release is 1 2 v2 for each block.

6 Test, Lesson 4 Energy-Work-Power- Answer Key Page 6 Since, fro the oentu equation, Vy equals negative MxVx over My, total potential energy PE equals ½ MxVx 2 + ½ My ties the square of negative MxVx over My. In other words, the kinetic energy of each block is inversely proportional to its ass, so yes, the saller block ends up having ore kinetic energy than the larger block. Algebraically, the total potential energy, PE, works out to be ½ MxVx 2 ties (1 + Mx/My). ½Mx Vx 2 is the kinetic energy of X. So the total potential energy of both blocks is the kinetic energy y of block X ties 1 plus the ass of X over the ass of Y. By algebra, the total potential energy works out to be the kinetic energy of ass X ties the ass of block Y squared plus the ass of block X ties the ass of block Y, all over the ass of block Y squared. 7. If a 140 g baseball is thrown at 20 and struck by a 900 gra bat for 10, and the ball leaves that bat at 50, how uch force did the bat strike the ball with? By solving this equation for the kinetic energy of block X, we get, the total potential energy of both blocks ties the ass of block Y divided by the total ass of both blocks. Going through the sae calculations for block X, we get the kinetic energy of block Y equaling the potential energy of both blocks ties the ass of block X over the total ass of both blocks. By dividing these two equations, we get the ratio of the kinetic energy of block X over the kinetic energy of block Y. That equals the ratio of the ass block Y over the ass of block X. (A) 880 newtons (B) 980 newtons (C) 1080 newtons (D) 1180 newtons (E) 1280 newtons Force ties distance is the definition of work. What is force ties tie? Force ties tie is ipulse.

7 Test, Lesson 4 Energy-Work-Power- Answer Key Page 7 The ipulse a baseball bat delivers to a baseball is the force of the bat ties the tie that the bat reains in contact with the baseball. An ipulse changes the speed and direction of the baseball. Speed and direction is just another way of saying velocity, so an ipulse changes the velocity of a oving ass. A ass ties its velocity is called oentu, so by changing the velocity of a ass, an ipulse changes a ass oentu. Matheatically, force ties tie, ipulse, equals the change in oentu. The change in oentu is the ass ties velocity after being ipulsed inus the ass ties velocity before being ipulsed. Since the ass doesn t change, F ties t equals ties the change in velocity. kg The force of the baseball bat is 980 k, 2 or 980 newtons. How does that force copare to the force of gravity? If a ass weighed 980 newtons, 980 kg its ass would be, or 100 kg. 2 A baseball bat weighs about a kilogra, so a batter swings with enough force to ake that one kilogra baseball bat appear to weigh 100 kg. Moreover, with a force of 980 newtons, the batter would be able to accelerate a stationary baseball 714 ties the acceleration of gravity. We know this because a force of 980 newtons divided by the ass of 140 gras = , which is 714 ties greater than the acceleration of gravity. The change in the baseball's velocity is the difference between its velocity headed toward hoe plate and its velocity leaving the bat. Since velocity is a vector, the direction of travel is iportant. If travel toward the plate is positive, travel in opposite direction is negative. The change in velocity for the baseball in this proble, then, is 20 inus a -50. Force ties one-one hundredth of a ond equals kg ties 70.

8 Test, Lesson 4 Energy-Work-Power- Answer Key Page 8 8. How uch energy did it take to hit the baseball and send it sailing off at? (A) 147 newton-eters (B) 247 newton-eters (C) 347 newton-eters (D) 447 newton-eters (E) 547 newton-eters The energy that went into hitting the ball is the change in the baseball s kinetic energy. Kinetic energy, like the change in oentu, depends on the baseball s velocity. Kinetic energy is one-half v squared. 9. If it requires 5 newtons of force to stretch a spring 0.5 eter beyond its resting length, how uch work will it take to stretch the spring 1.5 eters beyond its resting length? However, unlike change in oentu, kinetic energy is not a vector, so the direction of ball oveent is irrelevant. The kinetic energy of the ball before being hit is ½ x 0.14 kg x 20 2, or 28 newtoneters. The kinetic energy of the ball after being hit is ½ x 0.14 kg ties 50 2, or 175 newtoneters. This eans the batter supplied , or 147 newton-eters, or joules, of energy. Since energy is force ties distance, 147 newton eters is the equivalent of lifting a 15 kg ass 1 eter. (A) 8.25 newton-eters (B) 9.25 newton-eters (C) newton-eters (D) newton-eters (E) newton-eters The work needed to stretch or copress a spring away fro its resting position is force ties distance. The difficulty calculating the work perfored is that the force increases as the spring is stretched or copressed. The easure of how uch the force changes with distance is called the spring constant, K. The spring constant tells you how uch the force increases per distance stretched or copressed.

9 Test, Lesson 4 Energy-Work-Power- Answer Key Page 9 In this proble it takes 5 newtons of force to stretch the spring 0.5 eters. The spring constant easures the force increase for every 1 unit. Dividing a fraction always results in the denoinator being 1, so 5 newtons over 0.5 eters equals 10 newtons per 1 eter, which is the spring constant. Force over distance, which is the spring constant, allows you to calculate the force at any distance away fro the resting position, because force equals k ties the distance fro the resting position. In a force versus distance graph, the area under the graph line is a easure of the work perfored in stretching or copressing a spring. The slope of the graph line is the spring constant. Since K is 10 newtons, it takes newtoneters or joules to stretch the string eter 1.5 eters. 10. A 1.3 kg block of wood is kicked up a 37 degree incline at an initial velocity of 12, traveling 8.3 eters up the incline before it stops. What was the force of friction acting on the block of wood? Because the force starts out at zero at the resting position, the area under the graph line is ½ the force, which is k ties the distance, ties the distance, or ½ k ties distance squared. This is the work needed to stretch or copress a spring. It s also the potential energy gained by the spring as it s stretched or copressed. In this case, stretching the spring 1.5 eters beyond its resting length requires ½ k x newton-eters of energy. (A) 1.6 newtons (B) 2.6 newtons (C) 3.6 newtons (D) 4.6 newtons (E) 5.6 newtons

10 Test, Lesson 4 Energy-Work-Power- Answer Key Page 10 In a frictionless world, as a block of wood clibs an incline, all of its kinetic energy is converted to potential energy. In that case, kinetic energy goes on one side of the equal sign and potential energy on the other since they equal each other. In a friction world, additional work is needed to overcoe the force of friction as the wooden block slides up the incline. On which side of the equal sign does the force of friction ties distance traveled belong? Since kinetic energy is being used up in order to overcoe the force of friction and push the block of wood up the incline, the force of friction belongs on the sae side of the equation as potential energy, because they are both absorbing the block of wood s kinetic energy. In this case, the block of wood s kinetic energy is 1 one-half v squared, or ties 1.3 kg ties The block of wood starts out with a kinetic energy of 93.6 newton eters, or 93.6 joules. Part of these 93.6 newton eters of energy is converted into potential energy, equal to gh. Matheatically, 93.6 newton eters equals gh plus the force of friction ties 8.3 eters. The height that the block of wood reaches is 8.3 eters ties the sine of 37 degrees. Since the sine of 37 degrees is 0.6, the block of wood ended up 0.6 ties 8.3, or 5.0 eters off the ground. If h equals 5 eters, the force of friction works out to be 3.6 kg- squared, or 3.6 newtons. 11. If this sae block of wood now slides back down the incline, what will its speed be when it reaches the botto of the incline? The other part is used to overcoe the force of friction as the block of wood slides 8.3 (A) 3.2 eters up the incline. The energy needed to do this is the force of friction ties the (B) 4.2 distance traveled: 8.3 eters. (C) 5.2 (D) 6.2 (E) 7.2

11 Test, Lesson 4 Energy-Work-Power- Answer Key Page 11 This tie, as the block of wood slides down the incline, the block of wood s potential energy is giving up its energy to create kinetic energy and to overcoe the force of friction. So now the force of friction ties the distance traveled down the incline belongs on the sae side of the equal sign as the kinetic energy. What s the truth? Was one car speeding, both speeding, or neither speeding? The facts are that the ass of the car headed east was 1100 kg and the ass of the car headed north was 1300 kg. After the cars collide, they coe to a rest 18.7 eters fro the intertion in a 30 degree direction. The coefficient of friction for rubber on asphalt is 0.8. When the values for ass, the acceleration of gravity, height, the force of friction, and distance traveled are inserted into the equation, v 2 = , and v = 7.2, considerably slower than its speed up the incline. (A) The car headed north was speeding. (B) The car headed east was speeding. (C) Both were speeding (D) Neither was speeding. 12. Accident reconstruction is helpful in establishing whether one car was speeding. Two cars approach an intertion with no stoplight, and collide. The car headed north clais that the car headed east was exceeding the posted speed liit of 100 k hr and is therefore responsible for the accident. At the oent the two cars struck each other and began to slide as a single ass, their cobined energy was all kinetic energy: one-half their cobined ass, 1 plus 2, ties their cobined velocity, v 3, squared. The energy that went in to stopping the two vehicles was supplied by the force of friction ties the distance the two vehicles traveled after ipact, which we ll call d.

12 Test, Lesson 4 Energy-Work-Power- Answer Key Page 12 Therefore, the energy exerted by the force of friction ust equal the kinetic energy of the two cars at ipact. The force of friction is the weight of a ass ties the coefficient of friction, in other words, ass ties the acceleration of gravity ties μ. This force ties the distance traveled equals the kinetic energy of the two cars at the oent of ipact, which is 1 of their 2 cobined ass ties their cobined velocity squared. The asses cancel out, and when the values for u, gravity, and distance are inserted, v 3 becoes 17, Even in inelastic collisions, the total oentu before the collision has to equal the total oentu after the collision. The oentu before the collision is the ass of vehicle 1 ties its velocity, v 1, plus the ass of vehicle 2 ties its velocity, v 2. The oentu after the collision is the cobined ass, 1 + 2, ties their cobined velocity, v 3. This diagonal oentu vector is the su of a horizontal oentu vector,( ) v 3 cos 30 o, and a vertical oentu vector, ( ) v 3 sin 30 o. The horizontal oentu vector was provided only by the oentu of the car headed east, 1v1. Therefore, 1v1 can be substituted for x v3 x the cosine of 30 degrees, and 1v1 = (1 + 2) v3 cos 30 o.since the cosine of 30 o is 0.866, v1 works out to be 32 k, or 115 hr. Likewise, the vertical vector 1 plus 2 ties v3 ties the sine of 30 o cae exclusively fro the oentu of the car headed north, 2v2. So 2v2 = x v3 x the sine of 30 o. Since the sine of 30 o is 0.5, v2 works out to be 16 k, or 58. hr The car headed east was indeed exceeding the 100 k speed liit. hr The velocity vector representing the two cars is pointed in a 30 degree direction. The oentu vector for the two cars still points 30 degrees but its length is x v 3.

13 Test, Lesson 4 Energy-Work-Power- Answer Key Page A railroad car weighing 3000 kilogras breaks loose fro its train and sashes at 42.1 k hr into a 2000 kg railcar at rest. The railroad cars becoe attached and travel up a steep slope a distance, h. What is the oentu of the two railroad cars just before ipact, and what is the velocity of the two railroad cars iediately after ipact? 42.1 k hr is 11.7 /, to the total oentu just before ipact is 351,000 kg-/. The oentu of the two cars after the collision is unchanged, because oentu is always conserved after both elastic and inelastic collisions. The total oentu before ipact is 351,000 kg-/. After ipact, the two cars together, with a total ass of 5000 kg, travel at a velocity that akes their oentu, ties v, equal to 351,000 kg-/. Moentu Before Ipact (A) 50,000 kg-/ (B) 110,000 kg-/ (C) 245,000 kg-/ (D) 310,000 kg-/ (E) 351,000 kg-/ Velocity After Ipact ,000 kg-/ equals 5000 kg ties their cobined velocity. Their cobined velocity is 7 /. Moentu is ass ties velocity. The oentu of the two railroad cars before ipact is the ass ties velocity of each car, or 3000 kg ties 42.1 k. The car that s hr stationary has no oentu.

14 Test, Lesson 4 Energy-Work-Power- Answer Key Page What is the kinetic energy of the railroad cars after collision? How high up the slope do the two railroad cars travel? Traveling up the slope of the track converts all of the kinetic energy into potential energy. Kinetic energy, one-half v squared, becoes ass ties the acceleration of gravity ties the height the ass travels. Since ass is on both sides of the equal side, ass cancels out and height equals one-half velocity squared divided by the acceleration of gravity. With a velocity of 7, height works out to be 2.5 eters. Kinetic Energy After Ipact (A) 88,600 joules (B) 102,400 joules (C) 107,700 joules (D) 114,400 joules (E) 122,500 joules Height Attained 1.1 eters 1.5 eters 1.7 eters 2.1 eters 2.5 eters Because this collision is an inelastic collision, only oentu reains unchanged. Kinetic energy changes. The kinetic energy of the two railroad cars after ipact is one-half v squared. After ipact, the oving ass is 5000 kg and its velocity is 7 /. ½ ties 5000 kg ties the square of 7 / equals 122,500 kg- 2 / 2, or 122,500 joules. 15. A pile driver is a achine that slas a heavy weight onto a concrete or steel colun in order to drive the colun into the ground, for exaple, to serve as a support colun for a new building. This pile driver pounds with a 1000 kg weight positioned 5 eters above a concrete colun. When the pile drive is released, it drives the concrete colun 0.5 eters into the ground. What was the force delivered by the pile driver?

15 Test, Lesson 4 Energy-Work-Power- Answer Key Page 15 If 53,900 newton eters of energy was used to exert a force, F, over a distance of 0.5 eters, the pile driver exerted a force equal to the energy expended, 53,900 divided by the distance, 0.5 eters. The force works out to 107,800 kg 2. kg 2 2, kg (A) 38, (B) 138, 900 (C) 10,780 (D) 107,800 (E) 59,700 2 kg 2 kg 2 kg 2 kg 2 The pile driver begins with potential energy, gh. In this case, gh is 1000 kg x ties what? Does the pile driver drop 5 eters or 5.5 eters? 5.5 eters, because the pile driver dropped a total of 5.5 eters. Its potential energy, then, is 1000 kg x 9.8 kg x 5.5 eters, or 53, Does this answer ake sense? Had the pile driver driven the colun only 0.25 eters into the ground, it would have delivered 215,600 newtons of force, twice as uch force. You d think that with ore force, the pile driver would have driven the concrete colun further into the ground, not less. The pile driver can only deliver 53, 900 newton-eters of energy. If the pile driver was only able to drive the concrete colun one-tenth of a eter into the ground, it eans that the ground was very hard and ore force was needed, but the force ties distance delivered by the pile driver was still only 53, 900 newton-eters of energy. This 53,900 newton eters of energy was used to drive the colun 0.5 eters into the ground. Energy is force ties distance.

16 Test, Lesson 4 Energy-Work-Power- Answer Key Page What is the echanical advantage of each? The longer distance for the jack screw is the circuference turned by the handle, 2πl. The shorter path is the pitch, the vertical distance traveled with one coplete turn of the handle. (A) inclined plane: s/h; wedge: s/h; jack screw: l/h (B) inclined plane: s/h; wedge: l/h; jack screw: h/l (C) inclined plane: s/h ; wedge: l/h; jack screw: 2πl/h (D) inclined plane: s/l; wedge: s/h; jack screw: h/2πl (E) inclined plane: l/s; wedge: h/s; jack screw: 2πh/l 17. A coconut falls fro a height of 10 eters. What is its speed 1.5 eters fro the ground? Mechanical advantage is the ratio of greater force over the weaker force, which is also the ratio of the longer path over the shorter path. The only requireent is that force pushing the rock up the incline ust be parallel to the incline. The longer path for the incline is the length of the incline and the shorter one is vertical path of lifting. The longer path for the wedge is the distance it oves, l. (A) 12.9 (B) 13.1 (C) 13.5 (D) 14.2 (E) 15.6

17 Test, Lesson 4 Energy-Work-Power- Answer Key Page 17 One way to do this proble is to graph out the velocity of the coconut in a velocity versus tie graph, as we ve done in earlier probles. Another way is to calculate the total potential and kinetic energy of the coconut before it falls, and then when it s fallen to one and a half eters above the ground. The total energy of the coconut in the tree is all potential energy, because the coconut is not oving and has no kinetic energy. Its potential energy is gh, ass ties the acceleration of gravity ties its height, 10 eters. After the coconut has fallen, its total energy is still the su of its kinetic energy and potential energy, which has to equal the coconut s total energy before it fell. 18. When trying to sink a pool ball, you have to keep in ind where the shooting ball will end up after ipact. It s been said that if there s no English on the shooting ball, the shooting ball will always deflect away fro the ball it strikes at a particular angle. That angle is: ½ v 2 + gh before has to equal ½ v 2 + gh when h is 1.5 after is 1.5 eters. The ass of the coconut doesn t atter because ass cancels out on both sides of the equation. When the acceleration of gravity and the heights above the ground are inserted, the velocity of the coconut a eter and a half above the grounds works out to be 12.9 (A) 30 o (B) 45 o (C) 60 o (D) 75 o (E) 90 o

18 Test, Lesson 4 Energy-Work-Power- Answer Key Page 18 We know that in any collision, elastic or inelastic, the oentu of the two objects before collision equals the oentu of the two objects after collision. The ass ties velocity of the cue ball and the stationary pool ball before collision equals their ass ties velocity after collision. Before After cb v cb + pb v pb = cb v cb + pb v pb Since the velocity of the stationary pool ball before collision is zero, the oentu of the cue ball, alone, equals the oentu of the two pool balls after collision. And since the ass of cue ball is the sae as the pool ball, ass on either side of the equal sign cancels out, and we re left with a atheatical expression that says the velocities of the cue ball and the pool ball after collision add up to the velocity of the cue ball before collision. Before After v cb = v cb + v pb This atheatical equation eans if we add the two post-collision velocity vectors together, they will equal the velocity vector of the cue ball before collision. What we don t know is the angle between the two velocity vectors. That angle is derived fro the fact that the collision between the cue ball and the pool ball is an elastic collision, eaning that not only is the oentu before and after collision the sae, but so is the kinetic energy. 1 2 v2 of the cue ball and the stationary pool ball before collision equals one-half v squared of the cue ball and pool ball after collision. Before After ½ cb v cb + ½ pb v pb = ½ cb v cb + ½ pb v 2 pb Again, because the velocity of the pool ball before collision is zero, and because both balls have the sae ass, this equation siplifies to the velocity of the cue ball before collision, squared, equals its velocity squared after collision plus the velocity of the pool ball squared. Before After 2 v cb = 2 v cb + v 2 pb

19 Test, Lesson 4 Energy-Work-Power- Answer Key Page 19 That s the Pythagorean theore. So not only do the two velocity vectors after collision add up to the length of the cue ball s velocity vector before collision and allow us to ake the three vectors into a triangle, but the triangle they for is a right triangle. If there s no English on the cue ball, and if we ignore friction, the cue ball will always deflect away at a 90 degree angle fro whatever pool ball it strikes. 19. The efficiency of any engine is a easure of how uch electrical or cheical or heat or nuclear energy you put into the engine, and how uch work energy you get out. The energy that goes up the sokestack is wasted energy. If you know the aount of energy you put into the engine and the aount wasted, the energy ouput is siply energy in inus energy wasted. The forula for engine efficiency is work energy produced by the engine divided by the aount of work put into the engine. If a gasoline engine burns joules of energy and joules of energy is lost out the tail pipe, what is its efficiency? (A) 15% (B) 20% (C) 25% (D) 30% (E) 35% Engine efficiency is energy output divided by the energy put into the engine. Energy output is the energy put into the engine inus the energy wasted. If there s no English on the shooting ball, the shooting ball will always deflect away at a 90 degree angle fro the ball it strikes. If this gasoline engine burned joules of energy and joules of energy were lost out the tail pipe, inus 153.3, or 51.2 joules of energy, ust have been captured by the engine to push the pistons down.

20 Test, Lesson 4 Energy-Work-Power- Answer Key Page joules divided by the joules of energy put into the engine, joules, equals 0.25, or 25.0% efficiency. Nuclear power plants are about 33% efficient. Coal burning power plants are a little higher at 36-40% efficient. Solar panels have gotten up as high as 40-42% efficient, but ost are running below 20% efficiency.