UNIT 2 REVIEW. (Pages ) mv = 2 2 mv = p =

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1 UNI REVIEW (Pages 07 Understanding Concepts. One situation is a person carrying a book at a constant height across a level floor. A satellite in circular orbit or pushing on a brick wall are two other exaples.. he work done is transfored into another for of energy. One exaple is pushing a crate across a level floor. he work done is transfored into theral energy through friction.. Moentu and energy can be related as follows: p p E K If both the baseball and the shot have the sae kinetic energy, the shot will have a larger oentu because it has the larger ass.. If both objects have the sae kinetic energy, they both require the sae aount of work done on the to stop the. If the frictional force is the sae and the work done is the sae, the distance the frictional force acts is the sae. herefore, the two toboggans will have the sae stopping distance. 5. During the collision, the initial kinetic energy of the carts is stored as elastic potential energy in the spring. 6. Several coon devices that can store elastic potential energy are an elastic band, a bow for shooting arrows, golf balls, and a bungee cord. 7. Bupers ade of springs are ipractical because the kinetic energy of the collision would be stored in the spring, and then converted back into kinetic energy again. his last stage could project the car into oncoing traffic or concrete barriers and cause further collisions. 8. One possibility is: kg d 0.5 F N h? Copyright 00 Nelson Unit Review 9

2 Using conservation of energy: E E g y F d g( h F d F d ( h g F d h 0.50 g 5 (.5 0 N( (.5 0 kg(9.8 N/kg h he pile driver ust start fro a height of above the pile kg (a d b.00 W? he work done is equal to the area under the graph, therefore: W bh (.00 (0 N W.0 0 J he aount of work done is.0 0 J. (b d b.00 E K? he change in E K will be equal to the work done, therefore: W bh + lw (.00 (0 N + (0 N( E.0 0 J K he block s kinetic energy is.0 0 J, or 0 J. (c d.00 E K 0 J v? v (0 J (0.0 kg v.0 /s [W] he velocity at the.00- ark is.0 /s [W]. 9 Unit Energy and Moentu Copyright 00 Nelson

3 . W E K.5 0 J v 50.0 /s (a? (b d 5.00 F? v (.5 0 J (50.0 /s 0.0 kg W F d W F d.5 0 J 5.00 F.50 0 N [E] A constant force of.50 0 N [E] would give the object the sae final velocity kg E K J p? p p E (5.0 kg(5.0 0 J p 7 kg /s he oentu of the sled is 7 kg /s.. x 0.0 F 5.0 N.5 kg v.0 /s x? First we ust find the force constant of the spring: F kx F k x 5.0 N 0.0 k 50 N/ K Copyright 00 Nelson Unit Review 9

4 Using the law of conservation of energy: E E kx x k (.5 kg(.0 /s 50 N/ x 0.60 he axiu copression of the spring will be x.00 c (a E e? Ee kx (50.0 N/(0.000 E.00 0 J e he elastic potential energy is.00 0 J. (b x i x f 6.00 c E e? Ee kxf kxi ( kxf xi (50.0 N/ Ee J he change in elastic potential energy is J. (c.00 kg v? E E kx kx v (( ( (50.0 N/( kg v 0.67 /s he cart leaves the spring at a speed of 0.67 /s. 5. When a sall object bounces off a larger stationary object, the change in oentu is greater than if the object sticks to it. If the collision takes the sae aount of tie, the force applied will therefore have to be uch larger (twice as uch. When riot police use rubber bullets, the rubber bullet exerts a large force on the person or object it hits. 6. When the net force consists of only one force, ipulse is a very useful tool in analyzing the changes in otion produced by that force. 9 Unit Energy and Moentu Copyright 00 Nelson

5 7. 78 kg t in 60 s v /s v 0 /s F? ΣF t p ( v F t 78 kg(0 /s /s 60 s F N he agnitude of the force is N. 8. Let the subscript D represent the dog and W represent the wagon. D 9.5 kg v D. /s v W.0 /s W? Using the conservation of oentu: p p 0 Dv D+ Wv W DvD W v W (9.5 kg(. /s.0 /s W.8 kg he ass of the wagon is.8 kg. 9.. kg v.5 /s [W].6 kg (a E? E E E K (. kg(.5 /s.7 J he total energy of the syste before the syste is.7 J. (b v? p p At iniu separation, the two velocities of the carts will be the sae, therefore: ( + v v + (. kg(.5 /s. kg +.6 kg v 0.60 /s [W] he velocity of each cart will be 0.60 /s [W]. Copyright 00 Nelson Unit Review 95

6 (c E K? E E E K K K + (. kg(0.60 /s + (.6 kg(0.60 /s (. kg(.5 /s.6 J he change in total kinetic energy of the syste is.6 J. (d x c 0. k? Ee kx Ee k x (.6 J ( 0. k. 0 N/ he force constant of the spring is. 0 N/. 0.. kg.8 kg k. 0 N/ v.0 /s x? First we ust find the speed of the.-kg trolley: p p 0 + v (.8 kg(.0 /s (. kg v 8.0 /s We can now calculate the force constant: Ee + kx v + + x k (. kg( 8.0 /s + (.8 kg(.0 /s. 0 N/ x 0.0 he force constant of the spring is Choose right as positive. 5 kg v 6.0 /s 5 kg v.0 /s v? 96 Unit Energy and Moentu Copyright 00 Nelson

7 (a v 0.0 /s p p v + ( v v ( v v v + 5 kg( 6.0 /s ( 0.0 /s.0 /s + 5 kg v 0. /s [right], or 0. /s [left] he velocity of the 5-kg object would be 0. /s [left]. (b v 0.5 /s p p v + ( v v ( v v v + 5 kg( 6.0 /s 0.5 /s.0 /s + 5 kg v 0.87 /s [right], or 0.87 /s [left] he velocity of the 5-kg object would be 0.87 /s [left]. (c Since the objects stick together, their final velocities will be equal to each other, therefore: p p ( + v + v + (5 kg( 6.0 /s + (5 kg(.0 /s 5 kg + 5 kg v 0.8 /s [right], or 0.8 /s [left] he velocity of the 5-kg object would be 0.8 /s [left]... 0 kg v.5 0 /s v.0 0 /s First we ust calculate the ass of the sled and the rocket cobined: p p 0 + v (. 0 kg(50 /s.5 0 /s 8 kg he total ass required is 8 kg. At 0 kg/s, it will take s to burn that uch fuel. Copyright 00 Nelson Unit Review 97

8 .. kg v 8 /s [E] v 9 /s [W] v. /s? Choose east as positive. Using the conservation of oentu: p p + + v v v, therefore: Since it is a copletely inelastic collision ( v v ( v v ( v v ( v v (. kg(. /s 8 /s 9 /s. /s.6 kg he ass of the second bird is.6 kg.. v v v 0 v v 5? p p ( v v (Equation v v E E v v v v v ( v v ( v + v ( v v (Equation 98 Unit Energy and Moentu Copyright 00 Nelson

9 Set Equations and equal to each other: v v ( v+ v ( v v v v+ v v v+ v v v + 5 v v (Equation 5 Substitute Equation into Equation : v v v 5 v v 5 5 6v 5 he ass of the second nucleus is. 5. he total oentu before and after ust be the sae. he two initial oentu vectors will be the coponents of the final oentu vector. he dotted line on the diagra represents the direction after. A copletely inelastic collision eans they will stick together..0 0 kg v. 0 /s [E].6 0 kg v.0 0 /s [S] v? Copyright 00 Nelson Unit Review 99

10 First we ust calculate the initial oentu of each vehicle. For the truck: p v p (.6 0 kg(.0 0 /s.6 0 kg /s For the car: p p v (.0 0 kg(. 0 /s.8 0 kg /s he oentu of the car and truck coupled together after the collision: p p + p p (.6 0 kg /s + (.8 0 kg /s kg /s o calculate the final velocity of the car and truck coupled together: p v p v kg /s.6 0 kg kg 0.7 /s v /s p tanθ p p θ tan p.6 0 kg /s tan.8 0 kg /s θ 7 he final speed of the cars is /s [7º S of E]. 6. (a. 0 kg v 5 /s [5 S of W]. 0 kg v /s [5 S of W] 00 Unit Energy and Moentu Copyright 00 Nelson

11 θ + 9 θ 6 Units have been oitted until the final step for clarity. Using the cosine law: p p + p p p cosθ v v v ( ( v cosθ + v v ( ( cosθ v + v v (. 0 (5 + ( ( (. 0 (5( ( cos /s 9. /s sinφ sinθ p p (. 0 p sinθ φ sin p sinθ sin (. 0 kg(5 /s sin6 sin (. 0 kg(9.087 /s φ 6 Since 6º 5º 6º, the initial velocity of the second truck was 9. /s [6º N of W]. (b % lost? E E % lost 00% E % + 00% + + v v 00% + + ( v v ( v v + + (( ( ( ( ( (. 0 kg 5 /s /s + (. 0 kg /s /s 00% (. 0 kg(5 /s + (. 0 kg(9.087 /s % lost % he percentage of the initial kinetic energy lost is %. Copyright 00 Nelson Unit Review 0

12 7. θ 80º 5º 55º 8 kg v 8. /s [N] 95 kg v 6.7 /s [5 W of N] 85 kg v? + ( ( cosθ p p p ppcosθ + + v ( ( cosθ (8 kg (8. /s + (95 kg (6.7 /s (8 kg(8. /s(95 kg(6.7 /scos55 (8 kg + 95 kg + 85 kg v.9 /s sinφ sinθ p p p sinθ φ sin p sinθ sin (95 kg(6.7 /s sin55 sin (85 kg + 95 kg + 8 kg(.9 /s φ he final velocity of the cobined players is.9 /s [º W of N] 8. We are ost likely to see a coet when it is oving at its fastest speed. When a coet is within the solar syste, it is at its closest approach to the sun which eans ost of the coet s energy is in the for of kinetic energy. 0 Unit Energy and Moentu Copyright 00 Nelson

13 9. he farther an orbiting vehicle is fro Earth, the slower its speed. o increase the radius of orbit fro a particular point, the satellite would first need an increase in speed to increase its total energy. As it oved to a higher orbit, this additional kinetic energy would be converted into gravitational potential energy. At the higher orbit, the speed of the vehicle would be less than at the lower orbit. GM v r 0. he speed required to escape fro the surface of Earth is always. k/s. If a space craft is not on the surface of Earth, it would not need that uch speed. he ter escape speed is defined as the launch speed necessary to just escape Earth s gravitational pull.. G N /kg M kg r E g? First we ust calculate the radius of the satellite s orbit: r re k r o calculate the gravitational acceleration: GM g r ( N /kg ( kg 6 ( g 8.06 /s he agnitude of the gravitational acceleration is 8.06 /s.. Let the subscript X represent the unknown planet and E represent Earth. X 0.5M E r X 0.60r E g X? GM E ge re GM X gx rx G(0.5 ME (0.60 re 0.5 GM E 0.6 r E 0.5 ( ge 0.6 gx 0.69gE he surface gravitational acceleration of the unknown planet is 0.69g E. Copyright 00 Nelson Unit Review 0

14 . G N /kg M S kg r S r E.9 0 g? GM g r 0 ( N /kg (.99 0 kg 8 ( g N/kg [toward the Sun] he gravitational field of the Sun at the position of Earth is N/kg [toward the Sun].. Let the subscript M represent Mars, J represent Jupiter, and S represent the Sun. (a r M.8 0 r J M S kg? he radius of orbit will be: rm + rj r r o calculate the period: GM S r π π r GM S π ( ( N /kg (.99 0 kg s 8 hr d a s 6.6 a 600 s h 65.6 d he period of the asteroid s orbit in Earth years is 6.6 a. (b v? GM v r 0 ( N /kg (.99 0 kg v.6 0 /s he speed of the asteroid is.6 0 /s. 0 Unit Energy and Moentu Copyright 00 Nelson

15 60 in 60 s 5. h h s h in (a M N kg C? GM C π 6 ( N /kg (.0 0 kg π C.7 0 /s Kepler s third-law constant is.7 0 /s. (b r? r C r C (.7 0 /s (8.6 0 s 8 r.09 0 he satellite ust be fro Neptune to aintain its circular orbit. (c k 000 he altitude of the orbit is 8. 0 k. h 60 in 60 s d.77 d s d h in r. 0 8 M J? J r GM π π r M J G 8 π (. 0 ( N /kg (.59 0 s 7 M.90 0 kg J 5 he ass of Jupiter is kg. 7. (a kg v? E 0 + Eg 0 Eg GM r GM v r ( N /kg (.8 0 kg 6. 0 v. 0 /s In order to escape, the iniu speed of the olecule ust have been. 0 /s. Copyright 00 Nelson Unit Review 05

16 (b v? E Eg + Eg Eg Eg GM r GM v r ( N /kg (.8 0 kg 6 (. 0 v. 0 /s he speed of the olecule in an orbit around Mercury would be. 0 /s. (c Let the subscript M represent Mercury and O represent the orbit. v? E E + EgL EgO GM GM + rm ro GM GM v rm ro v GM r M r O GM r M r M GM rm ( N /kg (.8 0 kg 6 (. 0 v.67 0 /s he olecule would need to launch fro the surface at a speed of.67 0 /s. (d binding energy? binding energy E Eg GM r GM r 6 ( N /kg (.8 0 kg(5. 0 kg 6 (. 0 9 binding energy.9 0 J In this orbit, the olecule s binding energy would be J. 06 Unit Energy and Moentu Copyright 00 Nelson

17 8. (a Let the subscript J represent Jupiter, and B represent the black hole. M J kg M B 85M J 85( kg kg r? For a black hole, the escape speed is the speed of light, c. E 0 + Eg 0 Eg GM r GM r v 9 ( N /kg (.65 0 kg 8 (.00 0 /s r. 0 he Schwartzschild radius of the black hole is. 0. (b For a black hole, the escape speed is the speed of light, c /s. Applying Inquiry Skills Copyright 00 Nelson Unit Review 07

18 kg t 0.00 s (a k? Fro the diagra, there are dots one cycle, therefore: (0.00 s 0.8 s o calculate the force constant: π k k π π k π 0.55 kg ( ( 0.8 s k.8 0 N/ he force constant of the springs is.8 0 N/. (b he results would be the sae whether the paper was pulled quickly or slowly. he only value needed fro the paper is how any dots there are in each cycle. his can be deterined at alost any speed. he only restrictions on speed are that it ust be slow enough to get at least one full cycle on the paper, and fast enough that the dots don t overlap one another. (c he ost likely source of error would be the friction between the paper and the puck. (d Safety considerations include exercising caution when using an electrical spark tier, handling the vibrating spring coils carefully so as not to be pinched, and being careful not to receive a paper cut fro the sliding paper.. (a With no ass hanging fro the spring, ark that point as zero. Suspend one weight fro the acceleroeter and ark that point as g. Suspend a second weight and ark the botto of the spring as g. Measure the distances between 0g and g, and between g and g. hey should be the sae. Measure out the average of the two values beyond g and ark g and g. (b Soe possibilities are: spinning in a circle with the acceleroeter held horizontally landing after juping off a stool while holding the acceleroeter quickly lifting the acceleroeter into the air (c Soe possible answers are: the botto of a roller coaster ride during a large swing during a spin ride (if held horizontal kg k 6 N/ v 0. /s d.5 x.6 c 0.06 (a E e? E K? o calculate the elastic potential energy: Ee kx (6 N/( J E 0.0 J, or. 0 J e 08 Unit Energy and Moentu Copyright 00 Nelson

19 o calculate the kinetic energy: (0.88 kg(0. /s 0.00 J 0.0 J, or. 0 J he elastic potential energy was 0.0 J and the kinetic energy was 0.0 J. he discrepancy in values was due to energy lost in the for of sound and theral energy. (b F K? E E Eth 0 FK d FK d (0.88 kg(0. /s (.5 F N he average kinetic friction is N. (c Possible sources of error are: a desk or track that is not levelled properly the cart is not initially in contact with the wall incorrect release of the trigger K Making Connections. (a he jarring forces on your legs can be reduced when jogging by keeping your knees bent and allowing the to flex with each step. (b By reducing the jarring, you reduce the forces applied internally in your legs. his helps prevent painful daage to bones and tissue, such as cartilage, in the knee. 5. (a An air bag is used to reduce injury in two priary ways. Because it is a bag, the force that stops a passenger in the car is spread out over the entire surface of the body, not just a sall area such as the points the hands contact on the steering wheel. his reduces the pressure on the person s body. he second way the airbag reduces injury is by slowing the person down over a longer period of tie. his increase in stopping tie significantly reduces the total force required to bring the passenger to rest. (b Other safety devices that used to help are: cruple zones in a car snug fit and padding of any type of helet design of protective equipent (such as shin pads 6. r d h 600 s a.0 0 a s a d h (a M? GM r π π r M G 0 π ( ( N /kg (6. 0 s M.9 0 kg he total ass of the stars at the hub of our galaxy is.9 0 kg. Copyright 00 Nelson Unit Review 09

20 (b.9 0 kg kg he approxiate nuber of stars that are the size of our Sun is.5 0. Extension 7..5 kg v. /s.0 kg v? v? Choose right as positive. Using the conservation of energy (units are oitted: + +.5(..5v +.0v 5.9 v v (Equation Using the conservation of oentu (units are oitted: + (.5(..5v +.0v. v + 0.8v v. 0.8 v (Equation Substitute Equation into Equation : 5.9 (. 0.8 v + 0.8v 5.9 (5.9.68v v + 0.8v 0.68v +.v 0 v ( v v 0, or v 0 Sincev 0 is not valid because it represents no change in speed, we use: v 0.v.68 v.5556 v.6 /s [right] Substitute back into Equation : v. 0.8(.5556 v 0.6 /s [right] he velocity of ball is 0.6 /s [right], and the velocity of ball is.6 /s [right]. 8. Let the subscript b represent the bullet and w represent the block of wood. b.0 g.0 0 kg v b /s [N] v b.0 0 /s [N] w.0 kg d Unit Energy and Moentu Copyright 00 Nelson

21 (a v w? Choose north as positive. p p b b+ 0 b b + wvw b b b b v w w b( vb v b w (.0 0 kg(5.0 0 /s.0 0 /s.0 kg v 0.80 /s [N] w he wooden block oves at a velocity of 0.80 /s [N] just after the bullet exits. (b E K? (.0 kg(0.80 /s E 0.6 J K he axiu kinetic energy of the block is 0.6 J. (c F K? E E E E K th 0 FK d FK d (.0 kg(0.80 /s (0.0 FK.6 N [N], or.6 N [S] he average frictional force stopping the block is.6 N [S]. (d E K? ( v (.0 0 kg ((.0 0 /s ( /s.8 0 J he decrease in kinetic energy of the bullet is.8 0 J. (e he collision between the bullet and the block is not elastic, so ost of the energy is lost to theral energy. 9. Let the subscript P represent the plane, and B represent the barge. P.0 Mg.0 0 kg B.0 Mg.0 0 kg v P /s 50 /s v B 0.0 /s Copyright 00 Nelson Unit Review

22 For the landing of the plane: P P + B B ( P + B vpb P P + B B vpb P + B (.0 0 kg(50 /s kg kg v 6.67 /s o calculate the frictional force on the plane, and on the barge: FK g (.0 0 kg(9.8 N/kg F 50 N [backward] By Newton s third law, F K 50 N [forward] o calculate the acceleration of the plane: Σ F PaP ΣF ap P 50 N.0 0 kg a.5 /s o calculate the acceleration of the barge: Σ F BaB ΣF ab B 50 N.0 0 kg a.5 /s K P B o calculate the distance the plane will travel during the landing: vf vi + ap d vf vi d ap (6.67 /s (50 /s (.5 /s d 5.5 PB o calculate the distance the barge will travel during the landing: vf vi + ab d vf vi d ab (6.67 /s (0 /s (.5 /s d. herefore, the required length of the barge is Unit Energy and Moentu Copyright 00 Nelson

23 50. Let the subscript C represent the chair. (a C? k? For the epty chair: π k k π π k π ( C + 0 k For asses on the chair: π ( 0.90 s π C ( 0.90 s k k π π k π ( C + k o calculate the ass of the chair: π C π ( C + ( 0.90 s C C + ( 0.90 s C C ( 0.90 s C ( 0.90 s C ( 0.90 s. kg (.5 s ( 0.90 s 5. kg C Copyright 00 Nelson Unit Review

24 Siilarly,.9 kg.8 kg 5.0 kg C 5. kg C 5. kg C 5. kg (. s (.6 s (.79 s ( 0.90s ( 0.90s ( 0.90s 56. kg 67. kg C 5. kg C 5. kg (.9s (.09 s ( 0.90s ( 0.90s o calculate the average value for the ass of the chair: 5. kg + 5. kg + 5. kg + 5. kg + 5. kg + 5. kg C 6 5. kg o calculate the spring constant: C π k k π π C k π (5. kg ( 0.90 s k 7 N/ he ass of the chair is 5. kg, and the spring constant is 7 N/. (b f? Assuing a ass of 9 kg, we first calculate the period: π k 9 kg + 5. kg π 7 N/. s Since frequency is inversely proportional to period: f. s f 0. Hz he frequency is 0. Hz. Unit Energy and Moentu Copyright 00 Nelson

25 (c.0 s.98 s? π k k π k π k k π π k ( π 7 N/ ((.98 s (.0 s π. kg he astronaut loses. kg 5. (a Venus has the ost circular orbit because its eccentricity is closest to zero. Pluto has the ost elongated orbit because its eccentricity is the highest. (b (c (d he typical elliptical orbits shown for ost planets do not represent the true shape of the orbit. Most texts exaggerate the length of the orbit for effect. Even the ost elongated orbits have the Sun close to the centre of the ellipse. 5. Let the subscript X represent the planet and E represent Earth. X E 0 M r X E r F ge 6. 0 N F gx? Copyright 00 Nelson Unit Review 5

26 Half of the diaeter also eans half of the radius, therefore: F g ge F g ge E E Also, g g g E E re X X rx X GM GM G ME 0 re 0 GM E r E ( ge 0 0.g E Substituting the value of of the value of g X : F g F gx gx X FgE (0. ge ge 0.F ge 0.(6. 0 N. 0 N he astronaut weighs. 0 N on the new planet. 6 Unit Energy and Moentu Copyright 00 Nelson