Seat: PHYS 1500 (Fall 2006) Exam #2, V1. After : p y = m 1 v 1y + m 2 v 2y = 20 kg m/s + 2 kg v 2y. v 2x = 1 m/s v 2y = 9 m/s (V 1)

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1 Seat: PHYS 1500 (Fall 006) Exa #, V1 Nae: 5 pt 1. Two object are oving horizontally with no external force on the. The 1 kg object ove to the right with a peed of 1 /. The kg object ove to the left with a peed of 1 /. The two object collide and tick together. The cobined 3 kg object (a) ove to the left. (b) ove to the right. (c) i tationary. (d) ove in a way that can not be deterined fro the given inforation. 5 pt. Becaue of the Earth rotation about it axi, your weight i lightly different at the equator than at the North or South Pole. I your weight ore or le at the equator? Explain. Your weight i the noral force exerted by the floor to keep you in place. On the equator, you weigh lightly le than at the pole becaue the noral force inu g equal 0 at the pole. At the equator, the noral force inu g equal a all negative nuber (the centripetal force). 5 pt 3. Travelling with a velocity of.0 / in the x-direction, a 5.0 kg flugle explode into piece. After the exploion, piece 1 ha a a of 3.0 kg and a velocity with x- coponent of 4.0 / and y-coponent of 6.0 /. What i the velocity vector of piece after the exploion? (Ignore all external force.) Since there are no external force, the total oentu i conerved. The oentu before the exploion equal the oentu after. Before : p x = 10 kg / Before : p y = 0 kg / After : p x = 1 v 1x + v x = 1 kg / + kg v x After : p y = 1 v 1y + v y = 18 kg / + kg v y v x = 1 / v y = 9 / (V 1) Before : p x = 18 kg / Before : p y = 0 kg / After : p x = 1 v 1x + v x = 4 kg / + kg v x After : p y = 1 v 1y + v y = 1 kg / + kg v y v x = 3 / v y = 6 / (V ) Before : p x = 8 kg / After : p x = 1 v 1x + v x = 40 kg / + kg v x Before : p y = 0 kg / After : p y = 1 v 1y + v y = 0 kg / + kg v y v x = 6 / v y = 10 / (V 3)

2 10 pt 4. You are dragging a 0 kg box acro the floor with a rope tied around the box. The rope ake an angle of 37 with the vertical. The box tart fro ret. The tenion in the rope i 40 N. In dragging the box fro x = to x = 6, the box peed decreae fro 3 / to /. (a) How uch work doe the rope do on the box? (b) What i the force fro friction? (a) The work that the rope doe on the box can be found fro W = (F co θ) x. The angle θ i the angle between the force and the direction that the box ove. (b) The force fro friction can be found fro W fric + W rope = W net = KE; the work done by friction i the F fric x where F fric i the agnitude of the friction force and x i the ditance oved (the i becaue the friction force i in the oppoite direction that the box i oving). T = 40 N θ = 53 x = 4 W rope = 40 N co 53 4 = 96 J W net = KE = 1 ( [ 0 kg ] [ 3 ] ) = 50 J W fric = W net W rope = 50 J 96 J = 146 J = F fric 4 F fric = 36.5 N (V 1) T = 0 N θ = 37 x = 3 W rope = 0 N co 37 3 = 48 J W net = KE = 1 ( [ 10 kg 1 ] [ ] ) = 15 J W fric = W net W rope = 15 J 48 J = 63 J = F fric 3 F fric = 1 N (V ) T = 10 N θ = 53 x = 5 W rope = 10 N co 53 5 = 30 J W net = KE = 1 ( [ 40 kg 1 ] [ 3 ] ) = 160 J W fric = W net W rope = 160 J 30 J = 190 J = F fric 5 F fric = 38 N (V )

3 5 pt 5. Two ball, identical except for their color, are oving with equal peed on the ae horizontal urface. The red ball i rolling; the blue ball i liding o that it in t rotating at all. The energy of the red ball i (a) greater than that of the blue ball. (b) aller than that of the blue ball. (c) i the ae a that of the blue ball (d) relative to the blue ball can t be deterined fro the inforation given. 5 pt 6. A girl i tanding on a raft that i floating on a pond. Both the girl and the raft are initially tationary. The girl walk fro the end of the raft away fro the hore to the end of the raft cloet to the hore. Ignore the drag fro air and water. Doe the raft ove? If ye, what direction doe it ove? Explain. When the girl tep toward the hore, the boat exert a force on her toward the hore. By Newton 3rd law, he exert a force on the boat away fro the hore. Thu, the boat ove away fro the hore. 5 pt 7. A 10.0 kg vrabn i dropped fro a height of 1.5 above the floor. When it hit the floor, it ha a peed of 4.0 /. How uch work wa done by air reitance and other non-conervative force? The work done by all other force i W nc = W net W g where W g i the work done by gravity and W net i the total work done. The total work done can be found fro W net = KE. In thi proble, KE i = 0 and W g = gh becaue the vrabn ove down. W nc = 1 ( 10 kg 4 ) 10 kg = 80 J 150 J = 70 J (V 1) W nc = 1 ( 0 kg 3 ) 0 kg 10 3 = 90 J 600 J = 510 J (V ) W nc = 1 ( 30 kg ) 30 kg 10 3 = 60 J 900 J = 840 J (V 3)

4 10 pt 8. A 5 kg bucket hang fro a line that i wrapped around a cylinder that ha a a of 0 kg, oent of inertia of 10 kg, and a iu of. The cylinder rotate on a frictionle axle. If the bucket and cylinder tart fro ret, how far ha the bucket decended if it peed i 8 /? Since the only force that i acting in thi proble i gravity, then the total work done W net = KE = W g. Becaue the bucket i dropping, the work done by gravity i W g = gh. There are two ource of KE: tranlational kinetic energy of the bucket, KE = (1/)v, and the rotational kinetic energy of the cylinder, KE = (1/)Iω, with ω = v/r. ω = 8 / = 4 KE cyl = 1 10 kg ( 4 W g = 5 kg 10 h = 40 J ω = 6 / 3 = KE cyl = 1 15 kg ( W g = 10 kg 10 h = 10 J ω = 8 / 4 = KE cyl = 1 18 kg ( W g = kg 10 h = 100 J KE buc = 1 ( 5 kg 8 ) = 160 J ) = 80 J W net = 160 J + 80 J = 40 J h = 40 J 50 N KE buc = 1 ( 10 kg 6 ) = 180 J = 4.8 (V 1) ) = 30 J W net = 180 J + 30 J = 10 J h = 10 J 100 N KE buc = 1 ( kg 8 ) = 64 J =.1 (V ) ) = 36 J W net = 64 J + 36 J = 100 J h = 100 J 0 N = 5 (V 3)

5 5 pt 9. Circle all of the correct anwer. One J (Joule) of energy equal (a) 1 W/ (b) 1 N/ (c) 1 kg (d) 1 W (e) 1 N (f) 1 kg/ 5 pt 10. You throw 3 identical lead weight fro a height of. All three leave your hand at exactly the ae height and with the ae peed. Weight 1 i thrown at an angle 45 above the horizontal, weight i thrown horizontally, and weight 3 i thrown at an angle 45 below the horizontal. Neglecting air reitance, rank the peed of the weight jut before they reach the ground fro lowet to fatet. Explain. The change in kinetic energy i related to the total work done through W net = KE. In thi ituation, only gravity i doing work. Therefore, W net = gh which i the ae for all three weight. Becaue the change in kinetic energy i the ae for all three weight and they all tart with the ae kinetic energy, they all finih with the ae kinetic energy. Thi ean they all have the ae peed when they hit the ground. 5 pt 11. You have a 8 long, unifor etal rod of a 3 kg balanced on it idpoint. You hang a 8 kg a 1 fro one end and you pull down on a tring at the other end o that the rod i horizontal and otionle. What i the tenion in the tring? What i the force exerted on the rod by the wedge? Becaue the rod i otionle, the u of the torque ha to be 0 and the u of the force ha to be 0. I will take the axi of rotation to be at the wedge which ean only the a hanging fro the left and the tring give nonzero torque; the weight of the rod and the wedge act at the axi and give zero torque. The force are only in the y-direction o there i only one relation between the force. The force are fro the a, the tring, the weight of the rod, and the wedge. τ : 3 8 kg T = 0 F : 8 kg10 + T + 3 kg10 + F w = 0 T = 60 N F w = 170 N (V 1) τ : 4 15 kg T = 0 F : 15 kg10 + T + 7 kg10 + F w = 0 T = 10 N F w = 340 N (V ) τ : 9 kg T = 0 F : 9 kg10 + T + kg10 + F w = 0 T = 60 N F w = 170 N (V 3)

6 10 pt 1. Thi proble take place on earth. A 3 kg a i tied to a tring with a length of. The a i rotating clockwie in a vertical circle. When the tring i horizontal and the a i to the right of the axi of rotation, the a ha a peed of 8 /. At thi point, (a) what i the angular velocity? (b) what i the angular acceleration? (c) what i the net force acting on the a? (d) When the a i at the botto of the circle, what i it velocity? (a) Since the a i rotating clockwie, the angular velocity i ω = v/r. (b) The angular acceleration can be found fro α = a t /r. Gravity i the only force acting along the circle when the tring i horizontal. Therefore, the tangential acceleration ha a agnitude g. Becaue the torque i in the clockwie direction, the angular acceleration hould be negative. The angular acceleration i α = g/r. (c) The net force acting on the a i inu the tenion in the tring for the x-direction and g for the y- direction. The tenion in the tring i the only force that i aking the a go in a circle (when the tring i horizontal) therefore it ut equal v /r. (d) Since there i hardly any friction in thi ituation, the only force doing work on the a i gravity. Thi ean the aount of net work done i gr for thi ituation. Thi can be related to the change in kinetic energy to give KE f = KE i + gr. The final velocity i vf = vi + gr for thi proble. (a) ω = 8 / = 4 10 / (b) α = = 5 (c) (8 /) F x = 3 kg = 96 N F y = 3 kg 10 = 30 N ( (d) vf = 8 ) + 10 = 104 = 104 / (V 1) 1 / (a) ω = 4 = 3 10 / (b) α = 4 =.5 (c) (1 /) F x = kg = 7 N F y = kg 10 4 = 0 N ( (d) vf = 1 ) = 4 = 4 / (V ) 0 / (a) ω = 5 = 4 10 / (b) α = 5 = (c) (0 /) F x = 4 kg = 30 N F y = 4 kg 10 5 = 40 N ( (d) vf = 0 ) = 500 = 500 / (V 3)

7 Equation Baic Matheatic Forula in θ = h o /h co θ = h a /h tan θ = h o /h a h = h o + h a A circ = πr Circu of circ = πr V ph = 4π 3 r3 A ur of ph = 4πr co 37 = 4/5 in 37 = 3/5 co 53 = 3/5 in 53 = 4/5 Chapter Avg peed = ditance/elaped tie g = 10 / x = x f x i v = x/ t ā = (v f v i )/(t f t i ) = v/ t v = v 0 + at x = 1 (v 0 + v)t x = v 0 t + 1 at v = v 0 + a x Chapter 3 A x = A co θ A y = A in θ A = A x + A y tan θ = A y /A x r = r f r i v av = r/ t a av = ( v)/ t v x = v 0x + a x t x = 1 (v 0x + v x )t x = v 0x t + 1 a xt v x = v 0x + a x x v y = v 0y + a y t y = 1 (v 0y + v y )t y = v 0y t + 1 a yt v y = v 0y + a y y Chapter 4 F = a Fg = G 1 r 11 N G = kg w = g F 1 = F 1 f µ n f k = µ k n Chapter 5 W = (F co θ) x KE = 1 v W net = KE f KE 0 W g = g(y i y f ) PE = gy W nc = KE f KE i + PE f PE i P = W t P = F v Chapter 6 p = v I = F t F t = p = vf v i 1 v 1i + v i = 1 v 1f + v f

8 ω av = θ f θ i t f t i = θ t Chapter 7 α av = ω f ω i t f t i = ω t ω = ω i + αt θ = 1 (ω i + ω)t θ = ω i t + 1 αt ω = ω i + α θ v t = rω a t = rα a c = v r = rω T = 4π GM S r 3 Chapter 8 τ = rf in θ I r τ = Iα KEr = 1 Iω W nc = KE t + KE r + P E L Iω τ = L t

= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5.

= s = 3.33 s s. 0.3 π 4.6 m = rev = π 4.4 m. (3.69 m/s)2 = = s = π 4.8 m. (5.53 m/s)2 = 5. Seat: PHYS 500 (Fall 0) Exa #, V 5 pt. Fro book Mult Choice 8.6 A tudent lie on a very light, rigid board with a cale under each end. Her feet are directly over one cale and her body i poitioned a hown.