KEY. D. 1.3 kg m. Solution: Using conservation of energy on the swing, mg( h) = 1 2 mv2 v = 2mg( h)

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Phy 5 - Fall 206 Extra credit review eion - Verion A KEY Thi i an extra credit review eion. t will be worth 30 point of extra credit. Dicu and work on the problem with your group. You may ue your text book and your cla note. Name:. What i the magnitude of the momentum of a tenni ball with a diameter of 6.6 cm and a ma of 57 g that i traveling at it terminal velocity near the urface of the earth? The denity of air i.3 kg and, for a phere, c 0.50. m 3 A. 29 kg m B. 0.4 kg m C. 0.29 kg m D..3 kg m E. 0.63 kg m Solution: Terminal velocity i the velocity at which the force due to gravity i equal to the drag force. So, mg 2 ρcav2 v 2mg. Thu, p mv ρca m 2mg 2m 3 g ρca ρca 2(0.057 3 )g kgm.3.3(0.50)(π.033 2 ) The correct anwer i D. 2. A rope wing over a lake ha a vertical drop of 2.0 meter between where you (60.0 kg) tart and where the rope hang vertically. f you let go when the rope i vertical and you are.0 meter above the water, how far do you fly through the air before you hit the rock hiding jut below the urface of the water? A..4 m B..0 m C. 2.8 m D. 8.0 m E. 2.3 m Solution: Uing conervation of energy on the wing, mg( h) 2 mv2 v 2mg( h) 2g( h). That the horizontal velocity with which you fly off over the m water. t now become a kinematic problem. Horizontal launch with the velocity jut determined, at a height of h 2.0 m. t 2h 2, giving a horizontal ditance g d v x t 2g( h) 2h 2 2 ( h)h g 2 2 (2.0).0 2.8 m. The correct anwer i C.

Phy 5 - Fall 206 Verion A Page 2of 5 3. m going 35.0 m/ on the freeway (horizontal). The tire of the car in front of me pick up a rock and launche it into the air at an angle of 50 o. t reache a height of 2.0 meter before coming back down to.0 meter and triking my windhield, how fat i it going with repect to the windhield? A. 30.7 m B. 9.54 m C. 40.5 m D. 6.87 m E. 40.3 m Solution: vyf 2 vyi 2 + 2a( y) v yi vyf 0 2 2a( y) + 2g(2.0) 2.0 g. Uing thi initial y velocity and the launch angle, we can determine the x-component. v xi 2.0 g. We can alo determine the vertical final velocity, a the ame a an object dropped tan(50 o ) from ret, that ha fallen.0 m, vyf 2 vi 2 +2a( y) v yf 2g. We ve now determined that the final velocity of the rock with repect to the freeway i v F 2.0 g tan(50 o )î + 2gĵ. Due to the relative motion of the car, the velocity with repect to the windhield i v F (35 + 2.0 g )î + 2gĵ 40.3î + 4.43ĵ, giving a magnitude of 40.5 m. tan(50 o ) The correct anwer i C. 4. A 0.0 kg block i propelled up a lope with an angle of 30 o by a pring with a pring contant of 500 N m. f the pring i compreed by 0.500 m, and µ k 0.20, how far doe the block lide pat the pring equilibrium? A. 0.95 m B. 0.77 m C. 0.45 m D. 3.3 m E. Doen t move. Solution: Let be the ditance the pring i compreed and d be the ditance it lide pat ( equilibrium (the thing we re looking for). All the energy come from the pring U 2 k2), and it goe into friction over the total ditance (µ k mg co θ(d + )) and fighting gravity (mg in θ(d + )). So, 2 k2 µ k mg co θ(d + ) + mg in θ(d + ) d k 2 500(0.500 2 ) 2mg(µ k 0.500 0.45 m. co θ+in θ) 2(0)g(0.20 co 30 o +in 30 3 ) The correct anwer i C.

Phy 5 - Fall 206 Verion A Page 3of 5 5. A pace tation i made of a ring of uniformly ditributed ma (8000 kg) with a diameter of 5.00 km. The tation i tarted pinning by 4 thruter located along the ring. Each thruter can apply a contant 900 N force, tangential to the ring, cauing it to pin up. How long do the thruter need to fire o that omeone tanding on a cale on the inide of the ring meaure the ame weight a they did on the earth? A. 74 B. 244 C. 492 D. 348 E. 968 Solution: For the cale to read the ame a it doe on earth, the centripetal acceleration on the ring need to be g. g v2 t (ω)2 ω 2 ω g. Each thruter exert a torque of F in θ F, o Στ 4F. Στ α. r 2 dm 2π 0 ( 2 M 2π dθ) M 2. So, α Στ 8000 4(900) 4F M 2 4F M. ω αt t ω α (9.806)2500 348 The correct anwer i D. g 4F M g M M 4F 4F g 6. With the 2 vector, A and B, calculate A B and A B. A 5î 3ĵ + 2ˆk, B 2î + ĵ + 6ˆk A. A B 25, A B 20î 34ĵ ˆk B. A B 25, A B 20î 34ĵ ˆk C. A B, A B 20î 34ĵ ˆk D. A B 25, A B 20î + 34ĵ + ˆk E. A B, A B 20î + 34ĵ + ˆk Solution: A B ax b x + a y b y + a z b z 0 3 + 2, A B (a y b z b y a z )î (a x b z b x a z )ĵ + (a x b y b x a y )ˆk 20î 34ĵ ˆk The correct anwer i C.

Phy 5 - Fall 206 Verion A Page 4of 5 7. A pacehip i traveling at 500 m with repect to earth. There i an exploion which break the hip into 2 piece. The forward ection, which contain 40% of the ma i accelerated to 600 m. What i the peed at which the 2 piece are moving apart? A. 67 m B. 433 m C. 27 m D. 500 m E. 250 m Solution: The velocity of the forward ection increaed v 00 m. Uing conervation of momentum, we know that mv i.40mv f +.60mv 2f. Thi hould work in all inertial reference frame; why not chooe the one moving at 600 m with repect to the earth? That the frame in which we want the anwer anyway. n thi frame, v i 00 m and v f 0. We are looking for v 2f v i.40v f 00.40(0) 67 m. The correct anwer i.60.60 A. 8. The moment of inertia of a phere around any axi through it center i 2 5 m2. A 5.0 kg phere with a radiu of 50 cm roll 4.00 m without lipping down a ramp with a lope of 35 o. How long doe it take if it i releaed from ret? A. 0.764 B. 3.5 C..4 D. 2.34 E. 5.67 Solution: The height change while rolling down a length ramp with an incline of θ i h in θ, meaning that U mg( h) mg in θ. For rolling motion, K 2 ω2 + 2 mv2 cm, o mg in θ 2 ω2 + 2 mv2 cm. Since it in t lipping, v cm ω ω vcm. mg in θ 2 ( v cm ) 2 + 2 mv2 cm meaning that at the end, v cm 2g in θ m v v i+v f which mean that, ince v t 2 i 0, t 2 2 v f 2g in θ 2( 2 +) 5 2(4.00) 7 5.4. The correct anwer i C. g in θ g in 35 m 2 + 2 +. We know that 2( m 2 +) g in θ

Phy 5 - Fall 206 Verion A Page 5of 5 9. A particle velocity i decribed by the function V ηt 2 for ome contant η and time t. f the particle poition at t.0 i 4.0 m and the particle poition at t 2 3.0 i 2.0 m, what i the value of η? A. 2.0 m 3 B. 0.57 m 3 C. 0.3 m 3 D. 3.0 m 3 E. 0.92 m 3 Solution: V ηt 2 mean that x t 2 t V dt t 2 t ηt 2 dt η 3 (t3 ) t 2 t η 3 (t3 2 t 3 ) η 3 (3.03.0 3 ) 26η m We alo know that x (2.0 4.0) m 8.0 m. Thu, 3 26η 8.0 η 0.92 m where the unit came from jut looking at what they would 3 3 need to be o that η 3 (t3 ) end up in unit of meter. The correct anwer i E. 0. A map to Beardle Luke treaure ay to head 200 pace on a bearing 30 o wet of north, then 50 pace wet, followed by 300 pace outh to avoid walking through the forbidden meadow. You decide that the meadow doen t ound o bad and that you hould take the mot direct path. How far hould you go in which direction? A. 380 pace 32 o outh of wet B. 280 pace 63 o outh of wet C. 280 pace 27 o outh of wet D. 535 pace 62 o north of eat E. 380 pace 58 o outh of wet Solution: Firt off... don t cut through the forbidden meadow. That a horrible idea. But if you did, you d want your total diplacement D ( 200 in(30) 50) eat ˆ + (200 co(30) 300) north ˆ 250eat ˆ 27north. ˆ Which ha a magnitude of 2502 + 27 2 280 and a direction of tan (27/250) 27 o, outh of wet. The correct anwer i C.