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Unformatted text preview: 8-1MOMENTUM8Answers to Multiple-Choice Problems1.A, D2.D3.A, B, C4.C, D5.C6.B7.C8.A, D9.B10.C11.B, D12.A, D, E13.A, C, D14.A, B15.DSolutions to Problems8.1. Set Up:Use coordinates where is to the right. In (a)(c) denote the two objects as Aand B. In (d) call thethird object C.Solve: (a)(b)(c)(d)8.2. Set Up:The signs of the velocity components indicate their directions. Call the objects Aand B.Solve: (a)(b)(c)(d)8.3. Set Up:The signs of the velocity components indicate their directions.Solve: (a)(b)(c)Refect:Ahas no x-component of momentum so is the same in (b) and (c). Chas no y-component of momentum soin (c) is the sum of in (a) and (b).PyPyPxPy5pAy1pBy1pCy515.0 kg2 1211.0 m/s2116.0 kg2 110.0 m/ssin60215 23.0 kg#m/sPx5pAx1pBx1pCx5116.0 kg2 110.0 m/scos6021110.0 kg2 123.0 m/s25Py5pBy1pCy516.0 kg2 110.0 m/ssin6021552 kg#m/sPx5pBx1pCx516.0 kg2 110.0 m/scos6021110.0 kg2 123.0 m/s25Py5pAy1pCy515.0 kg2 1211.0 m/s215 255 kg#m/sPx5pAx1pCx51110.0 kg2 123.0 m/s25 230 kg#m/sPy5pAy1pBy518.0 kg2 125.0 m/scos3521110.0 kg2 125.0 m/scos3525 273.7 kg#m/sPx5pAx1pBx518.0 kg2 125.0 m/ssin3521110.0 kg2 15.0 m/ssin35255.7 kg#m/sPy5pAy1pBy516.0 kg2 125.0 m/s2118.0 kg2 112 m/ssin722561.3 kg#m/sPx5pAx1pBx5118.0 kg2 112 m/scos722529.7 kg#m/sPy5pAy1pBy519.0 kg2 13.0 m/s21527 kg#m/sPx5pAx1pBx5116.0 kg2 124.0 m/s25 224 kg#m/sPy5pAy518.0 kg2 145 m/ssin6025312 kg#m/sPx5pAx518.0 kg2 145 m/scos6025180 kg#m/sPx5pAx1pBx1pCx517.0 kg2 12.0 m/s2118.0 kg2 125.0 m/s2114.0 kg2 16.0 m/s25 22.0 kg#m/sPx519.0 kg2 14.0 m/s2116.0 kg2 128.0 m/s25 212 kg#m/sPx516.0 kg2 12.0 m/s2113.0 kg2 124.0 m/s25Px5pAx1pBx515.0 kg2 14.0 m/s2113.0 kg2 18.0 m/s2544 kg#m/s1x8.4. Set Up:Solve: (a)(b)The SUV has twice the mass and therefore twice the momentum and kinetic energy:8.5. Set Up:Solve: (a)(b)(i) so (ii) so Refect:Kdepends on and pdepends on so a smaller increase in is required for the lighter ball to have thesame kinetic energy.8.6. Set Up:and Solve: (a)so (b)so (c)(d)8.7. Set Up:From Problem 8.6,Solve: (a)and so (b)and so Refect:In (b),and When the mass is increased the speed must be decreased tokeep the same kinetic energy.8.8. Set Up:Consider the system consisting of the two skaters. Assume the heavier skater travels to the right and take this to be the direction.Solve: (a)There is no horizontal external force so The skaters are initially at rest so The lighter skater travels to the left at (b)The kinetic energy of the system was produced by the work the two skaters do on each other.8.9. Set Up:Consider the person and the earth to be an isolated system. Use coordinateswhere is upward, in the direction the person jumps.Solve:The earth recoils in the direction with speed so vE51mpersonmE2vperson5175 kg5.9831024kg212.0 m/s252.5310223m/s5mpersonvperson2mEvE.vE,2yPi,y50.Pi,y5Pf,y....
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- Summer '09