III MCCD canvas Loc X MA Hwy 13 PHYs 250 x c calculate The Amo x C what Is The E

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III MCCD canvas Loc X MA Hwy 13 PHYs 250 x c calculate The Amo x C what Is The Escap x C zero. Hypothetc x C Two Neutron stars x CI n The Figure Two x G mass of earth-co x C D www.webassign ne 61/2503 t/web/Student/Assignment-Responses/submit?dep 13. -12 points HRV 13 rn41 ssm My Note Two neutran stars are separated by a distance of 4.0 x 10to rn. They each have a mass at 3.5 x 1030 kg and a radius of 1.0 x 103 m. They are initially at rest with respect ta each other. As measured from that rest irarne, how tast are they moving the following positions? (a) Their separation has decreased one-half its initial value. duo b) They Additi al Mat section 13.5 0.5 2.5 points Pmlnus Answers HRW 10 13 POFD My Note In the figure below, twa satellites, A and B, both mass m 115 kg, mave in the same circular Drbit of radius r 8.13 x 106 m araund Earth but in opposite senses Di rotatian and therefore on a collision course. Earth (a) Find the total mechaninal energy Fs Fes of the two sarrilites Farth system hrtore the collision. I-566 10ea x 1 (b) If the collision is completely inelastic so that the wreckage remains as one piece of tangled material (mass 2m), find the total mechanical energy immediately after the collision. (c) Jusu uer une collisiuni is Lhe wreckage lalling din ully d Earth's center or bilirly arourd Earth? e Thle wreukay is falliriy directly uuward Earth's ceriler. The wreckage is orbiting around Larth. a V a Type here to search /18/20

Explanation / Answer

13 (a).

Initial separation is d = 4X1010 m

So initial mechanical energy is,

E1 = PE1 = -Gm2/d (kinetic energy is zero.)

Final separation is d/2. Let the speed be v for the stars, then mechanical energy is,

E2 = PE2 + KE = -2Gm2/d + 2(1/2)mv2

From conservation of energy we have,

E1 = E2, so,

-Gm2/d = -2Gm2/d + 2(1/2)mv2

or, v2 = Gm/d = (6.67X10-11 N-m2/kg2)(3.5X1030 kg)/( 4X1010 m)

or, v = 7.64X104 m/s

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13 (b).

When they are about to collide the separation between the centers is,

s = 2r = 2X105m

So, using conservation of energy,

-Gm2/d = -Gm2/(2r) + 2(1/2)mv2

or, Gm/(2r) - Gm/d = v2

or, v2 = Gm[1/2r - 1/d] = (6.67X10-11 N-m2/kg2)(3.5X1030 kg)[1/(2X105m) - 1/4X1010m]

or, v = 3.416X107 m/s

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14 (a).

mechanical energy before collision is,

U1 = -2GMm/r + 2.(1/2)mv2

Now, v = (GM/r)1/2

So, U1 = -2GMm/r + GMm/r = -GMm/r

or,  U1 = -(6.67X10-11 N-m2/kg2)(5.972X1024 kg)(115 kg)/(8.13X106 m)

or, U1 = -5.634X109 J

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14 (b).

If the collision is completely elastic, the final speed of the satellites after collision will be zero.

So mechanical energy is,

U2 = -2GMm/r = -2X(5.634X109 J)

or, U2 = -1.126X1010 J

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This concludes the answers. Check the answer and let me know if it's correct. If you need any more clarification, modification or correction, feel free to ask.....

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