Problem Set #8 Solutions

Ok at last we can put in numbers 7 ms this document

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Unformatted text preview: θB ) This last equality uses one of the trigonometric angle addition formulas; you’re not necessarily expected to know this, but it makes the final algebraic expression look a bit simpler. OK, at last we can put in numbers: ◦ (7 m/s) This document requires.1asin 85sin 25 = 3.012 m/s ≃ 3.01 PDF reader JavaScript-enabled m/s. vA = ◦ such as Adobe have an expression for v , we can solve for v : Reader. (b) Now that we A m B vB = m A vA B sin θA vi sin θB = mA sin θB sin(θA + θB ) or vB = sin θA sin θB mA vi sin θA . mB sin(θA + θB ) With the given numbers, vB = 18 (7.1 m/s) sin 60◦ = 2.057 m/s ≃ 2.06 m/s. 54 sin 85◦ (c) First let’s compute the initial and final kinetic energies: 1 1 2 Ki = mA vi = (0.018 kg)(7.1 m/s)2 = 0.45369 J. 2 2 1 1 1 1 2 2 Kf = mA vA + mB vB = (0.018)(3.012)2 + (0.054)(2.057)2 = 0.19594 J. 2 2 2 2 Now, the fraction of kinetic energy lost will be the amount lost, divided by the initial kinetic energy: Kf Ki − Kf =1− = 0.568. f= Ki Ki Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111 Spring 2012 PS #8 Solutions Page 6 of 9 Problem 7: (a) Momentum is conserved in explosions as well as collisions. The initial momentum of the object is purely in the +x direction: pix = (24.0 kg)(397 m/s) = 9528 kg m/s, piy = 0. The final momentum of the three pieces, combined, must equal the initial momentum: pf = pi =⇒ pf,1 + pf,2 + pf,3 = pi . We can calculate the momenta of two of the fragments from the given information: pf x,1 = 0 pf y,1 = (9.5 kg)(337 m/s) = 3201.5 kg m/s pf x,2 = (4.5 kg)(−433 m/s) = −1948.5 kg m/s pf y,2 = 0. So, momentum conservation in the x direction says 0 + (−1948.5 kg m/s) p x,3 = 9528 kg m/s =⇒ pf PDF reader This document requires+a fJavaScript-enabledx,3 = 11476.5 kg m/s, and in the y direction, we get such as Adobe Reader. (3201.5 kg m/s) + 0 + pf y,3 = 0 =⇒ pf y,3 = −3201.5 kg m/s. This gives us an overall momentum, for the third piece, of magnitude pf,3 = 11476.52 + (−3201.5)2 = 11915 kg m/s. Since this piece has a mass of 24.0 − 9.5 − 4.5 = 10.0 kg, this means its final speed is pf,3 vf,3 = = 1191.5 m/s ≃ 1.19 km/s. m3 (b) To find how much energy was released in the collision, we calculate the initial and final kinetic energies, and compare: 1 2 Ki = mtot vi = 1891 kJ 2 1 2 Kf,1 = m1 vf,1 = 539 kJ 2 1 2 Kf,2 = m2 vf,2 = 422 kJ 2 1 2 Kf,3 = m3 vf,3 = 7098 kJ 2 Kf = 539 + 422 + 7098 = 8059 kJ. So ∆K = 6.17 MJ. Copyright c 2012 University of Georgia. Unauthorize...
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