Problem Set #8 Solutions

Numerically this works out to be v 2070981342

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Unformatted text preview: cceleration kinematics lets us find their post-collision speed: 2 2 vf = vi + 2a(∆x) =⇒ v 2 = 2µk g (∆x). Numerically, this works out to be v= 2(0.70)(9.81)(34.2) = 21.67 m/s. From this, we get the post-collision momentum: pf = Mtot v = (1950.0 + 1413.0)(21.67) = 72885 kg m/s. Now, momentum is a vector quantity, and we just found the magnitude of the final momentum. Its components, given the direction of the cars is θ = 63.8◦ East of North, is pf x = pf sin θ = 65397 kg m/s pf y = pf cos θ = 32179 kg m/s. Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111 Spring 2012 PS #8 Solutions Page 4 of 9 Finally, we apply the conservation of momentum principle. Conveniently, the two cars were moving in the +x (cop) and +y (us) directions before the collision. This means pix = pf x =⇒ pcop = pf x piy = pf y =⇒ pus = pf y . and From this, we can calculate initial speeds: mcop vcop = pf x pf x vcop = mcop 65397 = 33.54 m/s. vcop = 1950.0 mus vus = pf y pf y vus = mus 32179 vus = = 22.77 m/s. 1413.0 After converting to more familiar units, this works out to be 75.0 mi/hr for the cop and This .document requires a JavaScript-enabled PDF reader 50 9 mi/hr for us. Yay law-abiding us! such as Adobe Reader. Problem 6: We begin by setting up the equations for this problem. In this 2-D collision, momentum is conserved in each coordinate direction. However, we can’t assume that energy is conserved. (In fact, reading ahead to part (c) should persuade us that this is an inelastic collision). In the x direction, the conservation of momentum equation can be written as pix = pf x =⇒ mA vi = mA vA cos θA + mB vB cos θB . In the y direction, there was no momentum to start with, so there’s no final momentum either: piy = pf y =⇒ 0 = mA vA sin θA − mB vB sin θB . A vi B A vA θA vB θB B We write the second term with a negative sign because puck B heads off below the x axis. (a) We have two momentum conservation equations, and two unknowns. To find vA , we first solve the y equation for vB (actually for mB vB ): sin θA . mB vB sin θB = mA vA sin θA =⇒ mB vB = mA vA sin θB Copyright c 2012 University of Georgia. Unauthorized duplication or distribution prohibited. Physics 1111 Spring 2012 PS #8 Solutions Page 5 of 9 Now we can substitute this into the x equation: mA vA cos θA + mA vA sin θA cos θB = mA vi . sin θB Canceling out the common mass mA and combining terms, we get vA cos θA + sin θA cos θB sin θB = vi =⇒ vA cos θA sin θB + sin θA cos θB sin θB = vi . Solving for vA gives vA = vi sin θB vi sin θB . = cos θA sin θB + sin θA cos θB sin(θA +...
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