Physics 141 Second Mid-Term Examination Spring 2015 March 31, 2015

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1 Physics 141 Second Mid-Term Examination Spring 2015 March 31, 2015 Your Number Solutions LAST Name (print) FIRST Name (print) Signature: UIN #: Your Section: Barkan 9 am Barkan 11 am Goeckner 3 pm PROBLEM POINTS SCORE S1. & S2. 10 S3. & S4. 10 L1. 20 L2. 25 L3. 20 L4. 15 Total Textbooks, cell phones, or any other forms of wireless communication are strictly prohibited in this or any exam. Giving or receiving aid in an examination is cause for dismissal from the University. Any other violation of academic honesty can have the same effect. 2. Perform the necessary calculations in the space provided. If additional space is needed, use the back of the question sheets. 3. ALL WORK MUST BE SHOWN IN ORDER TO RECEIVE FULL CREDIT. 4. All answers should include appropriate units. 5. Clearly circle or box your answers. Advice: Read each a question completely before doing any part of it. Total of 10 pages, double-sided on 5 sheets (plus equation sheet)

2 SHORT PROBLEMS (4 problems 20 points total) Partial credit may be given. S1. (6 points) An object is placed on the cylindrical inner surface of an object lying on its side, so that the direction of gravity in the image is down. If the coefficient of friction between that object and the cylindrical surface is µ S, find the maximum value of the angle θ (that is, find θ MAX ) for which the object won t slide. (For your convenience, in the drawing, the hypotenuse of the triangle outside the circle is tangent to the circle at the point of contact.) S2. (4 points) You throw a ball with a mass of kg against a brick wall. It hits the wall moving horizontally to the left at 30.0 m / s and rebounds horizontally to the right at 20.0 m / s. Find the impulse of the net force on the ball during its collision with the wall. 2 10

3 SHORT PROBLEMS (4 problems 20 points total) Partial credit may be given. S1. (6 points) An object is placed on the cylindrical inner surface of an object lying on its side, so that the direction of gravity in the image is down. If the coefficient of friction between that object and the cylindrical surface is µ S, find the maximum value of the angle θ (that is, find θ MAX ) for which the object won t slide. (For your convenience, in the drawing, the hypotenuse of the triangle outside the circle is tangent to the circle at the point of contact.) S2. (4 points) You throw a ball with a mass of kg against a brick wall. It hits the wall moving horizontally to the left at 30.0 m / s and rebounds horizontally to the right at 20.0 m / s. Find the impulse of the net force on the ball during its collision with the wall. Initial: Final: pp ii = (0.400 kkkk)(30.0 mm ss)( xx ) pp ff = (0.400 kkkk)(20.0 mm ss)(+xx ) JJ = Δpp = pp ff pp ii = (0.400 kkkk)(20.0 mm ss)(+xx ) (0.400 kkkk)(30.0 mm ss)( xx ) = ( kkkk mm ss)xx 2 10

4 S3. (4 points) Determine the (x,y) coordinates of the center-ofmass for the system of four masses shown in the figure. 4 y [m] 6 kg (8kkkk)(1mm) + (9kkkk)(2mm) + (7kkkk)(4mm) + (6kkkk)(4mm) XX CCCC = 9kkkk + 8kkkk + 7kkkk + 6kkkk kg 7 kg = 2.6 m 1 9 kg x [m] (8kkkk)(2mm) + (9kkkk)(1mm) + (7kkkk)(2mm) + (6kkgg)(4mm) YY CCCC = 9kkkk + 8kkkk + 7kkkk + 6kkkk = 2.1 m (2.6 m, 2.1 m) S4. (6 points) A block is sliding down an incline at 30 from the horizontal. At one point it is going 3 m / s, and after sliding 4 meters along the incline it is going 5 m / s. Use work-energy methods to find the coefficient of kinetic friction, µ k. You may use the fact that the normal force has magnitude equal to m g cos(30 ). For each force, calculate WW = FF dd = FFFF cos φφ FFFF dd W N = [mgcos(θ)] d cos(90º) = 0 W f = [µ k mgcos(30 )] d cos(180º) = - µ k mgd cos(30 ) W g = [ mg ] d cos(90º-θ) = mgd sin(30 ) ff kk dd NN dd Next, use the work-kinetic energy theorem. FF gg dd WW TTTTTT = ΔKK = KK ff KK ii 0 μμ kk mmmmmm cos(30 ) + mmmmmm sin(30 ) = 1 mmvv 2 ff 2 1 mmvv 2 ii 2 μμ kk gggg(0.866) + 1 gggg = 1 vv 2 2 ff 2 1 vv 2 ii 2 μμ kk = or with only 1 significant figure μμ kk =

5 L1. (20 points) A car of mass M = 1000 kg travels at constant speed v around a banked turn. The road is banked at an angle θ and has a radius of curvature r = 83.0 m. Some partial credit can be earned by completing the Identify and Setup steps like on the written homeworks. (hint: try keeping things symbolic until the last step.) a) (8 pts) First consider an icy day, so there is no friction between the car s tires and the road. The driver finds that at exactly v = 25.0 m/s the car can stay on the banked road. Find the angle of the embankment θ in degrees. (hint: this part is slightly easier if you use the usual up/down/left/right coordinate system.) Known: M, g, v, r, ƒ ss = 0, aa = vv 2 / Unknown: NN, θ Target Variable: θ NN yy θ NN NN xx aa FF gg y-direction: x-direction: NN yy + FF gg = 0 NN cos θθ MMMM = 0 NN = MMMM/ cos θθ NN xx = MMaa NN sin θθ = MM vv2 vv2 sin θθ = MM tan θθ = vv2 gggg θθ = tan 1 vv2 = gggg tan 1 MMMM cos θθ 625mm 2 /ss 2 (9.81 mm/ss^2)(83.0mm) =

6 b) (12 pts) Next consider after the ice has melted, so that µ s causes a static friction force parallel to the surface of the road (i.e. pointing downwards and inwards). If the driver can now go v = 41.0 m/s at most without the car slipping, what is the value of µ s? (hint: Either choice of a coordinate system equally good, equal in workload; either choice requires two vectors to be split into components.) Known: M, g, v, r, aa = vv 2 /, θ from Part (a) Unknown: NN, ƒ, ss µ s, Target Variable: µ s The normal force and the weight are still in the same directions, and the radial acceleration too. Add the friction, and choose a coordinate system. Either way, break two vectors into components: NN NN yy NN vv θ yy xx ƒ ss,mmmmmm,yy FF gg θ NN xx ƒ ss,mmmmmm,xx aa ƒ ss,mmmmmm FF gg,vv uu θ FF gg FF gg,uu aa uu aa θ ƒ ss,mmmmmm aa vv y-direction: NN yy + FF gg + ƒ ss,mmmmmm,yy = 0 NN cos θθ MMMM μμ ss NN sin θθ = 0 NN = MMMM/(cos θθ μμ ss sin θθ) v-direction: NN + FF gg,vv = MMaa vv NN MMMM cos θθ = MM vv2 sin θθ NN = MMMM cos θθ + MM vv2 sin θθ x-direction: NN xx + ƒ ss,mmmmmm,xx = MMaa NN sin θθ + μμ ss NN cos θθ = MM vv2 MMMM (sin θθ + μμ (cos θθ μμ ss sin θθ) ss cos θθ) = MM vv2 (sin θθ + μμ ss cos θθ) = vv2 (cos θθ μμ gggg ss sin θθ) u-direction: ƒ ss,mmmmmm + FF gg,uu = MMaa uu μμ ss NN + MMMM sin θθ = MM vv2 cos θθ μμ ss MMMM cos θθ + MM vv2 vv2 sin θθ += MM cos θθ MMMM sin θθ μμ ss = vv^22 cccccc θθ ssssss θθ gggg cccccc θθ+ vv22 gggg ssssss θθ =

7 L2. (25 points) A block of mass M slides from rest down an incline from a height H; then on a horizontal surface it hits a mass m which is attached to a spring (and the other end of the spring is fixed); together they compress the spring of constant k; and then M returns back up the incline alone. All answers may include H, M, m, k, and g. There is no friction anywhere. a) (5 pts) Find the speed v 1 of M just before it hits m. b) (5 pts) Find the speed v 2 of the two masses together just after they begin to move together. c) (5 pts) Find the maximum distance x max that the masses together compress the spring. 6

8 d) (5 pts) After compressing the spring as in Part (c), the spring pushes the two masses back again, until x = 0, at which point M again separates from m. Find the final height h of the mass M when it returns up the incline. e) (5 pts) Meanwhile, mass m oscillates alone on the spring. Find the maximum compression x max of the spring for that motion. 7 25

9

10 L4. (15 points) A flywheel (a uniform disk of mass M and radius R that spins freely on its central axis) is initially spinning with a period T 1. A brake is applied briefly so that it slows at a constant rate as it turns thru an angle β. Afterward its period of rotation is found to be T 2 (T 2 > T 1 ). For this problem, keep all angular units in radians. All answers may include T 1, T 2, β, M, R, and g. (Okay, probably not g.) a) (5 pts) Find the angular acceleration a of the flywheel during the braking. b) (5 pts) Find the length of time t that the brake is applied. c) (5 pts) Find the work done by the brake

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