chp11 - " Problem 3 A 62-kg parachutist hits the ground...

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Unformatted text preview: ," Problem 3. A 62-kg parachutist hits the ground moving at 35'km/hand come~ to a stop in 140 IUS.Find the ' average impulsive force on the chutist, and compare 1with the chutist's weight. Problem 17. In a railroad switchyard, a 56-ton freight car is se'ntat 7.0 mi/h toward a 31-ton car that is moving in the same direction at 2.6 mi/h: ' .' - . (a) What is the speed of the pair after they couple together? (b) What fraction of the initial kinetic energy was lost in the collision? Solution ' I The average impulsive force (from Equations 11-1 and 2) is Fav = Ap/f),t = (0 - mVi)/D.t = -(62 kg)x (35 m/3.6 s)/(0.14 s) = -4.31 kN (opposite to the di;rectionof Vi, or upward for most jumps under calm' wind conditions). The magnitude is slightly over seven times the parachutist's weight, or 7.09 mg. _ Solution ! "'----_._-------~-,~ (a) If we assume the switchyard track is straight and level, the collision is one-dimensional, totally i~elasticl and Equation 11-4 applies: vf = [(56T){7.0 mi/h) + ~ (31 T)(2.6 mi/h)]/(56 + 31)T = 5.43 mi/h. (b) The Problem -..-7. Safety st~ndards call for a 1900-kgc~r colliding at J 12 m/s WIth a concrete wall to expenence an _/ ~ -- average impulsive force not greater than 50 kN. What is the minimum permissible time for the car to come to a stop during such a collision?~ " initial and final kinetic energies are Ki = ![(56 T)x ~(7.0mi/h)2 + (31 T)(2.6 mifh)2] = 1477T(mi/h)2j Kf = ~(56 + 31)T (5.43 mi/h)2 = 1284 T(ini/hf The fraction lost is (Ki - Kf )/Ki = 13.1%. (Note: It was . not necessary to change to standard units to answer ~is question.) ../" Problem - - 33; While'playing ball in the street, a child ,Sol,ution Frotn Equation 11-2, if Fav = D.p/ D.t ::; 50 kN, then D.t ~ D.p/50 kN. When the car is brought to rest in a collision with the wall, the magnitude of its momentum change is D.p = 10- mVil = (1900 kg) x -(12:mjs), so D.t ~ 0.456 s. (Note: ifthe car had bounced off the wall during the collis, D.p would ;' ion, have been larger, as well as D.t.) / accidentally tosses a 18 m/s toward the' front of a car moving toward him at 14 m/s. ' What is the speed of the ball after it rebounds elastically from the car? Solution In a head-on elastic collision, the relative velocity of separation is equal to the negative of the relative velocity of approach (see Equation 11-8). If m2 ~ mi • (a car versus a ball), then v2f Rl V2i = 14 mis, and , VI! = V2! + ~2i - VIi = 14 IIl/s+ 14 m/s ! ('-18 m/s) =46 m/s. (We chose positive velocities in • . the direction of the car.) _ ••••• :f Problem . 14. An object like a ball used, in a sporting event can , be characterized by its coefficient of " restitution, defined as the ratio of outgoing to incident speed when the ball collides with a rigid; surface. The coefficientof restitution of a typical' tennis ball is about 0.7. What fraction of the ball's kinetic energy is lost at each bounce? I Problem Solution Thecoetficient of restitution given implies v f , -0. 7v i for each bounce (perpendicular to the rigid surface). The fractional change in kinetic energy is " (Vf Vi ~ 1 2/ 2) !:J.l(/Ki=(Kf-Ki)/Ki= = l \ Solution, I 1 I 35. A proton moving at 6.9 Mm/s collides elastically and head-on with a second proton moving in the opposite direction at 11 Mm/s. Find their velocities after the collision. (o:tf-l 1) = An elastic, head-on collision between two protons is described by Equations 11-9a and b with ml = m2' Therefore, vif = V2i = -11 Mm/s, and v2f = Vii :::: \ 6.9 Mm/s; i.e., the protons simply exch~e places. ~ = -0.51, or a loss of 51%. . II' -, 1; Pro~lem , 37. Two objects, one initially at rest, undergo a one.. _dimensional elastic collision. If half the kinetic energy of the initially moving object is-transferredto the other object, what is the ratio of their masses? Problem 65. A 1400-kg car moving at 75 km/h runs into a -1200=kgcar moving in the same direction at 50 kril/h (Fig. 11-26). The two cars lock together .' and both drivers immediately slam on their brakes. If the cars come to rest in a distance of 18 m, what is the coefficient of friction? Solution If one sets V2i = 0 in Equations 1l-9a and b, one _'_ - obtains -vlf = (ml - m2)vli/(ml + m2) and v2f = 2mlvli/(ml + m2)' (This describes a one..dimensional , elastic collision between two objects, one initially at rest.) If halfthe kinetic energy of the first object is transferred to the second, ~Kli = ~mlv~i = K2! = !m2[2mlvl'i/(ml + m2)j2, or 8mlm2 = (ml + m2)2 . .The resulting quadratic equation, m~ - 6mlm2 + m~ = 0, has two solutions, ml = (3 :f: y'8)m2 = 5.83m2 or (5.83)-lm2' Since the quadratic is symmetric in ml and m2, one solution equals the other with nil and m2 interchanged. Thus, one object- • is 5.83 times more massive than the other. ...--// FIGURE 11-26 Problem 65. '---.._---_......:..-_----Problem Solution . Tb.,i$isa one..dimensional, totally inelastic collision, so . the final velocity (positive in the initial direction of ! motion) is ' VI = .45. Two pendulums of equal length l = 50 cm are suspended from the same point. The pendulum bobs are steel spheres with masses of 140 and 390 g. The more massive bob is drawn back to make a 15° angle with the vertical (Fig. 11-21). When it is released, the bobs collide elastically. What is the maximum angle made by the less massive pendulum? (1400 kg)(75 km/h) + (1200 kg)(50 km/h) " 1400 kg+ 1200 kgI' = 63.5km/h = 17.6 m/s. I \ IOn a horizontal road (normal force equals weight, and' 1 no change in-gravitational potential,energy), the I work-energy theorem implies Wnc = -J1.kmgx = \ilK = therefore -4mvli I '~29X _ V]. ~ (17.6 m/s)2 _ 0 88 2)(18 m) - . . - 2(9',8m/s FIGURE 11-21 Problem 45. Solutfon The speed of the larger bob before the collision is (froriJ. onservation of energy) Vii = J2g£(1 - cos 15°) c (see Example 8-6), while for the smaller bob, V2i = O. After the collision, the smaller bob will reach a I maximum angle given by (again from conservation of , energy) v~! = 2g£(1 - cos O2), where V2! can be found from Equation 11-9b. Therefore, v2f - 2 _ ( ,2ml ,)2 v2. = (2(390»)2 2g£(1 _ cos 150).' ml + m2 11 530 ( = 2g£(I-cos02), 'cosO2 - or ~ '0.926, and 02 - 22.2°. ...
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This note was uploaded on 07/27/2009 for the course PHYSICS 101 taught by Professor Wormer during the Spring '08 term at NYU Poly.

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