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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ball.at 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°. ...
View Full Document

Ask a homework question - tutors are online