ee. in:.-"}$ .»s 3.5 m/S. Solution. In determining the velocity of recoil, consider only the horizontal component of the velocity of the cannon-ball, .since the recoil caused by the vertical component of this velocity will be damped by the reaction forces of the earth. The horizontal component of the velocity of the ball v = vo cos oi. According to the law of conservation of momentum, we have: M v, L-. = mvx, hence U _ mvx _ mvocosoi *“ M"* “M _° 97 _ P cos ot . v--—————P+Q . Note. See the solution to Problem 96. 7. Work. Energy. Power 2 2 98. F=-%;§ z 1,250 kgf. Solution. At the moment of firing the velocity of the barrel u is determined from the law of conservation of momentum and will be equal to ,,-2 _ M where rn and M are the masses of the shell and the gun barrel. The 2 kinetic energy @ imparted to the barrel at the moment of firing will be completely expended on the work against the braking force A = FS, i.e., the following equality will be true M uz 1-—¤¤——§ S 2 F Hence, M uz mzvz F 2S 2MS i’ 5 kg! 99. F:-é'? (vg-p2gh):1,25o kgr. Solution. At the end of fall the kinetic energy of the body will 2 be E =%-|- rngh where rn is the mass of the body.
170 ANSWERS AND SOLUTIONS The mean resistance of the sand can be determined from the equa- tion of the law of conservation of energy FS = E. Hence _ E mv§ mgh The problem can also be solved with the aid of Newton’S second law and by calculating the acceleration of the body as it penetrates into the sand. The velocity of the body as it hits the earth is v=`\/vg-{-2gh. . . . vz 12%-I-2gh The accelerat1on of the body in the sand is a=Tz§=--2f and the sand resistance F: nzazing- (03 -|-2gh) Since the weight mg is small compared with F, its action during the motion of the body in the sand is neglected. h h Z im kl"", ···"**') • Solution. The sledge at the top of the hill has a potential energy E = mgh. During motion this energy is expended on the work A, to overcome the forces of friction over the path DC and on the work A2 to overcome these forces over the path CB, i.e., E = mgh = A, + A2 The force of friction F, over the path DC Fi=kmg ——·L1 ‘\/g2.}.h2 where l is the length of AC. The work will be Ai = F,DC = klmg For the path CB the force of friction F2 = kmg and the work is Ag = FZCB = kmg (S -— l) Hence, and mgh=Ai-|-A2=mgkS h k==-S- The equation of Newton’s second law for the motion of the sledge over the path DC will be mg —·—·i...;—....··· — Fg = may 1/g2.|.h2
CHAPTER 1. MECHANICS 171 and therefore ap:—-;}.3;-..- (I--L) . Sirwe J-<1 a >0 and 1/[2+hZ S S ’ * the sledge will move over the path DC with a uniformly accelera- ted motion. The acceleration over the path CB is oz == -—kg and the sledge moves with a uniformly retarded motion. 101. (1) When the coefiicient of friction is constant the value of S will not change if the hill is more sloping. (2) The sledge will not move. With an angle of inclination tan on = h . . . =-3,- = k, the force of friction equal1zes the component of the force of gravity directed along the inclined surface, 102. The potential energy En of the parallelepiped in various positions will be: E, = 2mgl when it rests on the small side. E2 = mgl when it rests on the middle side. E3 = gg! when it rests on the large side. The most stable position corresponding to the body’s minimum potential energy will be in the last of the three cases. 103. A z 375 kgf-m. Solution. The work done against the air resistance is equal to the loss of the kinetic energy of the bullet: 2A:-{-7%--—£l-*5-:-[§(v§ -»¤) ,*:*,375 kgs-m 104. Solution. ln the first case the boy throwing the stone does an amount of work _ mv? Ar 2 If, in the second case, he throws the stone with the same force, he will do the same work but this work is now expended to 1mpart kinetic energy to the stone and to the boy. Therefore, mv? _ nw§ M uz 1 2 __ 2 J'- 2 ( ) whe1·e u is the velocity with which the boy moves. By the law of conservation of momentum, ITLU2 '—: Mu Solving these exguations it is possible to find the velocity of the stone in the secon case MU2=U1 • U2 <v4
172 ANSWERS AND ··SOLUT10NS and the velocity of the boy after he has thrown the stone 1/ ”‘” u=v -——-——-— 1 M (M —|-m) The velocity of the stone relative to the boy is v=v2—|—u=v, I/-1%*;,1-, v> v, Since the power N = Fv and the stone moves faster relative to the boy in the Second case than in the first (v > v,), the boy should develop a greateipfower in the second case. 105. v, = 1 s; A, = 15 kgf-m; A2 = 37.5 kgf-m; N, = = 10 kgf—m/S and N2 = 25 kgf-m/S. Solution. In both cases the man imparts the same acceleration a = Tg- to the boat A and therefore the velocity of the boat A (v, = 1 -.: at = % t = 1 m/S) will be the Same in both cases. In the first 1 2 case the work done by the man is A, = % = 15 kgf—m and in the second case 2 2 A2 :2%+2%;:37.5 kgf-m where v2 = -’§·t is the velocity of the second boat by the end of the z third second. The power developed by the man at the end of the third second in the first case is N, = Fv, = 10 kgf-m/S and in the second N2 '—= F (U1 + U2) = kgf*H]./S if N2 > N,. 106. Solution. The velocity of the body will be less in the second case since the body’s potential energy at a height h is expended in the first case to impart kinetic energy to the body alone, whilst in the se- cond case it IS used to impart kinetic energy to the body and the prism at the same time. 107. tan Br-= tan oz. Solution. Let us denote the velocity of the prism (Fig. 214) by u, the horizontal and vertical components of the load velocity v relative to the earth by vx and vy and the angle between the direction of motion of thB load and the h0l'iZOIlt.&l B, assuming t Dy 1 an B- vx < >
CHAPTER I. MECHANICS 173 ` . Since the prism is acted upon in a vertical direction by the reaction of the support in addition to the load, the law of conservation of momentum may be applied only to the horizontal components of the velocity of the load and the prism when the behaviour of the "load- p)rism" system is considered. The velocities u and vx will obviously e linked by the ratio • Mu =mvx (2) Let the load be situated at the point A of the prism at a certain moment of time (Fig. 215). During the first second a ter this the prism v RZ `~—` · = n \\ ua A ve ”é I Y (y vg iv l \ UV “ ¥ IL x J /1'///.;./·’>/i," 7///71 /1 . /,11, /1/ ///2//1 /// Fig. 214 Fig. 215 has moved u cm to the left and the load has been displaced vx cm horizontally to the right and by vy cm vertically. All these displace- ments should be such as to return the load again to the prism at a cer- tain point B. Therefore, the velocities u, vx and vy should satisfy the laws of conservation of energy and momentum and also the ratio Uy =-‘ t3I1 CZ This ratio expresses the condition that the moving load is always located on the prism. We fund from (2) that u : § vx. Inserting the value of u into equation (3), utilizing (1) and perform- ing simple transformations we get Uy m-{-M Z **' ;‘*'1""" t tan B vx M an or As we would expect, tan B > tan on and B_> oc. _ _ The velocity of the load down the moving prism is directed at a larger angle to the horizontal than during descent from a stationary prism. Using the law of conservation of energy and knowing the height of the initial position of the load it is possible to calculate the velocities u and v. _ _ 108. After the impact the balls will exchange their velocities. Solution. If the masses of the balls are denoted by mi and mz and the velocities after impact by as and y, we can obtain from the law of conservation of momentum maui + mzvz = mw? + mzy (1)
{74 ANSWERS AND SOLUTIONS Applying the law of conservation of energy and assuming that the total kinetic energy of the balls does not change after the impact we may write m,v§ m2v§ __ m1.1:2 mzyz 2" ‘+"_‘2 ‘ _f +" 2 ‘ (2) Solving equations (1) and (2) simultaneously and using mi = = mz = m, we get y = v, and a: = v2, i.e., perfectly elastic balls of equal mass will exchange their velocities after impact. If the first ball moves from the left to the right with velocity v, before impact, it will move in the opposite direction with velocity vz after impact. 109. M2 = 300 kg. Solution. Momentum before meeting ’ after meeting First boat (M 1-|-m) v M iv Second boat -M2v -(M2-|-m)v2 The momentum of the two boats before the load is shifted is equal to their momentum after _the transfer of the load, i.e., (M1+ m)v···MzU= M1v“(M2+m)v2 Hence, M2= =300 kg. U—U2 The energy of the boats before they meet is E (M1+Mz+m)v2 1:-__—'-2_-—— and after reshifting the load the energy of the boats is __ M10? (Mz+m) vi E2 ··~ T + -5* _ The energy has diminished due to the transfer of a part of the energy 1Dto heat when the velocities of the load and the second boat become the same. h 110. N=—”%-==49>< 1011 erg/s:/490 kW. 111. F = 4,896 kgf. Solution. The useful power N2 = kN, = Fu. Hence, F N I Ng=—i;Z— and F:-§;%4,896 kgf
CHAPTER I. MECHANICS 175