Sol5 - Massachusetts Institute of Technology Physics Department Physics-8.01 Fall 1999 Solutions for Assignment 5 by Dru Renner Through out these

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: Massachusetts Institute of Technology - Physics Department Physics-8.01 Fall 1999 Solutions for Assignment # 5 by Dru Renner October 22, 1999 Through out these solutions, the following quantities will b e used: MS ME RE G g = 1:99 1030 kg = 5:98 1024 kg = 6:4 106 m = 6:67 1011 N m2=kg2 = 9:8 m=s2 is the mass of the sun is the mass of the earth is an approximate radius of the earth is the gravitational constant is an approximate acceleration of gravity at the surface E = K + U = mv + U 2 2 Please see the Figure 8.14 on page 205. First consider E = E1. (a) The turning points occur when v = 0 ) E = U . For E1 there is only one turning point, a left turning point at x 0:2. (b) The speed is maximum when U is minimum. For E1 the speed is maximum for x 1:0. The speed is a local minimum when U is a local maximum. For E1 the speed is local a minimum for x 1:6. Also for E1 there is a turning point at which v = 0 thus the minimum speed occurs for x 0:2. (c) The orbit is only bound if there are both left and right turning points. For E1 there is only a left turning point, so the orbit is unbound. Now consider E = E2 . (a) For E2 there is a left turning point at x 0:3 and a right turning point at x 3:0. 1 Problem 5.1 (Ohanian, page 205, problem 22) The law of conservation of mechanical energy states that (b) For 2 the speed is maximum for 1 0. For 2 the speed is a local minimum for 1 6. Also for 2 there are two turning points at which = 0 thus the minimum speed occurs for 0 3 and 3 0. (c) For 2 there is both a left and right turning point, so the orbit is bound. E x : E x : E v x : x : E Problem 5.2 Now consider = 3 . (a) For 3 there is a left turning point at 0 5 and a right turning point at 1 3. (b) For 3 the speed is maximum for 1 0. For 3 the speed is a local minimum for 1 6. Also for 3 there are two turning points at which = 0 thus the minimum speed occurs for 0 5 and 1 3. (c) For 3 there is both a left and right turning point, so the orbit is bound. E E E x : x : E x : E x : E v x : x : E (Ohanian, page 205, problem 25) The Table 8.1 on page 195 lists the two quantities you need \Yearly energy expenditure of the United States" 8 1019 J \Combustion of 1 gal. of gasoline" 1 3 108 J : We can use these quantities to make various conversion factors. The amount of gasoline required by the United States per year would be 8 1019 8 1019 J per year = 1 3 108 gal of gasoline per year = : : = 6 2 1011 gal of gasoline per year : : = 1 6 2 1011 gal of gasoline per year = 6 2 1011 365 gal of gasoline per day : : = : : Problem 5.3 (a) 1 7 109 gal of gasoline per day : : = (Ohanian, page 207, problem 43) The total force that 6000 Egyptians can move is F = 6000 360 N = 2 16 106 N : The maximum weight F Mg they can move is given by =) Mg = kM g = F k = 2 160 10 = 7 2 106 N 3 : 6 : : 2 The amount of gasoline required by the United States per day would be = M = Mg = 7:2 10 7:3 105 kg g g (b) The total power that 6000 Egyptians can deliver is P = 6000 0:20 hp = 1:2 103 hp Power in terms of force and velocity is given by the relation We need to be careful because the unit \hp" is not a metric unit. The conversion is 1 hp = 745:7 W Thus the velocity is given by P = 1:2 103 745:7 W 4:1 101 m=s v = F 2:16 106 N Problem 5.4 Once we convert from \hp" to the metric unit \W," we are assured the the answer has the units of \m=s." Let L be the length of the unstretched cord, h = 100 m, k = 100 N=m, m = 50 kg, and let x denote the distance the bungee jumper has fallen from the bridge. Bridge L x h d Water 3 P = Fv =) 3 2 v = P = 21::16 10 6 1 hp F 10 1 N The corresponding maximum mass M is 6 (a) Mechanical energy is conserved if we ignore air drag and if we assume that no heat is dissipated in the string. The relevant forces are then only gravity and the spring force, and these are conservative forces thus mechanical energy is conserved. The cord will not stretch until she has fallen a distance L. Thus for x < L her mechanical energy is given by Ex<L = mg(h x) + mv 2 2 2 For x > L she begins to stretch the cord, and her mechanical energy is given by Ex>L = mg(h x) + mv + k(x L) 2 2 E = mgh 2 She starts from rest on the bridge, so her mechanical energy is given by We want to choose L so that she comes to a stop just above the water thus her mechanical energy at the point is given by E = k(h L) 2 Mechanical energy conservation gives 2 mgh = k(h L) 2 2 =) L = h 2mgh 69 m k The bungee jumper will hang freely at the distance d above the water at which gravity and the spring force balance. This is given by (b) s k(h L d) = mg =) Problem 5.5 d = h L mg 26 m k If the rotation rate was too fast, i.e. the gravitational force was not strong enough to provide for the centripetal force needed, the material at the equator would begin to move outwards, hence the planet would not be stable. 4 (a) Section 9.1 beginning on page 212 calculates the acceleration at the surface of a planet (assuming a spherical planet). The result is given in Equation (6) on page 214. With the substitutions ME 7! M and RE ! R, the equation is 7 g = GM R2 where M and R are the mass and radius of the planet respectively. If the planet has a uniform density then the mass is given by 4 M = 3 R3 =) g = 4 GR 3 By the argument above, the maximum rotational rate of the planet is given when this acceleration is precisely the centripetal force for the motion of the material at the surface. This corresponds to a minimum period T . Circular motion gives the relation 2 g=a= v R =) v = gR = 4 G R R = R 4 G 3 3 q s s The period of motion is then given by 3 2 T = 2 vR = q 4 R = G R 3 G (b) s For the value = 3:0 g=cm3 = 3:0 103 kg=m3, the minimum period is Problem 5.6 T = 6:67 10311 3 103 6:9 103 s 1:9 hour 7:9 102 day s (Ohanian, page 239, problem 15) Please see Figure 9.42 on page 239. The center of mass, located at rCM = 0 in the gure, is given as rCM = 0 = m2 r2 m1 r1 =) m1 r1 = m2 r2 Now we suppose that the orbit of each star is a circle centered on rCM = 0. Then the centripetal acceleration for the motion of m1 is given by 1 1 a1 = m F = m (Gm1m22 = (r Gmr2 )2 1+ 2 1 1 r1 + r2 ) Similarly, 5 a2 = (r Gmr1 )2 + 1 2 The angular frequency for circular motion is given by !=a r For mass m1 the angular frequency is, G !1 = m2 (r + r )2 r1 1 2 Similarly for mass m2 , G !2 = m1 (r + r )2 r2 1 2 We can check that using m1 r1 = m2 r2 ) !1 = !2 = ! where ! is the period of the binary system. We would like to make this equality, i.e. symmetry between m1 and m2 , more apparent. a1 = (r Gmr2 )2 = !2r1 + 1 2 (1) (2) a2 = (r Gmr1 )2 = !2r2 + 1 2 Adding equation (1) and (2) gives G !2(r1 + r2) = (m1 + m2 ) (r + r )2 1 2 =) !2 = G((rm1 + m32 ) +r ) 1 2 The period is then given by 2 2 (r + r 3 T 2 = 4!2 = 4 (m 1 + m2)) G 1 2 6 Problem 5.7 2 (Ohanian, page 239, problem 16) We need to use the result from problem 5.6, which is 2 T = G(m4 + m ) (r1 + r2)3 1 2 =) 2 + r1 + r2 = G(m1 4 2m2)T !1=3 Let Msun be the mass of the sun and m1 = 10Msun and m2 = 25Msun be the masses of the black hole and supergiant respectively. The period is T = 5:6 days 4:84 105 s, and the distance between the stars is ! ! G(m1 + m2 )T 2 1=3 = 35 GMsun(4:84 105)2 1=3 3:0 1010 m d = r1 + r2 = 4 2 4 2 Problem 5.8 (a) (Ohanian, page 241, problem 28) The mechanical energy of the projectile just after the gun shot is E = mv 2 2 GM R E m E where ME and RE are the mass and radius, respectively, of earth m is the projectile mass and v is the \muzzle speed" of the gun. If the projectile is to just barely reach the distance of the moon, then the mechanical energy at that point is E = GME m D where D is the distance between the moon and the center of earth. Using the previous two equations and the conservation of mechanical energy gives mv2 GME m = GME m 2 RE D s v = 2GME (b) =) 4 1 1 RE D 1:1 10 m=s The gun must deliver the appropriate kinetic energy to the projectile. 2 1 1 K = mv = mGME R D 2 E 11 1:2 10 11 J 1:2 1011 J = 1:2 10 9 tons of TNT 29 tons of TNT 4:2 10 7 (c) The equations of motion for constant acceleration are v = at and x = at 2 L T = 2a 2 The time required for the projectile to traverse the length L = 500 m of the barrel is s For the projectile to achieve the velocity v = 1:1 104 m=s at that point, the acceleration must be Problem 5.9 v2 a = 2L 1:2 105 m=s2 (Ohanian, page 241, problem 30) (a) The orbital mechanical energy is given in equation (27) on page 226. Using the replacement MS ! ME and the values m = 3500 kg and r = 100 km + RE = 105 m + RE 7 gives E E1 = GMr m 1:07 1011 J 2 For the satellite at rest on the surface of the planet, the mechanical energy is E2 = GME m 2:18 1011 J R E The mechanical energy change is E = E2 E1 = 1:11 1011 J (b) Energy is required to raise the temperature of a material and energy is required to cause a material to pass from one phase to another phase. (This latter process occurs at one temperature.) Energy is often expressed as a calorie for such considerations. The conversion to Joules is 1 cal = 4:187 J The \heat of fusion" for aluminum is 95:3 kcal=kg 3:99 102 J=kg. This is the amount of energy required to cause one kilogram of aluminum to melt. Thus the energy required to melt the entire satellite is 8 Emelt = 3:99 102 3500 1:4 106 J (Actually another smaller amount of energy is required to raise the temperature of aluminum to its melting temperature.) The \heat of vaporization" for aluminum is 2520 kcal=kg 1:06 104 J=kg. This is the amount of energy required to cause one kilogram of aluminum to vaporize (boil). Thus the energy required to vaporize the entire satellite is Evapor = 1:06 104 3500 3:71 107 J Problem 5.10 The total energy for both processes is 3:85107 J. The energy change above is su cient to cause both melting and vaporization. (Ohanian, page 267, problem 12) Let m1 = 1500 kg and m2 = 3500 kg be the mass of the car and truck respectively. Label the north direction by the y-axis and the east direction by the x-axis. Then the car has velocity v1 = 80 km=h = 22:2 m=s along the y direction, and the truck has velocity v2 = 50 km=h = 13:9 m=s along the x direction. The momentum for the each is given by ^ p1 = m1v1 y = 3:330 104 y ~ ^ p ~2 = m2 v2 x = 4:865 104 x ^ ^ where the units for p are kg m=s2. ~ (a) The momentum after the collision is ~ ~ ~ P = (m1 + m2 )V = 5000 V Momentum conservation gives ~ ~ ~ P = p1 + p2 = 4:865 104 x + 3:330 104 y ^ ^ Thus the velocity of the two cars after the collision is 1 ~ V = 5000 (4:865 104 x + 3:330 104 y) = 9:73 x + 6:66 y ^ ^ ^ ^ The magnitude is 9 jV j = (9:73)2 + (6:68)2 11:8 m=s The direction is given by q = tan1 6:66 9:73 34 This means that both cars move 34 north of east. (b) The kinetic energy before the collision is 1 2 Kbef ore = m2v1 + m2v2 2 2 7:1 105 J The kinetic energy after the collision is Kaf ter = (m1 +2m2 )V The change in kinetic energy is 2 3:5 105 J K = Kaf ter Kbef ore 3:6 105 J Problem 5.11 Thus 3:6 105 J of kinetic energy is lost during the collision. (Ohanian, page 267, problem 13) The force of impact of the hydrogen atoms on the satellite is, as always, just ~ p F = d~ dt For convience, label the area of the satellite A = 1:0 m2, the density of ions = 107 cm3, the mass of each ion m = 1:7 1027 kg, and the speed of each ion v = 4 105 m=s. Consider a small time interval, t. Then the ions impart a small amount of momentum, p, to the satellite. This gives the relationship ~ ~ F ~ p t which will become exact in the limit t ! 1. (It turns out that if the solar wind is assumed to be uniform in time, then the relationship written is exactly true even for nite t.) The volume of solar wind that strikes the satellite during the time interval t is 10 V = Av t The number of ions in this volume is N = V = Av t If all the ions stick to the satellite, then the momentum delivered to the satellite, along the direction of the solar wind, is p = mvN = m Av2 t Therefore the force is 2 ~ ~ F p = m Avt t = m Av2 t which is exact in the limit t ! 0. Therefore the force is exactly ~ F = m Av2 = 2:72 109 N Problem 5.12 where the direction is given by the solar wind. (Ohanian, page 268, problem 27) As in Example 6 on page 254, we will use Equation (26) on page 254 which gives 1Z 1 Z z dV + Z z dV zCM = M z dV = M semi rod where we have broken the integral into two pieces: one piece, \semi", includes all the mass along the semicircular section, the other piece, \rod", includes all the mass along the straight section. The \semi" integral was worked out in Example 6. The integral is then given as Z semi z dV = 2 AR2 (Be careful. The M here is the mass of the entire section the M for Example 6 is the mass for only the \semi" section.) Now we need to calculate the remaining \rod" integral. Z rod z dV But this integral is clearly proportional to the z component of the center of mass for the rod, which is 0. We can also see this because z = 0 along the straight section. Therefore 11 Z rod z dV = 0 Therefore i 1 h zCM = M 2 AR2 + 0 The mass is given by M = Msemi + Mrod = A R + A2R = AR ( + 2) and then Problem 5.13 0 xCM = 0 (Ohanian, page 270, problem 45) Let v = 5:0 103 m=s and h = 2:5 104 m and m be the mass of the ballistic missile. Consider the equations of motion for an object of mass m with only horizontal speed v at an altitude of h. The equations of motion give 0 z = gt and x = v t 2 2 0 This object will strike the ground at the time tland = 2gh and a location s xland = v 2gh 0 s For the piece with mass m = m that falls immediately downward, v = 0, so 2 0 0 xland = 0 12 As in Example 6, the symmetry of the masses indicates that zCM = 22R + For the piece with mass m = m that does not fall immediately downward, momentum 2 conservation gives m 0 + m v = mv =) v = 2v 2 2 0 0 0 The subsequent motion gives xland = 2v 2gh s 7:1 105 m 0 0 The center of mass moves as if the missile never exploded, i.e. m = m and v = v. Then xland = v 2gh s Problem 5.14 which is clearly halfway between the landing points for the two fragments. (Ohanian, page 270, problem 50) This problem refers to problem 5.10. (a) The translational kinetic energy of the center of mass before the collision is 2 2 ~ (m1 + m2 )vCM = P2 = (m1 +2m2)V 2 2(m1 + m2 ) ~ where P is the total momentum and V is the velocity of the center of mass, both calculated in problem 5.10. This form illustrates that the translational kinetic energy is conserved if momentum is conserved. Thus the value is given by 2 (m1 + m2)vCM = 5000 (11:8)2 2 2 3:5 105 J The internal kinetic energy satis es equation (34) on page 260 K = Kint + (m1 + m2 )vCM 2 2 =) 2 Kint = K (m1 + m2 )vCM 2 So before the collision, Kint = Kbef ore 3:5 105 7:1 105 3:5 105 3:6 105 J where the value of Kbef ore was calculated in problem 5.10. (b) The translational kinetic energy of the center of mass after the collision is the same as before by the argument above. The value is given by 13 2 (m1 + m2)vCM = 5000 (1:18 102)2 = 3:5 105 J 2 2 After the collision, both the car and truck move as one object, so there is no internal kinetic energy. Kint = 0 14 ...
View Full Document

This note was uploaded on 11/23/2009 for the course 640 373 taught by Professor Staff during the Spring '08 term at Rutgers.

Ask a homework question - tutors are online