Solutions to Physics: Principles with Applications, 5/E, Giancoli Chapter 7

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1 CHAPTER 7 1. p = mv = (0.0 kg)(8.1 m/s) = 0.18 kg m/s.. During the throwing we use momentum conservation for the one-dimensional motion: 0 = (m boat + m child )v boat + m package v package ; 0 = (55.0 kg kg)v boat + (5.40 kg)(10.0 m/s), which gives v boat = 0.70 m/s (opposite to the direction of the package). 3. We find the force on the epelled gases from F = Æp/Æt = (Æm/Æt)v = (1300 kg/s)(40,000 m/s) = N. An equal, but opposite, force will be eerted on the rocket: N, up. 4. For this one-dimensional motion, we take the direction of the halfback for the positive direction. For this perfectl inelastic collision, we use momentum conservation: M 1 + M = (M 1 + M )V; (95 kg)(4.1 m/s) + (85 kg)(5.5 m/s) = (95 kg + 85 kg)v, which gives V= 4.8 m/s. 5. For the horizontal motion, we take the direction of the car for the positive direction. The load initiall has no horizontal velocit. For this perfectl inelastic collision, we use momentum conservation: M 1 + M = (M 1 + M )V; (1,500 kg)(18.0 m/s) + 0 = (1,500 kg kg)v, which gives V= 1.3 m/s. 6. For the one-dimensional motion, we take the direction of the first car for the positive direction. For this perfectl inelastic collision, we use momentum conservation: M 1 + M = (M 1 + M )V; (9500 kg)(16 m/s) + 0 = (9500 kg + M )(6.0 m/s), which gives M = kg. 7. We let V be the speed of the block and bullet immediatel after the embedding and before the two start to rise. For this perfectl inelastic collision, we use momentum conservation: mv + 0 = (M + m)v; (0.01 kg)(10 m/s) = (0.01 kg kg)v, which gives V = 3.10 m/s. For the rising motion we use energ conservation, with the potential reference level at the ground: KE i + PE i = KE f + PE f ;!(M + m)v + 0 = 0 + (m + M)gh, or h = V /g = (3.10 m/s) /(9.80 m/s ) = m. v M m V m+m Page 7 1

2 8. On the horizontal surface after the collision, the normal force is F N = (m + M)g. We find the common speed of the block and bullet immediatel after the embedding b using the work-energ principle for the sliding motion: W fr = ÆKE; µ k (m + M)gd = 0!(M + m)v ; 0.5(9.80 m/s )(9.5 m) =!V, which gives V = 6.8 m/s. For the collision, we use momentum conservation: mv + 0 = (M + m)v; (0.015 kg)v = (0.015 kg kg)(6.8 m/s), which gives v = m/s. m v M F fr V F N (m+m)g 9. The new nucleus and the alpha particle will recoil in opposite directions. Momentum conservation gives us 0 = MV m α v α, 0 = (57m α )V m α ( m/s), which gives V = m/s. 10. Because mass is conserved, the mass of the new nucleus is M = u 4.0 u = 18 u. Momentum conservation gives us M 1 V 1 = M V + m α v α, ( u)(40 m/s) = (18 u)(350 m/s) + (4.0 u)v α, which gives v α = m/s. 11. Momentum conservation gives us m + M = m + M, (0.013 kg)(30 m/s) + 0 = (0.013 kg)(170 m/s) + (.0 kg), which gives = 0.39 m/s. m = 0 M M m 1. (a) With respect to the Earth after the eplosion, one section will have a speed and the other will have a speed = + v relative. Momentum conservation gives us mv =!m +!m, or v =! +!( + v relative ) = +!v relative ; m/s = +!( m/s), which gives = m/s. The other section will have = + v relative = m/s m/s = m/s. (b) The energ supplied b the eplosion increases the kinetic energ: E = ÆKE = [!(!m) +!(!m) ]!m = [!(!975 kg)( m/s) +!(!975 kg)( m/s) ]!(975 kg)( m/s) = J. Page 7

3 13. If M is the initial mass of the rocket and m is the mass of the Before epelled gases, the final mass of the rocket is = M m. Because the gas is epelled perpendicular to the rocket in the rocket s frame, it will still have the initial forward velocit, so the velocit of the rocket in the original direction will not change. We find the perpendicular component of the rocket s velocit after firing from = v 0 tan θ = (115 m/s) tan 35 = 80.5 m/s. Using the coordinate sstem shown, for momentum conservation in the perpendicular direction we have = m, or (M m ) = m ; (3180 kg m )(80.5 m/s) = m (1750 m/s), which gives m = 140 kg. v 0 After gas θ 14. We find the average force on the ball from F = Æp/Æt = (Æmv/Æt) =[( kg/s)(65.0 m/s) 0]/( s) = 130 N. Because the weight of a 60-kg person is 600 N, this force is not large enough. 15. We find the average force on the ball from F = Æp/Æt = m Æv/Æt =(0.145 kg)[(5.0 m/s) ( 39.0 m/s)]/( s) = N. 16. (a) We find the impulse on the ball from Impulse = Æp = m Æv =(0.045 kg)(45 m/s 0) =.0 N s. (b) The average force is F = Impulse/Æt = (.0 N s)/( s) = N. 17. The momentum parallel to the wall does not change, therefore the impulse will be perpendicular to the wall. With the positive direction toward the wall, we find the impulse on the ball from Impulse = Æp = m Æv = m[( v sin θ) (v sin θ)] = mv sin θ = (0.060 kg)(5 m/s) sin 45 =.1 N s. The impulse on the wall is in the opposite direction:.1 N s. v v θ 18. (a) With the positive direction in the direction of the fullback (East), the momentum is p = m fullback v fullback = (115 kg)(4.0 m/s) = kg m/s (East). (b) We find the impulse on the fullback from Impulse fullback = Æp fullback = kg m/s = kg m/s (West). (c) We find the impulse on the tackler from Impulse tackler = Impulse fullback = kg m/s (East). (d) We find the average force on the tackler from F tackler = Impulse tackler /Æt = ( kg m/s)/(0.75 s) = N (East). Page 7 3

4 19. (a) The impulse is the area under the F vs. t curve. The value of each block on the graph is 1 block = (50 N)(0.01 s) = 0.50 N s. We estimate there are 10 blocks under the curve, so the impulse is Impulse = (10 blocks)(0.50 N s/block) 5.0 N s. (b) We find the final velocit of the ball from Impulse = Æp = m Æv; 5.0 N s = (0.060 kg)(v 0), which gives v = 83 m/s. F (N) t (s) TZ 0. The maimum force that each leg can eert without breaking is ( N/m )( m ) = N, so, if there is an even landing with both feet, the maimum force allowed on the bod is N. We use the work-energ principle for the fall to find the landing speed: 0 = ÆKE + ÆPE; 0 =!mv land 0 + (0 mgh ma ), or v land = gh ma. The impulse from the maimum force changes the momentum on landing. If we take down as the positive direction and assume the landing lasts for a time t, we have F ma t = m Æv = m(0 v land ), or t = mv land /F ma. We have assumed a constant force, so the acceleration will be constant. For the landing we have = v land t +!at = v land (mv land /F ma ) +!( F ma /m)(mv land /F ma ) =!mv land /F ma = mgh ma /F ma ; 0.60 m = (75 kg)(9.80 m/s )h ma /( N), which gives h ma = 69 m. 1. For the elastic collision of the two balls, we use momentum conservation: + m = + m ; (0.440 kg)(3.70 m/s) + (0.0 kg)(0) = (0.440 kg) + (0.0 kg) Because the collision is elastic, the relative speed does not change: = ( ), or 3.70 m/s 0 = Combining these two equations, we get = 1.3 m/s, and = 4.93 m/s.. For the elastic collision of the two pucks, we use momentum conservation: + m = + m ; (0.450 kg)(3.00 m/s) + (0.900 kg)(0) = (0.450 kg) + (0.900 kg) Because the collision is elastic, the relative speed does not change: = ( ), or 3.00 m/s 0 = Combining these two equations, we get = 1.00 m/s (rebound), and =.00 m/s. 3. For the elastic collision of the two billiard balls, we use momentum conservation: + m = + m ; m(.00 m/s) + m( 3.00 m/s) = m + m Because the collision is elastic, the relative speed does not change: = ( ), or.00 m/s ( 3.00 m/s) = Combining these two equations, we get = 3.00 m/s (rebound), and =.00 m/s. Page 7 4

5 Note that the two billiard balls echange velocities. Page 7 5

6 4. For the elastic collision of the two balls, we use momentum conservation: + m = + m ; (0.060 kg)(.50 m/s) + (0.090 kg)(1.00 m/s) = (0.060 kg) + (0.090 kg) Because the collision is elastic, the relative speed does not change: = ( ), or.50 m/s 1.00 m/s = Combining these two equations, we get = 0.70 m/s, and =.0 m/s. 5. (a) For the elastic collision of the two balls, we use momentum conservation: + m = + m ; (0.0 kg)(5.5 m/s) + m (0) = (0.0 kg)( 3.7 m/s) + m Because the collision is elastic, the relative speed does not change: = ( ), or 5.5 m/s 0 = ( 3.7 m/s), which gives = 1.8 m/s. (b) Using the result for in the momentum equation, we get m = 1.1 kg. 6. (a) For the elastic collision of the two bumper cars, we use momentum conservation: + m = + m ; (450 kg)(4.50 m/s) + (550 kg)(3.70 m/s) = (450 kg) + (550 kg) Because the collision is elastic, the relative speed does not change: = ( ), or 4.5 m/s 3.7 m/s = Combining these two equations, we get = 3.6 m/s, and = 4.4 m/s. (b) r the change in momentum of each we have Æp 1 = ( ) = (450 kg)(3.6 m/s 4.50 m/s) = 396 kg m/s; Æp = m ( ) = (550 kg)(4.4 m/s 3.70 m/s) = As epected, the changes are equal and opposite kg m/s. 7. (a) For the elastic collision of the two balls, we use momentum conservation: + m = + m ; (0.80 kg) + m (0) = (0.80 kg) + m (! ). Because the collision is elastic, the relative speed does not change: = ( ), or 0 =!, which gives =!. Using this result in the momentum equation, we get m = kg. (b) The fraction transferred is fraction = ÆKE /KE 1 =!m ( )/! =!m [(! ) 0]/! = (m / = ((0.840 kg)/(0.80 kg) = Page 7 6

7 8. We find the speed after falling a height h from energ conservation:!m = Mgh, or v = (gh) 1/. The speed of the first cube after sliding down the incline and just before the collision is = [(9.80 m/s )(0.0 m)] 1/ = 1.98 m/s. For the elastic collision of the two cubes, we use momentum conservation: M + m = M + m ; M(1.98 m/s) +!M(0) = M +!M Because the collision is elastic, the relative speed does not change: = ( ), or 1.98 m/s 0 = Combining these two equations, we get = m/s, and =.64 m/s. Because both cubes leave the table with a horizontal velocit, the will fall to the floor in the same time, which we find from H =!gt ; 0.90 m =!(9.80 m/s )t, which gives t = 0.49 s. Because the horizontal motion has constant velocit, we have 1 = t = (0.660 m/s)(0.49 s) = 0.8 m; = t = (.64 m/s)(0.49 s) = 1.1 m. h v H 9. (a) For the elastic collision of the two masses, we use momentum conservation: + m = + m ; + 0 = + m Because the collision is elastic, the relative speed does not change: = ( ), or 0 = If we multipl this equation b m and subtract it from the momentum equation, we get ( m ) = ( + m ), or = [( m )/( + m )]. If we multipl the relative speed equation b and add it to the momentum equation, we get = ( + m ), or = [ /( + m )]. (b) When «m we have = [( m )/( + m )] [( m )/(m )] = ; = [ /( + m )] [( /m ) 0; so, 0; the small mass rebounds with the same speed; the large mass does not move. An eample is throwing a ping pong ball against a concrete block. (c) When» m we have = [( m )/( + m )] [( )/( )] = ; = [ /( + m )] ( / ) = ; so, ; the large mass continues with the same speed; the small mass acquires a large velocit. An eample is hitting a light stick with a bowling ball. (d) When = m we have = [( m )/( + m )] = 0; = [ /( + m )] = ( / ) = ; so 0, ; the striking mass stops; the hit mass acquires the striking mass s velocit. An eample is one billiard ball hitting an identical one. Page 7 7

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9 30. We let V be the speed of the block and bullet immediatel after the collision and before the pendulum swings. For this perfectl inelastic collision, we use momentum conservation: mv + 0 = (M + m)v; (0.018 kg)(30 m/s) = (0.018 kg kg)v, which gives V = 1.14 m/s. Because the tension does no work, we can use energ conservation for the swing:!(m + m)v = (M + m)gh, or V = (gh) 1/ ; 1.14 m/s = [(9.80 m/s )h] 1/, which gives h = m. We find the horizontal displacement from the triangle: L = (L h) + ; (.8 m) = (.8 m m) +, which gives = 0.61 m. m v L M θ L h m+m V m+m 31. (a) The velocit of the block and projectile after the collision is v = m /(m + M). The fraction of kinetic energ lost is fraction lost = ÆKE/KE = [!(m + M)v!m ]/!m = {(m + M)[m /(m + M)] m }/m = [m/(m + M)] + 1 = + M/(m + M). (b) For the data given we have fraction lost = M/(m + M) = (14.0 g)/(14.0 g g) = Momentum conservation gives 0 = + m ; 0 = + 1.5, or = 1.5 The kinetic energ of each piece is KE =!m ; KE 1 =! =!(m /1.5)( 1.5 ) = (1.5)!m = 1.5KE. The energ supplied b the eplosion produces the kinetic energ: E = KE 1 + KE =.5KE ; 7500 J =.5KE, which gives KE = 3000 J. For the other piece we have KE 1 = E KE = 7500 J 3000 J = 4500 J. Thus KE(heavier) = 3000 J; KE(lighter) = 4500 J. 33. On the horizontal surface after the collision, the normal force on the joined cars is F N = (m + M)g. We find the common speed of the joined cars immediatel after the collision b using the work-energ principle for the sliding motion: W fr = ÆKE; µ k (m + M)gd = 0!(M + m)v ; 0.40(9.80 m/s )(.8 m) =!V, which gives V = 4.68 m/s. For the collision, we use momentum conservation: mv + 0 = (m + M)V; ( kg)v = ( kg kg)(4.68 m/s), which gives v = 15 m/s (54 km/h). Page 7 9

10 34. (a) For a perfectl elastic collision, we use momentum conservation: + m = + m, or ( ) = m ( ). Kinetic energ is conserved, so we have! +!m =! +!m, or ( ) = m ( ), which can be written as ( )( + ) = m ( )( + ). When we divide this b the momentum result, we get + = +, or =. If we use this in the definition of the coefficient of restitution, we get e = ( )/( ) = ( )/( ) = 1. For a completel inelastic collision, the two objects move together, so we have =, which gives e = 0. (b) We find the speed after falling a height h from energ conservation:!m = mgh, or = (gh) 1/. The same epression holds for the height reached b an object moving upward: = (gh ) 1/. Because the steel plate does not move, when we take into account the directions we have e = ( )/( ) = [(gh ) 1/ 0]/{0 [ (gh) 1/ ]}, so e = (h /h) 1/. 35. Momentum conservation for the eplosion gives us 0 = + m ; 0 = + 3, or = 3 On the horizontal surface after the collision, the normal force on a block is F N = mg. We relate the speed of a block immediatel after the collision to the distance it slides from the work-energ principle for the sliding motion: W fr = ÆKE; µ k mgd = 0!m, or d =! /µ k g. If we use this for each block and form the ratio, we get d 1 /d = ( / ) = ( 3) = 9, with the lighter block traveling farther. Page 7 10

11 36. For the momentum conservation of this one-dimensional collision, we have + m = + m (a) If the bodies stick together, = = V: (5.0 kg)(5.5 m/s) + (3.0 kg)( 4.0 m/s) = (5.0 kg kg)v, which gives V = = = 1.9 m/s. (b) If the collision is elastic, the relative speed does not change: = ( ), or 5.5 m/s ( 4.0 m/s) = 9.5 m/s = The momentum equation is (5.0 kg)(5.5 m/s) + (3.0 kg)( 4.0 m/s) = (5.0 kg) + (3.0 kg), or (5.0 kg) + (3.0 kg) = 15.5 kg m/s. When we combine these two equations, we get (c) If comes to rest, = 0. = 1.6 m/s, = 7.9 m/s. (5.0 kg)(5.5 m/s) + (3.0 kg)( 4.0 m/s) = 0 + (3.0 kg), which gives = 0, = 5. m/s. (d) If m comes to rest, = 0. (5.0 kg)(5.5 m/s) + (3.0 kg)( 4.0 m/s) = (5.0 kg) + 0, which gives = 3.1 m/s, = 0. (e) The momentum equation is (5.0 kg)(5.5 m/s) + (3.0 kg)( 4.0 m/s) = (5.0 kg)( 4.0 m/s) + (3.0 kg), which gives = 4.0 m/s, = 1 m/s. The result for (c) is reasonable. The 3.0-kg bod rebounds. The result for (d) is not reasonable. The 5.0-kg bod would have to pass through the 3.0-kg bod. To check the result for (e) we find the change in kinetic energ: ÆKE = (! +!m ) (! v 1 +!m ) =![(5.0 kg)(5.5 m/s) + (3.0 kg)( 4.0 m/s) ]![(5.0 kg)( 4.0 m/s) + (3.0 kg)(1 m/s) ] = J. Because the kinetic energ cannot increase in a simple collision, the result for (e) is not reasonable. 37. Because the initial momentum is zero, the momenta of the three products of the deca must add to zero. If we draw the vector p neutrino θ diagram, we see that p nucleus = (p electron + p neutrino ) 1/ = [( kg m/s) + ( kg m/s) ] 1/ = kg m/s. We find the angle from tan θ = p neutrino /p electron = ( kg m/s)/( kg m/s) = 0.581, so the angle is 30.1 from the direction opposite to the electron s. p electron θ p nucleus Page 7 11

12 38. For the collision we use momentum conservation: -direction: + 0 = ( + m )v cos θ; (4.3 kg)(7.8 m/s) = (4.3 kg kg)v cos θ, which gives v cos θ = 3.39 m/s. -direction: 0 + m = ( + m )v sin θ; (5.6 kg)(10. m/s) = (4.3 kg kg)v sin θ, which gives v sin θ = 5.77 m/s. We find the direction b dividing the equations: tan θ = (5.77 m/s)/(3.39 m/s) = 1.70, so θ = 60. We find the magnitude b squaring and adding the equations: v = [(5.77 m/s) + (3.39 m/s) ] 1/ = 6.7 m/s. p 1 p p θ Page 7 1

13 39. (a) Using the coordinate sstem shown, for momentum conservation we have -momentum: m A v A + 0 = m A v A cos θ A + m B v B cos θ B ; -momentum: = m A v A sin θ A m B v B sin θ B (b) With the given data, we have : (0.400 kg)(1.80 m/s) = (0.400 kg)(1.10 m/s) cos 30 + (0.500 kg)v B cos θ B, which gives v B cos θ B = m/s; : 0 = (0.400 kg)(1.10 m/s) sin 30 (0.500 kg)v B sin θ B, which gives v B sin θ B = m/s. We find the magnitude b squaring and adding the equations: v B = [(0.440 m/s) + (0.678 m/s) ] 1/ = m/s. We find the direction b dividing the equations: tan θ B = (0.440 m/s)/(0.678 m/s) = 0.649, so θ B = v A Before After v A θ A θ B v B 40. (a) Using the coordinate sstem shown, for momentum conservation we have -momentum: mv + 0 = 0 + m cos θ, or cos θ = v; -momentum: = m + m sin θ, or sin θ = If we square and add these two equations, we get + = 4. For the conservation of kinetic energ, we have!m + 0 =!m +!(m), or =. When we add this to the previous result, we get = 3. Using this in the -momentum equation, we get cos θ = Ã3/, or θ = 30. (b) From part (a) we have = v/ã3. Using the energ result, we get = = /3 or = v/ã3. (c) The fraction of the kinetic energ transferred is fraction = KE /KE 1 =!(m) /!m = m( /3)/!m = %. p 1 p 1 θ p Page 7 13

14 41. Using the coordinate sstem shown, for momentum conservation we have -momentum: mv sin θ 1 + mv sin θ = 0, or θ 1 = θ. -momentum: mv cos θ 1 + mv cos θ = m ; mv cos θ 1 = mv/3; cos θ 1 or θ 1 = 70.5 = θ. The angle between their initial directions is φ = θ 1 + θ = (70.5 ) = 141. θ 1 θ 1 p 1 p p Page 7 14

15 4. Using the coordinate sstem shown, for momentum conservation we have -momentum: m + 0 = m cos α + M cos θ ; 5M(1.0 m/s) = 5M cos α + M cos 80, or 5 cos α = cos m/s. -momentum: 0 = m sin α + M sin θ ; 0 = 5M sin α + M sin 80, or 5 sin α = cos 80. For the conservation of kinetic energ, we have!mv =!m +!M ; 5M(1.0 m/s) = 5M + M, or 5 + = 70 m /s. We have three equations in three unknowns: α,, We eliminate α b squaring and adding the two momentum results, and then combine this with the energ equation, with the results: (a) = 3.47 m/s. (b) = 11.9 m/s. (c) α = 3.9. α m m M θ m = 5M M 43. Using the coordinate sstem shown, for momentum conservation we have : m n v n + 0 = m n v n cos θ 1 + m He v He cos θ ; m n ( m/s) = m n v n cos θ 1 + 4m n v He cos 45, or p n v n cos θ 1 = ( m/s) 4v He cos 45. p n θ 1 : = m n v n sin θ 1 + m He v He sin θ ; 0 = m n v n sin θ 1 + 4m n v He sin 45, or θ v n sin θ 1 = 4v He sin 45. For the conservation of kinetic energ, we have p He!m n v n + 0 =!m n v n +!m He v He ; m n ( m/s) = m n v n + 4m n v He, or v n + 4v He = m /s. We have three equations in three unknowns: θ 1, v n, v He We eliminate θ 1 b squaring and adding the two momentum results, and then combine this with the energ equation, with the results: θ 1 = 76, v n = m/s, v He = m/s. 44. Using the coordinate sstem shown, for momentum conservation we have Page 7 15

16 : 0 + m = m cos α + 0; 3.7 m/s = cos α; : m + 0 = m sin α + m ;.0 m/s = sin α +, or sin α =.0 m/s For the conservation of kinetic energ, we have!m +!m =!m +!m ; (.0 m/s) + (3.7 m/s) = +. We have three equations in three unknowns: α,, We eliminate α b squaring and adding the two momentum results, and then combine this with the energ equation, with the results: α = 0, = 3.7 m/s, =.0 m/s. The two billiard balls echange velocities. m m m m α Page 7 16

17 45. Using the coordinate sstem shown, for momentum conservation we have : m + 0 = m cos θ 1 + m cos θ, or = cos θ 1 + cos θ ; : = m sin θ 1 m sin θ, or 0 = sin θ 1 sin θ. For the conservation of kinetic energ, we have!m + 0 =!m +!m ; = +. We square each of the momentum equations: = cos θ 1 + cos θ 1 cos θ + cos θ ; 0 = sin θ 1 sin θ 1 sin θ + sin θ. If we add these two equations and use sin θ + cos θ = 1, we get = + (cos θ 1 cos θ sin θ 1 sin θ ) +. If we subtract the energ equation, we get 0 = (cos θ 1 cos θ sin θ 1 sin θ ), or cos θ 1 cos θ sin θ 1 sin θ = 0. We reduce this with a trigonometric identit: cos θ 1 cos θ sin θ 1 sin θ = cos(θ 1 + θ ) = 0, which means that θ 1 + θ = 90. m m θ 1 m θ m 46. We choose the origin at the carbon atom. The center of mass will lie along the line joining the atoms: CM = (m C C + m O O )/(m C + m O ) = [0 + (16 u)( m)]/(1 u + 16 u) = m from the carbon atom. 47. We choose the origin at the front of the car: CM = (m car car + m front front + m back back )/(m car + m front + m back ) = [(1050 kg)(.50 m) + (140 kg)(.80 m) + (10 kg)(3.90 m)]/(1050 kg kg + 10 kg) =.74 m from the front of the car. 48. Because the cubes are made of the same material, their masses will be proportional to the volumes:, m = 3 = 8, m 3 = 3 3 = 7. From smmetr we see that CM = We choose the -origin at the outside edge of the small cube: 0 CM = ( 1 + m + m 3 3 )/( + m + m 3 ) = { (! 0 ) + 8 [ 0 +!( 0 )] + 7 [ !(3 0 )]}/( ) = /36 = from the outer edge of the small cube. Page 7 17

18 49. We choose the origin at the center of the raft, which is the CM of the raft: CM = (M raft m + m 3 3 )/(M + + m + m 3 ) = [0 + (100 kg)(9.0 m) + (100 kg)(9.0 m) + (100 kg)( 9.0 m)]/[600 kg + 3(100 kg)] = 1.10 m (East). CM = (M raft m + m 3 3 )/(M + + m + m 3 ) = [0 + (100 kg)(9.0 m) + (100 kg)( 9.0 m) + (100 kg)( 9.0 m)]/[600 kg + 3(100 kg)] = 1.10 m (South). m 3 M m 50. We choose the coordinate sstem shown. There are 10 cases. CM = (5 + 3m + m 3 )/(10m) = [5(! ) + 3( +! ) + ( +! )]/(10) = 1.. CM = (7 + m + m 3 )/(10m) = [7(! ) + ( +! ) + ( +! )]/(10) = 0.9. The CM is 1. from the left, and 0.9 from the back of the pallet. 51. We know from the smmetr that the center of mass lies on a line containing the center of the plate and the center of the hole. We choose the center of the plate as origin and along the line joining the centers. Then CM = 0. A uniform circle has its center of mass at its center. We can treat the sstem as two circles: a circle of radius R, densit ρ and mass ρ¹(r) with 1 = 0; a circle of radius R, densit ρ and mass ρ¹r with = 0.80R. We find the center of mass from CM = ( 1 + m )/( + m ) = [4ρ¹R (0) ρ¹r (0.80R)]/(4ρ¹R ρ¹r ) = 0.7R. The center of mass is along the line joining the centers 0.07R outside the hole. R CM R C C 0.80R 5. If we assume a total mass of 70 kg, for one leg we have m leg = m bod!( )/100 = (70 kg)!(34.5)/100 = 1 kg. Page 7 18

19 53. If we measure from the shoulder, the percentage of the height to the center of mass for each of the segments is upper arm: = 9.5; lower arm: = 5.9; hand: = Because all masses are percentages of the bod mass, we can use the percentages rather than the actual mass. Thus we have CM = (m upper upper + m lower lower + m hand hand )/(m upper + m lower + m hand ) = [!(6.6)(9.5) +!(4.)(5.9) +!(1.7)(38.1)]/[!(6.6) +!(4.) +!(1.7)] = 19. The CM of an outstretched arm is 19% of the height. 54. We choose the shoulder as the origin. The locations of the centers of mass for each of the segments are hand upper arm: upper = [( ) /100] (155 cm) = 14.7 cm; upper = 0; lower arm: lower =[ (81. 6.) /100] (155 cm) = 9.5 cm; lower arm lower = [( ) /100] (155 cm) = 10.7 cm; hand: hand = [(81. 6.) /100] (155 cm) = 9.5 cm; upper arm hand = [( ) /100] (155 cm) = 9.6 cm. Because all masses are percentages of the bod mass, we can use the percentages rather than the actual mass. Thus we have CM = (m upper upper + m lower lower + m hand hand )/(m upper + m lower + m hand ) = [!(6.6)(14.7 cm) +!(4.)(9.5 cm) +!(1.7)(9.5 cm)]/[!(6.6) +!(4.) +!(1.7)] = cm. CM = (m upper upper + m lower lower + m hand hand )/(m upper + m lower + m hand ) = [!(6.6)(0) +!(4.)(10.7 cm) +!(1.7)(9.6 cm)]/[!(6.6) +!(4.) +!(1.7)] = 7.6 cm. 55. We use the line of the torso as the origin. The vertical head trunk and neck locations of the centers of mass for each of the segments, as a percentage of the height, are torso and head: 0; upper arms upper legs upper arms: ua = ( ) = 9.5; lower arms: la = ( ) = 5.9; lower arms hands: h = ( ) = 38.1; lower legs upper legs: ul = ( ) = 9.6; feet lower legs: ll = ( ) = 33.9; feet: f = ( ) = Because all masses are percentages of the bod mass, we can use the percentages rather than the actual mass. Thus we have CM = (m ua ua + m la la + m h h + m ul ul + m ll ll + m f f )/m bod = [(6.6)( 9.5) + (4.)( 5.9) + (1.7)( 38.1) + (1.5)( 9.6) + (9.6)( 33.9) + (3.4)( 50.3)]/100 = 9.4. The CM will be 9.4% of the bod height below the line of the torso. For a height of 1.8 m, this is about 17 cm, so es, this will most likel be outside the bod. Page 7 19

20 56. (a) If we choose the origin at the center of the Earth, we have CM = (m Earth Earth + m Moon Moon )/(m Earth + m Moon ) = [0 + ( kg)( m)]/( kg kg) = m. Note that this is less than the radius of the Earth and thus is inside the Earth. (b) The CM found in part (a) will move around the Sun on an elliptical path. The Earth and Moon will revolve about the CM. Because this is near the center of the Earth, the Earth will essentiall be on the elliptical path around the Sun. The motion of the Moon about the Sun is more complicated. 57. We choose the origin of our coordinate sstem at the woman. (a) For their CM we have CM = (m woman woman + m man man )/(m woman + m man ) = [0 + (90 kg)(10.0 m)]/(55 kg + 90 kg) = 6. m. (b) Because the CM will not move, we find the location of the woman from CM = (m woman woman + m man man )/(m woman + m man ) 6. m = [(55 kg) woman + (90 kg)(10.0 m.5 m)]/(55 kg + 90 kg), which gives woman = 4.1 m. The separation of the two will be 7.5 m 4.1 m = 3.4 m. (c) The two will meet at the CM, so he will have moved 10.0 m 6. m = 3.8 m. 58. Because the two segments of the mallet are uniform, we know that the center of mass of each segment is at its midpoint. We choose the origin at the bottom of the handle. The mallet will spin about the CM, which is the point that will follow a parabolic trajector: CM = (md + ML)/(m + M) = [(0.500 kg)(1.0 cm) + (.00 kg)(4.0 cm cm)]/ (0.500 kg +.00 kg) = 4.8 cm. L m d M 59. The CM will land at the same point, D from the launch site. If part I is still stopped b the eplosion, it will fall straight down, as before. (a) We find the location of part II from the CM: CM = (m I I + m II II )/(m I + m II ) D = [m I D + 3m I II ]/(m I + 3m I ), which gives II = 7D/3, or D/3 closer to the launch site. (b) For the new mass distribution, we have CM = (m I I + m II II )/(m I + m II ) D = [3m II D + m II II ]/(3m II + m II ), which gives II = 5D, or D farther from the launch site. Page 7 0

21 60. The forces on the balloon, gondola, and passenger are balanced, so the CM does not move relative to the Earth. As the passenger moves down at a speed v relative to the balloon, the balloon will move up. If the speed of the balloon is v relative to the Earth, the passenger will move down at a speed v v relative to the Earth. We choose the location of the CM as the origin and determine the positions after a time t: CM = (m balloon balloon + m passenger passenger )/(m balloon + m passenger ) 0 = [Mv t + m(v v )t]/(m + m), which gives v = mv/(m + m) up. If the passenger stops, the gondola and the balloon will also stop. There will be equal and opposite impulses acting when the passenger grabs the rope to stop. 61. We find the force on the person from the magnitude of the force required to change the momentum of the air: F = Æp/Æt = (Æm/Æt)v = (40 kg/s m )(1.50 m)(0.50 m)(100 km/h)/(3.6 ks/h) = N. The maimum friction force will be F fr = µmg (1.0)(70 kg)(9.80 m/s ) = N, so the forces are about the same. 6. For the sstem of railroad car and snow, the horizontal momentum will be constant. For the horizontal motion, we take the direction of the car for the positive direction. The snow initiall has no horizontal velocit. For this perfectl inelastic collision, we use momentum conservation: M 1 + M = (M 1 + M )V; (5800 kg)(8.60 m/s) + 0 = [5800 kg + (3.50 kg/min)(90.0 min)]v, which gives V= 8.16 m/s. Note that there is a vertical impulse, so the vertical momentum is not constant. 63. We find the speed after being hit from the height h using energ conservation:!mv = mgh, or v = (gh) 1/ = [(9.80 m/s )(55.6 m)]1/ = 33.0 m/s. We see from the diagram that the magnitude of the change in momentum is Æp = m( + v ) 1/ = (0.145 kg)[(35.0 m/s) + (33.0 m/s) ] 1/ = 6.98 kg m/s. We find the force from F Æt = Æp; F( s) = 6.98 kg m/s, which gives F = N. We find the direction of the force from tan θ = v /v = (33.0 m/s)/(35.0 m/s) = 0.943, θ = mv ² p θ mv 64. For momentum conservation we have : mv 0 = %mv, which gives v = 8v 0 ; : 0 0 ) %mv, which gives v = v 0. The rocket s forward speed increases because the fuel is shot backward relative to the rocket. Page 7 1

22 65. From the result of Problem 45, the angle between the two final directions will be 90 for an elastic collision. We take the initial direction of the cue ball to be parallel to the side of the table. The angles for the two balls after the collision are tan θ 1 = 1.0/Ã3.0, which gives θ 1 = 30 ; tan θ = ( )/Ã3.0, which gives θ = 60. Because their sum is 90, this will be a scratch shot θ 1 θ In the reference frame of the capsule before the push, we take the positive direction in the direction the capsule will move. (a) Momentum conservation gives us mv astronaut + Mv satellite = mv astronaut + Mv satellite, = (140 kg)(.50 m/s) + (1800 kg)v satellite, which gives v satellite = m/s. (b) We find the force on the satellite from F satellite = Æp satellite /Æt = m satellite Æv satellite /Æt =(1800 kg)(0.194 m/s 0)/(0.500 s) = 700 N. There will be an equal but opposite force on the astronaut. 67. For each of the elastic collisions with a step, conservation of kinetic energ means that the velocit reverses direction but has the same magnitude. Thus the golf ball alwas rebounds to the height from which it started. Thus, after five bounces, the bounce height will be 4.00 m. 68. For the elastic collision of the two balls, we use momentum conservation: + m = + m ; m + 0 = m( /4) + m, or m = 5m /4. Because the collision is elastic, the relative speed does not change: 0 = ( ); = ( /4), or = 3 /4. Combining these two equations, we get m = 5m/ On the horizontal surface, the normal force on a car is F N = mg. We find the speed of a car immediatel after the collision b using the work-energ principle for the succeeding sliding motion: W fr = ÆKE; µ k mgd = 0!m. We use this to find the speeds of the cars after the collision: 0.60(9.80 m/s )(15 m) =!v A, which gives v A = 13.3 m/s; 0.60(9.80 m/s )(30 m) =!v B, which gives v B = 18.8 m/s. For the collision, we use momentum conservation: m A v A + m B v B = m A v A + m B v B ; (000 kg)v A + 0 = (000 kg)(13.3 m/s) + (1000 kg)(18.8 m/s), which gives v A =.7 m/s. We find the speed of car A before the brakes were applied b using the work-energ principle for the preceding sliding motion: W fr = ÆKE; Page 7

23 µ k m A gd =!m A v A!m A v A0 ; 0.60(9.80 m/s )(15 m) =![(.7 m/s) v A0 ], which gives v A0 = 6.3 m/s = 94.7 km/h = 59 mi/h. Page 7 3

24 70. We choose the origin of our coordinate sstem at the initial position of the 75-kg person. For the location of the center of mass of the sstem we have CM = ( 1 + m + m boat boat )/( + m + m boat ) = [(75 kg)(0) + (60 kg)(.0 m) + (80 kg)(1.0 m)]/ (75 kg + 60 kg + 80 kg) = 0.93 m. Thus the CM will be.0 m 0.93 m = 1.07 m from the 60-kg person. When the two people echange seats, the CM will not move. The end where the 75-kg person started, which was 0.93 m from the CM, is now 1.07 m from the CM, that is, the boat must have moved 1.07 m 0.93 m = 0.14 m toward the initial position of the 75-kg person. m m boat CM L m boat CM L m 71. (a) We take the direction of the meteor for the positive direction. For this perfectl inelastic collision, we use momentum conservation: M meteor v meteor + M Earth v Earth = (M meteor + M Earth )V; (10 8 kg)( m/s) + 0 = (10 8 kg kg)v, which gives V= m/s. (b) The fraction transformed was fraction = ÆKE Earth /KE meteor =!m Earth V /!m meteor v meteor = ( kg)( m/s) /(10 8 kg)( m/s) = (c) The change in the Earth s kinetic energ was ÆKE Earth =!m Earth V =!( kg)( m/s) = 0.19 J. 7. Momentum conservation gives 0 = + m, or / = /m. The ratio of kinetic energies is KE /KE 1 =!m /! = (m / )( / ) =. When we use the result from momentum, we get (m / )( /m ) =, which gives /m =. 73. (b) The force would become zero at F (N) t = 580/( ) = s. 580 At t = s the force is 580 ( )( s) = + 40 N. The impulse is the area under the F vs. t curve, and consists of a triangle and a rectangle: 40 Impulse =!(580 N 40 N)( s) + 0 (40 N)( s) = 0.93 N s. (c) We find the mass of the bullet from Impulse = Æp = m Æv; 0.93 N s = m(0 m/s 0), which gives m = kg = 4. g. t (ms) 3.0 Page 7 4

25 74. We find the speed for falling or rising through a height h from energ conservation:!m = mgh, or = gh. (a) The speed of the first block after sliding down the incline m and just before the collision is = [(9.80 m/s )(3.60 m)] 1/ = 8.40 m/s. h For the elastic collision of the two blocks, we use momentum conservation: m + M = m + M ; (.0 kg)(8.40 m/s) + (7.00 kg)(0) = (.0 kg) + (7.00 kg) Because the collision is elastic, the relative speed does not change: = ( ), or 8.40 m/s 0 = Combining these two equations, we get = 4.38 m/s, = 4.0 m/s. (b) We find the height of the rebound from = gh ; ( 4.38 m/s) = (9.80 m/s )h, which gives h = m. The distance along the incline is d = h /sin θ = (0.979 m)/sin 30 = 1.96 m. θ M 75. Because energ is conserved for the motion up and down the incline, mass m will return to the level with the speed For a second collision to occur, mass m must be moving faster than mass M: ³ In the first collision, the relative speed does not change: 0 = ( ), or =, so the condition becomes ³, or ³ For the first collision, we use momentum conservation: m + 0 = m + M, or = (M/m) When we use the two versions of the condition, we get ³ 3, so we need (M/m) ³ 3, or m ² M/3. Page 7 5

26 76. (a) If the skeet were not hit b the pellet, the horizontal distance it would travel can be found from the range epression for projectile motion: R = ( 0 /g) sin θ = [(30 m/s) /(9.80 m/s )] sin (30 ) = 79.5 m. At the collision the skeet will have the -component of the initial velocit: = v 0 cos θ = (30 m/s) cos 30 = 6.0 m/s. We use energ conservation to find the height attained b the skeet when the collision occurs:!mv 0 =!Mv 1 + Mgh; h v 0 h!(30 m/s) =!(6.0 m/s) + (9.80 m/s )h, which gives h = 11.5 m. Using the coordinate sstem shown, for momentum conservation of the collision we have : M + 0 = (M + m)v ; (50 g)(6.0 m/s) = (50 g + 15 g)v, which gives V = 4.5 m/s; : 0 + m = (M + m)v ; (15 g)(00 m/s) = (50 g + 15 g)v, which gives V = 11.3 m/s. We use energ conservation to find the additional height attained b the skeet after the collision:!(m + m)(v + V )=!(M + m)v + (M + m)gh ;![(4.5 m/s) + (11.3 m/s) ]=!(4.5 m/s) + (9.80 m/s )h, which gives h = 6.54 m. (b) We find the time for the skeet to reach the ground from the vertical motion: = 0 + V t +!( g)t ; θ total 11.5 m = 0 + (11.3 m/s)t!(9.80 m/s )t. The positive solution to this quadratic equation is t = 3.07 s. The horizontal distance from the collision is = V t = (4.5 m/s)(3.07 s) = 75 m. The total horizontal distance covered is total =!R + =!(79.5 m) + 75 m = 115 m. Because of the collision, the skeet will have traveled an additional distance of Æ = total R = 115 m 79.5 m = 35 m. ² 77. Obviousl the spacecraft will have negligible effect on the motion of Saturn. In the reference frame of Saturn, we can treat this as the equivalent of a small mass bouncing off a massive object. The relative velocit of the spacecraft in this reference frame will be reversed. The initial relative velocit of the spacecraft is v SpS = v Sp v S = 10.4 km/s ( 9.6 km/s) = 0.0 km/s. so the final relative velocit is v SpS = 0.0 km/s. Therefore, we find the final velocit of the spacecraft from v SpS = v Sp v S ; 0.0 km/s = v Sp ( 9.6 km/s), which gives v Sp = 9.6 km/s, so the final speed of the spacecraft is 9.6 km/s. Page 7 6