Tài liệu Physics exercises_solution: Chapter 03 doc

3.1: a) ,sm4.1 )s0.3( )m1.1()m3.5( ave,    x v .sm3.1 )s0.3( )m4.3()m5.0( ave,    y v b) sm9.1or ,sm91.1)sm3.1()sm(1.4 22 ave v to two significant figures,     43arctan 1.4 1.3 θ . 3.2: a) andm6.45)s0.12)(sm8.3(Δ)( ave,  tvx x m.8.58)s0.12)(sm9.4(Δ)( ave,  tvy y b) m.4.74)m8.58()m6.45( 2222  yxr 3.3: The position is given by jir ˆ )scm0.5( ˆ ])scm5.2(cm0.4[ 22 tt   . (a) i ˆ ]cm0.4[)0( r , and jiji ˆ )cm0.10( ˆ )cm0.14( ˆ s)2)(scm0.5( ˆ ]s)2)(scm(2.5cm0.4[)s2( 22  r . Then using the definition of average velocity, .jiv ji ˆ )scm5( ˆ )scm5( s2 ˆ )0cm10( ˆ )cm4cm14( ave    scm1.7 ave v at an angle of 45 . b) jijiv ˆ )scm5( ˆ )scm5( ˆ )scm5( ˆ )scm5.2)(2(  tt dt rd   . Substituting for s1,0t , and 2 s, gives: jivjv ˆ )scm5( ˆ )scm5(s)1(, ˆ )scm5()0(   , and jiv ˆ )scm5( ˆ )scm10(s)2(   . The magnitude and direction of v  at each time therefore are: scm0.5:0t at 90 ; scm1.7:05.1t at scm11:05.2;45  t at 27 . c) 3.4: jiv ˆ 3 ˆ 2 2 ctbt   . This vector will make a 45 -angle with both axes when the x- and y-components are equal; in terms of the parameters, this time is cb 32 . 3.5: a) b) ,sm7.8 )s0.30( )sm90()sm170( 2 ave,    x a .sm3.2 )s0.30( )sm110()sm40( 2 ave,    y a c)   .1951808.14arctan ,sm0.9)sm32()sm7.8( 7.8 3.2 22 2 2 2    . 3.6: a) 2222 sm23.00.31sin)sm45.0(,sm39.00.31cos)sm45.0(  yx aa , so sm5.6)s0.10)(sm39.0(sm6.2 2  x v and sm52.0)s0.10)(sm23.0(sm8.1 2  y v . b) sm48.6)sm52.0()sm5.6( 22 v , at an angle of    6.4arctan 5.6 52.0 above the horizontal. c) 3.7: a) b) . ˆ )sm4.2( ˆ 2 ˆ ])sm4.2[( ˆ )sm4.2( ˆ 2 ˆ 2 2 jja jijiv       tt c) At s0.2t , the velocity is jiv ˆ )sm8.4( ˆ )sm4.2(   ; the magnitude is sm4.5)sm8.4()sm4.2( 22  , and the direction is     63arctan 4.2 8.4 . The acceleration is constant, with magnitude 2 sm4.2 in the y -direction. d) The velocity vector has a component parallel to the acceleration, so the bird is speeding up. The bird is turning toward the y -direction, which would be to the bird’s right (taking the z - direction to be vertical). 3.8: 3.9: a) Solving Eq. (3.18) with 0y , 0 0  y v and s350.0t gives m6000 0 .y  . b) m385.0tv x c)  2.72,sm60.3,sm43.3s,m10.1 0 vgtvvv yxx below the horizontal. 3.10: a) The time t is given by s82.7 2  g h t . b) The bomb’s constant horizontal velocity will be that of the plane, so the bomb travels a horizontal distance m470)s82.7)(sm60(  tvx x . c) The bomb’s horizontal component of velocity is 60 m/s, and its vertical component is sm7.76 gt . d) e) Because the airplane and the bomb always have the same x-component of velocity and position, the plane will be 300 m above the bomb at impact. 3.11: Take y to be upward. Use Chirpy’s motion to find the height of the cliff. s50.3,,sm80.9,0 0 2 0  thyyav yy m0.60gives 2 2 1 00  htatvyy yy Milada: Use vertical motion to find time in the air. ?,sm80.9m,0.60,0.32sin 2 000  tayyvv yy s55.3gives 2 2 1 00  ttatvyy yy Then s55.3,0,0.32cos 00  tavv xx gives m86.2 0  xx . 3.12: Time to fall 9.00 m from rest: 2 2 1 gty  22 )sm8.9( 2 1 m00.9 t s36.1t Speed to travel 1.75 m horizontally: tvx 0  )s36.1(m75.1 0 v sm3.1 0 v 3.13: Take +y to be upward. Use the vertical motion to find the time in the air: ?,m5.19m)8.1m3.21(,sm80.9,0 0 2 0  tyyav yy s995.1gives 2 2 1 00  ttatvyy yy Then ?,s995.1,0m,0.61 00  xx vtaxx s.m6.30gives 0 2 2 1 00  xxx vtatvxx b) sm6.30 x v since 0 x a sm6.19 0  tavv yyy sm3.36 22  yx vvv 3.14: To make this prediction, the student needs the ball’s horizontal velocity at the moment it leaves the tabletop and the time it will take for the ball to reach the floor (or rather, the rim of the cup). The latter can be determined simply by measuring the height of the tabletop above the rim of the cup and using 2 2 1 gty  to calculate the falling time. The horizontal velocity can be determined (although with significant uncertainty) by timing the ball’s roll for a measured distance before it leaves the table, assuming that its speed doesn’t change much on the hard tabletop. The horizontal distance traveled while the ball is in flight will simply be horizontal velocity  falling time. The cup should be placed at this distance (or a slightly shorter distance, to allow for the slowing of the ball on the tabletop and to make sure it clears the rim of the cup) from a point vertically below the edge of the table. 3.15: a) Solving Eq. (3.17) for 0 y v , with  0.45sin)sm0.15( 0 y v , s.08.1 sm80.9 45sin)sm0.15( 2   T b) Using Equations (3.20) and (3.21) gives at m)52.4,m18.6(),(, 1 yxt : )m52.4,m8.16(,:)m74.5,m5.11(, 32 tt . c) Using Equations (3.22) and (3.23) gives at ),sm9.4,sm6.10(:)0,sm6.10(,:)sm9.4,sm6.10(),(, 321  ttvvt yx for velocities, respectively, of sm7.11 @ 24.8, sm6.10 @ 0 and sm7.11 @ 24.8. Note that x v is the same for all times, and that the y-component of velocity at 3 t is negative that at 1 t . d) The parallel and perpendicular components of the acceleration are obtained from .,, )( |||| 2 || aaa va a vva a              vv For projectile motion, y gvg  vaja   so, ˆ , and the components of acceleration parallel and perpendicular to the velocity are 22 1 sm9.8,sm1.4: t . 2 2 sm8.9,0:t . 22 3 sm9.8,sm1.4:t . e) f) At t 1 , the projectile is moving upward but slowing down; at t 2 the motion is instantaneously horizontal, but the vertical component of velocity is decreasing; at t 3 , the projectile is falling down and its speed is increasing. The horizontal component of velocity is constant. 3.16: a) Solving Eq. (3.18) with m75.0,0 0  yy gives s391.0t . b) Assuming a horizontal tabletop, 0 0  y v , and from Eq. (3.16), sm58.3/)( 00  txxv x . c) On striking the floor, sm83.32 0  gygtv y , and so the ball has a velocity of magnitude sm24.5 , directed 9.46 below the horizontal. d) Although not asked for in the problem, this y vs. x graph shows the trajectory of the tennis ball as viewed from the side. 3.17: The range of a projectile is given in Example 3.11, gvR 0 2 0 2sin   . a) km38.1)sm80.9(110sin)sm120( 2 2  .b) km4.8)sm6.1(110sin)sm120( 2 2  . 3.18: a) The time t is s63.1 2 0 sm80.9 sm0.16  g v y . b) m1.13 2 0 2 1 2 2 1 2 0  g v y y tvgt . c) Regardless of how the algebra is done, the time will be twice that found in part (a), or 3.27 s d) x v is constant at sm0.20 , so m365s)27.3)(sm0.20( . . e) 3.19: a) sm0.189.36sin)sm0.30( 0  y v ; solving Eq. (3.18) for t with 0 0 y and m0.10y gives s992s,68.0 sm80.9 )m0.10)(sm80.9(2)sm0.18()sm0.18( 2 2 2 .t    b) The x-component of velocity will be sm0.249.36cos)sm0.30(  at all times. The y-component, obtained from Eq. (3.17), is sm3.11 at the earlier time and sm3.11 at the later. c) The magnitude is the same, sm0.30 , but the direction is now 9.36 below the horizontal. 3.20: a) If air resistance is to be ignored, the components of acceleration are 0 horizontally and 2 sm80.9 g vertically. b) The x-component of velocity is constant at sm55.70.51cos)sm0.12(  x v . The y-component is sm32.90.51sin)sm0.12( 0  y v at release and sm06.11)s08.2)(sm80.9()sm57.10( 2 0  gtv y when the shot hits. c) m715s)08.2)(sm55.7( 0 .tv x  . d) The initial and final heights are not the same. e) With 0y and v 0y as found above, solving Eq. (3.18) for m81.1 0 y . f) 3.21: a) The time the quarter is in the air is the horizontal distance divided by the horizontal component of velocity. Using this time in Eq. (3.18), rounded.m53.1 60cos)sm4.6(2 m)1.2)(sm80.9( m)1.2(60tan cos2 tan 2 22 2 2 0 22 0 2 0 2 0 2 0 00        v gx x v gx v x vyy x x y b) Using the same expression for the time in terms of the horizontal distance in Eq. (3.17), .sm89.0 60cos)sm4.6( )m1.2)(sm80.9( 60sin)sm4.6( cos sin 2 00 00      v gx vv y [...]... 2v0 sin  0 gt  ( gt ) 2 2  v0  2 g (v0 sin  0 t  1 2 gt ) 2 2  v0  2 gy, where Eq (3.21) has been used to eliminate t in favor of y This result, which will be seen in the chapter dealing with conservation of energy (Chapter 7), is valid for any y, positive, negative or zero, as long as v 2  0 For the case of a rock thrown from the roof of a building of height h, the speed at the ground is... from the cliff to the plain Use this vertical motion to find the time: v0 y  0, a y  9.80 m/s 2 , y  y0  45 m, t  ? y  y 0  v0 y t  1 a y t 2 gives t  3 .03 s 2 During this time the rock travels horizontally x  x0  v0 xt  (49 m/s)(3 .03 s)  150 m The rock lands 50 m past the foot of the dam 3.68: (a) When she catches the bagels, Henrietta has been jogging for 9.00 s plus the time for the bagels... 1.4 h 9.80 m/s 2 3.30: 550 rev/min  9.17 rev/s , corresponding to a period of 0.109 s a) From Eq (3.29),  v  2TR  196 m/s b) From either Eq (3.30) or Eq (3.31), arad  1.13  104 m/s 2  1.15  103 g 3.31: Solving Eq (3.30) for T in terms of R and arad , a) 4 2 (7.0 m)/(3.0)(9.80 m/s2 )  3.07 s 3.32: a) Using Eq (3.31), arad  5.91  10 3 m/s 2 2R T b) 1.68 s  2.97  10 4 m/s b) Either... atan  0.5 m/s2 So, a  ((0.643 m/s 2 ) 2  (0.5 m/s2 ) 2 )1 / 2  0.814 m/s 2 , 37.9 to the right of vertical b) 3.35: b) No Only in a circle would arad point to the center (See planetary motion in Chapter 12) c) Where the car is farthest from the center of the ellipse 3.36: Repeated use of Eq (3.33) gives a) 5.0  m/s to the right, b) 16.0 m/s to the left, and c) 13.0  m/s to the left 3.37: a)... component of velocity divided by the acceleration –g; the distance is the constant horizontal component of velocity multiplied by this time, or (23.7 m/s  ((30.0 m/s)sin33.0)) x  (30.0 m/s)cos33.0  103 m (9.80 m/s 2 ) d) 3.24: a) v0 cost  45.0 m cos  45.0 m  0.600 (25.0 m/s)(3.00 s)   53.1 b) v x  (25.0 m/s) cos 53.1  15.0 m/s vy 0 v  15.0 m/s a  9.80 m/s 2 downward c) Find y when t... involves decreasing the period by a factor of 3 , so that the new period T  is given in terms of the previous period T by T   T / 3 3.29: Using the given values in Eq (3.30), 4 2 (6.38  10 6 m)  0 .034 m/s 2  3.4  10 3 g 2 ((24 h)(3600 s/h)) (Using the time for the siderial day instead of the solar day will give an answer that differs in the third place.) b) Solving Eq (3.30) for the period T . 3/TT   . 3.29: Using the given values in Eq. (3.30), .104.3m/s 034 .0 s/h))h)(360024(( m) 1038 .6(4 32 2 62 rad ga      (Using the time for the siderial. the constant horizontal component of velocity multiplied by this time, or m. 103 )m/s80.9( ))0m/s)sin33.((30.0m/s23.7( 0m/s)cos33.0.30( 2    x d)