chap6 solutions - CHA'PT—ER fl SOLUTIONS CHAPTER SIX...

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Unformatted text preview: CHA'PT—ER fl SOLUTIONS CHAPTER SIX SOLUTIONS Chapter Six Readings 0rdway, F., "Principles of Rocket Engines," Sky and Te1e8c0pe, 14, 1954, p. 48. Resh, R.E., "Air Bags," Scientific American, v. 274 (June '96), p 116. 6.1 6.2 6.3 6.4 6.5 6.6 6.7 -l 2-191313 ii KE‘zm" ‘2 m =2m Use p=mv: (a) p = (1.67x10‘27 kg)(5.00 x 106 m/s) = 8.35 x 10-21 kg m/s (b) p = (150x 10-2 kg)(3.00 x 102 m/s) = 4.50 kg m/s (c) p = (75.0 kg)(1o.o m/s) = 750 kgm/s (d) p = (5.98 x1024 kg)(2.98 x 104 m/s) e 1.78 x 1029 kg m/s (a) Since the ball was thrown straight upward, it is at rest momentarily (v = 0) at its maximum height. Therefore,p=0. (b) The maximum height is found from V2 = V92 + Zay with_v=0. 0 = (15 m/s)2 -2(9.80 m/szmm Thus, hm = 11.5 m. h We need the velocity at w = 5.7-5 111, thus v2 = V02 + Zay gives 2 v2 = (15 m/s)2 - 2(9.80 m/s2)(5.75 m), or v = 10.6 m/s. Therefore, p n (0.10 kg)(10.6 m/s) = leg m/s- (a) The momentum ofthe bulletis (0.003)(1500) = 4.5‘kg In/s. The momentum of the baseball is pr=.(0.145)v, where v is the speed of the baseball. Therefore, if the two momenta are equal, we must have (0.145}“V =45, so 1? = 3.1.0 11115. (b) The kinetic energl of the. bullet is %EG£034(15:00)2 = 33751, while the kinetic energy of the baseball is %(o.145)(31.0)2 = 69.7 J. The bullet has the larger kinetic energy by a factor of 48.4. Use, FAt=Ap=me — mvi. F(0.30 s) = 0 -(1500 kg)(15 m/s) F=—7.5 x 104 N (the negative sign indicates the direction of the force is opposite to that of the car's original motion. impulse =47 t=Ap= m(AV). Thus, Impulse = (70.0Jggfl520 m/s - 0} = 364 kg III/S and 172111123186 = 363-83271931/3 ___= 438 kgm/sz 1 or F= 438 N. (a) First, find thé Madeline—red to the-ball. FAt=Ap=me - mvi. = O - (0.500 kg)(15.0 m/s) = - 7.5 kg m/s. (b) FAt=- 7.5 kg m/s, and At=0.0208,so 72 c'H" APT'E' R 's_fi s‘ OLUTIONS PM = —6.3 kg m/s. The negative sign indicates that the impulse is in the direction opposite to the initial velocity. (b) F = (i111pulse)/At= -(6.3 kg m/s)/2.0 x 10-3 s} = -3.2 x 103 N 6.14 Only the component of the hall's velocity perpendicular to the wall will change. This velocity component before hitting the wall is (10.0)sin60° = 8.66 m/s. After hitting the wall, this component is «8.66 m/s, because the rebound angle is also 60°. The change in momentum during contact with the wall is therefore Ap = (3.0)(8.66) — (3.00)(—8.66) = 51.96 kg m/s. The average force on ball is then 1:: 95% = 51.96/0.200 = 260 N directed away from the wall. 6.15 From conservation of momentum (after = before), we find the velocity of the man: (74.5 ngVman +(1.20 kg)(5.00 m/s) = 0, or Vman = -8.05 x 10'2 m/s (direction is southward) The time to travel 5.00 m to shore is t= 5.00 m/8.05 x 10'2 m/s = 62 s. 6.16 Before the collision, we have the total momentum being (0.20)(55) + 0 = 11 kg m/s. After the collision, we have for total momentum (0.20)(40) + (0.046)v = 8 + 0.046v. Conservation of momentum gives 8.0 + 0.046v = 11, which yields v = 65 m/s. 6.17 (a) From conservation of momentum, we have, choosing the direction of the bullet's motion as positive, mm + mbvb = 0, which gives ml, 5.0 x 10-3 VR=-fi =-—"W300m/s=-0.49m/s. (b) The mass of the man plus rifle is 74.5 kg. We use the same approach —3 as in (a), to fmd v = - 5%31-0— (300 m/s) = -2.0 x 10-2 m/s. 6.18 (a) Let vg and vp' be the velocity of the girl and the plank relative to the ice surface. Then we may say that vg - vp is the velocity of the girl relative to the plank, so that V3 - Vp = 1.50. (1} But also we must have mgvg + mpvp = 0, since total momentum of the girl-plank system is zero relative to the ice surface. Therefore 45.0vg + 150vp = 0, or vp = 0300 V3. (2) Putting this into the equation (1) above gives Vg — (-0.300 vg) = 1.50, or vg = 1.15 m/s. (h) Then, using (2) above: VI, = —0.300(1.15)= -0.346 m/s. 6.19 Consider the thrower first, with velocity after the throw of Vt momentum after = momentum before (65.0 kg)vt + (0.045 kg)(30.0 m/s) = (65.045 kg)(2.50 m/s), or Vt = 2.48 m/s. Now, consider the (catcher + ball), with velocity of vc after the catch: (60.045 kg) vc = (0.045 kg)(30.0 m/s) + (60 kg)(0), or vc = 2.25 x 10-2 m/s. 74 CHAPTER EX SOLUTIONS 20.0 cm/s + V1f= VZf. (2) Solve (1) and (2) simultaneously: V1f= -6.67 cm/s, and V2f = 13.3 cm/s. (b) KEhefore =% (5.00 x 10-3 kg)(20.0 x 10-2 m/s)2 = 10-4 J, KEsecond object=%(10.0 x 10’3 kg)(13.3 x 102)2 = 8.89 x 10'5 J, so KEsecond object W20” 6.30 m1V11+ m2V21=m1V1f+ mzv'zf. (10.0 g)(20.0 cm/s) + (15.0 g)(-30.0 cm/s) = (10.0 g)v1f+ (15.0 g)v2f, (1) and V11+ V1f= V2i + vzf, which becomes 20.0 (rm/3+ V1f= ~30.0 cm/s+ V2f. (2) Solve (1) and (2) simultaneously: v1 f = 40.0 cm/s; V2f = 10.0 cm/s. 6.3 1 Momentum conservation gives: m1 v11 + m2 vzi = m1V1f + 1112 V; f, or (25.0 g)(20.0 cm/s) + (10.0 g)(15.0 cm/s) = (25.0 g)v1_f+ (10.0 g)vzf. (1) For head—on elastic collisions, we know that v1i+ V1f= v2 i + V2f. Thus, 20.0 cm/s+v1f=15.0 cm/s+ sz. (2) SeclyingU) and (2) yields V1f= 17.1 cm/s, and V2f= 22.1 cm/s. 6.32 First, consider causewation of momentum, m1V1i+ m2V21= mLV1f+ mZVZf- Both bails have the-same mass, so this equation, becomes V11+ V2i=V1f+ sz- . (1) For an elastic head—on collision, we also have Vli - V21= -(V1r- V2f) Let us solve this equation for Vlf. V1r= V2f+VZi'V1i (2) Use (2) to eliminate V1f from (1). The result is V2f= Vli (3) Now, eliminate V2f from (1) by use of (3). We find V1f= V2i (4) Thus, equations (3) and (4) show us that, under the conditions of equal mass, objects striking one another in a head-on elastic collision, the two objects exchange velocities. Thus, we may write the results of the various collisions as (a) V1f=0, sz= 1.50 m/s (b) V1f= -1.00 m/s, V2f= 1.50 m/s (C) V1f= 1.00 m/s, v2f= 1.50 m/s 6.33 (a) First, we conserve momentum in the X direction (the direction of travel of the fullback): (90 kg)(5.0 m/s) + 0 = (185 kg) Vcose where 9 is the angle between the direction of the final velocity V and the X axis. We find: Vcose = 2.43 m/s (1) Now consider conservation of momentum in the y direction (the direction of travel of the opponent. (95 kg)(3.0 m/s) + O = (185 kg)(Vsine) which gives, Vsine = 1.54 m/s (2) Divide equation (2) by (1): ...
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