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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 2191313 ii
KE‘zm" ‘2 m =2m Use p=mv:
(a) p = (1.67x10‘27 kg)(5.00 x 106 m/s) = 8.35 x 1021 kg m/s (b) p = (150x 102 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 = 5.75 111, 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 = 36383271931/3 ___= 438 kgm/sz 1 or F= 438 N. (a) First, ﬁnd thé Madeline—red to theball.
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 103 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 103 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%310— (300 m/s) = 2.0 x 102 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 girlplank 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 102 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 103 kg)(20.0 x 102 m/s)2 = 104 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 thesame 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 headon 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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 Fall '01
 Shapiro
 Physics

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