Chapter 6 Solutions C.W. Fay August 23, 2011 College College Physics Physics 7th Edition Edition By Wilson, Wilson, Buffa and Lou. Lou. Numbers Numbers as follo follows ws
edition.chapter.number, MC mulitple choice, CQ concept question.
7.6.5, 6.6.12 A 0.150kg baseball traveling with a horizontal speed of 4.50m/s is hit by a bat and then moves moves with a speed of 34.7m/s in the opposite direction. direction. What is the change in the ball’s momentum? 150kg v1 = 4.50m 50m/s v2 = −34. 34.7m/s Given: m = 0.150kg ∆ p = m2 v2 − m1 v1 = (0. (0.150kg 150kg)( )(−34. 34.7m/s) (0.150kg 150kg)(4 )(4..50m/s 50m/s)) = −5.88kgm/s 88kgm/s (1) m/s) − (0.
7.6.6 A 15.0g rubber bullet hits a wall with a speed of 150m/s, if the bullet bounces straight back with a speed of 120m/s, what is the change in momentum of the bullet? Given:: m = 15. 15.0g = 0.0150kg 0150kg vi = 150m 150m/s vf = −120m/s 120m/s ∆ p = (0.0150kg 0150kg)( )(−120m/s 120m/s)) − (0. (0.0150kg 0150kg)(150 )(150m/s 50m/s p = mv mvf − mvi = (0. m/s)) = 4.50m/s
(2)
7.6.10, 6.6.16 Two runners of mass 70kg and 60kg, respectively, have a total linear momentum of 350kgm/ 350kgm/s. s. The heav heavier ier runner runner is running running at 2.0m/s. 2.0m/s. Determ Determine ine the possibl possiblee velocities of the lighter runner.
1
Given:
m1 = 70kg m2 = 60kg pT OT = 350kgm/s v2 =?
v1 = 2.0m/s
pT OT = p1 + p2 p2 = P T OT − p1 P T OT − m1 v1 = 3.5m/s v2 = m2 or pT OT = p2 − p1 p2 = P T OT + p1 P T OT + m1 v1 = 8.2m/s opposite of v 1 v2 = m2
(3) (4) (5) (6) (7) (8)
7.6.13 A loaded tractor-trailer with a total mass of 5000kg traveling at 3.0km/h hits a loading dock and comes to a stop in 0.64s. What is the magnitude of the average force exerted on the truck by the dock? Given:: m = 5000kg vi = 3.0km/h = 0.833m/s vf = 0m/s t = 0.64s ∆ p mvf − mvi mvi = = ∆t ∆t ∆t (5000kg)(0.833m/s) = = 6.5 × 103 N 0.64s
F ave =
(9)
(10)
7.6.14, 6.6.21 A 2.0kg mud ball drops from rest at a height of 15m. If the impact between the ball and the ground lasts 0.50s, what is the average net force exerted by the ball on the ground? m = 2.0kg h = 15m t = 0.50s vf = 0 Given: F =? F ∆t = ∆ p
(11) (12)
(13)
use the conservation of energy to find v, 1 2 mv = mgh 2 2gh v = 2
(14)
Now place it with the impulse equation, F ∆t = mvf − mvi √ −mvi = − m 2gh = −69N F = ∆t ∆t
(15)
(16)
7.6.20, 6.6.33 An automobile with a linear momentum of 3.0 × 104 kgm/s is brought to a stop in 5.0s. Given: pi = 3.0 × 104 kgm/s pf = 0kgm/s ∆t = 5.0s F =? Impulse = F ∆t = ∆ p 4 ∆ p 0kgm/s − 3.0 × 10 kgm/s = = −6.0 × 103 N F = ∆t 5.0s
(17)
(18)
7.6.21 A pool player imarts an impulse oof 3.2Ns to a stationary 0.25kg cue ball with a cue stick. What is the speed of the ball just after impact? Given:: I = 3.2Ns m = 0.25kg v0 = 0m/s vf =? I = ∆ p = mvf − mvi
vf =
3.2Ns I = = 12.8m/s m kg
(19) (20)
7.6.26, 6.6.39 A volleyball is traveling toward you. (a) Which action will require a greater force on the volleyball, your catching the ball or your hitting the ball back? why? (b) A 0.45kg volley ball travels with a horizontal velocity 0f 4.0m/s over the net. You jump up and hit the ball back with a horizontal velocity of 7.0m/s. If the contact time is 0.040s, what was the average force on the ball? m = 0.45kg vi = 4.0m/s vf = −7.0m/s ∆t = 0.040s Given: F =? a) hitting it back b) F ∆t = ∆ p
3
(21)
(22) F ∆t = mvf − mvi (4.0m/s) − (−7.0m/s) vf − vi = (0.45kg) = 1.2 × 102 N (23) F = m ∆t 0.040s
—–, 6.6.40 A 1.0kg ball is thrown horizontally with a velocity of 15m/s against a wall. If the ball rebounds horizontally with a velocity of 13m/s and the contact time is 0.020s, what force is exerted on the ball by the wall? m = 1.0kg vi = 15m/s vf = −13m/s Given: ∆t = 0.020s F =? ∆ p = m∆v ∆ p m∆v (1.0)(15 − (−13)) = = F = ∆t ∆t 0.020 3 = 1.4 × 10 Npointingoutofthewall
(24)
(25)
(26)
7.6.29, 6.6.43 A 0.45kg piece of putty is dropped from a height of 2.5m above a flat surface. When it hits the surface, the putty comes to rest in 0.30s. What is the average force exerted on the putty by the surface? Given: m = 0.45kg h = 2.5m ∆t = 0.30s F =? 1 2 mv = mgh 2 2gh v = F =
(27)
√
mv m 2gh = = 10.5N up ∆t ∆t
(28) (29)
7.6.31 An incoming 0.14kg baseball has a speed of 45m/s. The batter hits the ball, giving it a speed of 60m/s. If the contact time is 0.040s, what is the average force of the bat on the ball? Given: m = 0.14kg vi = 45m/s vf = −60m/s t = 0.040s F ave =
∆ p mvf − mvi = ∆t ∆t
4
(30)
=
(0.14kg)(45m/s) − (0.14kg)(−60m/s) = 370N 0.040s
(31)
7.6.33, 6.6.44 If the billiard ball in Fig 6.27 is in contact with the rail for 0.010s, what is the magnitude of the average force exerted on the ball? Given: t = 0.010s θ = 60 v = 15.0m/s m = 0.20kg Since the force from the bumper is in the y direction the only change in the moment is in the direction. ◦
∆ p ∆ py = t t mv cos θ − (−mv cos θ) 2mv cos θ = = t t 2 = 3.0 × 10 N
F =
(32)
(33) (34)
—-, 6.6.55 A 60kg astronaut floating at rest in space outside a space capsule throws his 0.50kg hammer such that it moves with a speed of 10m/s relative to the capsule, what happens to the astronaut? Given:: m1 = 60kg m2 = 0.50kg vH = 10m/s using the conservation of momentum, pi = pf 0 = pman − phammer pman = phammer m1 vm = m2 vH
vm =
m2 vH = 0.083m/s(west) m1
(35) (36) (37) (38) (39)
7.6.38, 6.6.56 In pairs figure-skating competition, a 65kg man and his 45kg female partner stand facing each other on skates on the ice. If they push apart and the woman has a velocity of 1.5m/s eastward, what is the velocity of her partner? (neglect friction) m1 = 65kg m2 = 45kg v2 = 1.5m/s Given: v1 = pi = 0 pf = 0 pi = p f = 0 = m 1 v1 − m2 v2 5
(40)
v1 =
m2 v2 = 1.0m/s west m1
(41)
7.6.39, 6.6.57 To get off a frozen, frictionless lake, a 65.0kg person takes off a 0.150kg shoe and throws it horizontally, directly away from the shore with a speed of 2.00m/s. If the person is 5.00m from the shore, how long doesh he take to reach it? mm = 60kg ms = 0.50kg vs = 10m/s ∆x = 5.00m Given: t =? using the conservation of momentum, pi = pf 0 = pman − pshoe pman = pshoe m1 vm = ms vs ms vs vm = mm
(42) (43) (44) (45)
(46) (47)
since there is no acceleration in the x-direction (frictionless) Given:: ax = 0m/s2 v0 = v m x = x0 + v0 t + 1/2ax t2 ∆x = v0 t ∆x t = vm mm ∆x = = 1.08 × 103 s ms vs
(48) (49) (50)
(51)
7.6.41 Consider two sting-suspended balls, both with a mass of 0.15kg. One ball is pulled back in line with the other so it has a vertical height of 10cm, and then is released. (a) What is the speed of the ball just before hitting the stationary one? (b) If the collision is completely inelastic to what height do the balls swing? Given: m = 0.15kg hi = 0.10m The problem can be broken into 3 parts. Part I the fall of the ball to just before the collision, Part II the collision, and Part III the two balls rising to a height of hf . Part 6
I, III can be analyzied by the conservation of energy. Part II by the conservation of momentum. Conservation of Energy, ∆K + ∆U ∆K 1 2 1 2 mv − mv 2 f 2 i vf 2 − vi2
= 0 = −∆U
(52) (53)
= mghi − mghf
= 2mg(h1 − h2 )
(54) (55) (56)
Conservation of momentum, ∆ p pf − pi pi mv
= = = =
0 0 pf (m + M )V
(57) (58) (59) (60)
Now we apply what we know. v the initial velocity from the conservation of momentum is the final velocity for the fall of the first ball. V the final velocity from the conservation of momentum for the collision is the initial velocity for the two balls raising. The velocity of any ball at its highest point is 0 and the height at the lowest point is set to 0. v = V =
m( 2ghi )
=
2gh = 1.4m/s i 2gh f (m + M )( 2gh f )
m 2ghi = 2m 2ghf 2ghi = 2ghf 4 hi = 0.025m hf = 4
(61) (62) (63)
(64)
(65)
(66)
—-, 6.6.61 Two identical cars hit each other and lock bumpers. In each of the following cases, what are the speeds of the cars immediately after coupling bumpers? (a) A car moving with a speed of 90km/h approaches a stationary car; (b) the two cars approach 7
each other with speeds of 90km/h and 120km/h respectively; (c) two cars travel in the same direction with speeds of 90km/h and 120km/h, respectively. a) Given: m1 = m2 v1 = 90km/h v2 = 0 m1 v1 + m2 v2 = (m1 + m2 )v3 (m1 v1 + m2 v2 ) v3 = (m1 + m2 ) (mv1 ) v1 = = 45km/h = 12.5m/s v3 = (2m) 2 b) Given: m1 = m2
v1 = 90km/h v2 =
(67) (68)
(69)
−120km/h
(m1 v1 + m2 v2 ) (m1 + m2 ) (mv1 + mv2 ) v1 + v2 = = = −15km/h = (2m) 2
v3 = v3
c) Given: m1 = m 2
(70)
−4.17m/s
v1 = 90km/h v2 = 120km/h
(m1 v1 + m2 v2 ) (m1 + m2 ) (mv1 + mv2 ) v1 + v2 = = = 105km/h = 29.2m/s (2m) 2
v3 = v3
(71)
(72)
(73)
7.6.46, 6.6.63 A 10g bullet moving horizontally at 400m/s penetrates a 3.0kg wood block resting on a horizontal surface. If the bullet slows down to 300m/s after emerging from the block, what is the speed of the block immediately after the bullet emerges? Given: m1 = 10g m2 = 3.0kg v1 = 400m/s v1f = 300m/s p1 = p2 m1 v1 = m1 v1f + m2 v2f m1 v1 − m1 v1f v2f = m2
8
(74) (75) (76)
7.6.50 For a movie scene, a 75kg stuntman drops from a tree onto a 50kg sled that is moving onn a frozen lake with a velocity of 10m/s toward the shore. (a) What is the speed of the sled after the suntman is on board? (b) If the sled hits the bank and stops, but the stuntman keeps on going, with what speed does he leave the sled?(neglect friction) Given:: mm = 75kg ms = 50kg vs = 10m/s a) ms vs = (mm + ms )vf ms vs = 4.0m/s vf = (mm + ms )
mm vm = (mm + ms )vf (mm + ms )vf = 6.7m/s vm = mm
(77)
(78)
b) (79)
(80)
7.6.52, 6.6.69 A projectile that is fired from a gun has an initial velocity of 90km/h at an angle of 60.0 above the horizontal. When the projectile is at the top of its trajectory, an internal explosion causes it to separate into two fragments of equal mass. One of the fragments falls straight downward as though it has been released from rest. How far from the gun does the other fragment land? Given: v = 90.0km/h θ = 60.0 We apply the conservation of momentum, the moment before the explosion and we know that the momentum is p b = mv0x , ◦
◦
= pb pf mv0x = m2 0 + m2 v2x = 2v0x v2x
(81) (82) (83)
We need only concern ourselves with the x-component just before it explodes the y-component is equal to zero so the fall time is the same as the rise time. we find the rise time as, vytop = v0y − gt rise v0y v0 sin θ = trise = = g g 9
(84)
(85)
where v ytop = 0. Now we write, vo2 cos θ sin θ = 82.8m x = 3 g 3 vo2 = sin2θ 2 g or 3/2 times the range equation.
(86)
(87)
7.6.53, 6.6.70 A moving shuffleboard puck has a glancing collision with a stationary puck of the same mass as shown in Fig 6.32. If friction is negligible, what are the speeds of the pucks after the collision? m1 = m2 = m v1 = v1 x = 0.95m/s θ1 = 50 θ2 = 40 Given: v1oy = 0m/s v2 = 0m/s v1 =? v2 =? ◦
0
◦
0
0
the conservation of momentum in the x-direction, mv1
0x
= mv1x + mv2x
(88)
(89)
the conservation of momentum in the y-direction, mv1
0y
= mv1y + mv2y
to simplify the equations we can write the velocities in terms of v 1 , v 2 , θ 1 , and θ 2 , v1x v1y v2x v1y
= = = =
v1 cos θ1 v1 sin θ1 v2 cos θ2 v2 sin θ2
(90) (91) (92) (93)
Simplifying our equations, we write. mv1 = mv1 cos θ1 + mv2 cos θ2 0 = mv1 sin θ1 − mv2 sin θ2 0
(94) (95) (96)
Now we have 2 equations and 2 unknowns, so the solution just remains to do some algebra, mv1 sin θ1 − mv2 sin θ2 = 0 v1 sin θ1 = v2 sin θ2 sin θ2 v1 = v2 sin θ1 10
(97) (98) (99)
Now put this into the equation for momentum in the x-direction, v1
0
v1
0
v1
0
v1
0
= v1 cos θ1 + v2 cos θ2 sin θ2 = v2 cos θ1 + v2 cos θ2 sin θ1 sin θ2 = v2 cos θ1 + cos θ2 sin θ1 v2 = (sin θ2 cos θ1 + cos θ2 sin θ1 ) sin θ1
(100)
(101)
(102) (103)
using the trig identity, sin(θ1 + θ2 ) = sin θ2 cos θ1 + cos θ2 sin θ1 We can write, v1 = 0
v2 sin(θ1 + θ2 ) sin θ1
(104) (105)
since θ 1 + θ2 = 90 and sin 90 = 1, ◦
◦
v1
0
v2 similarly, v1 = v2
v2 sin θ1 = v1 sin θ1 = 0.73m/s =
(106)
0
(107)
sin θ2 = v1 sin θ2 = 0.61m/s sin θ1
0
(108)
7.6.58, 6.6.82 A 4.0kg ball with a velocity of 4.0m/s in the +x-direction collides head-on elastically with a stationary 2.0kg ball. What are the velocities of the balls after the collision? Given: m1 = 2m2 v1i = 4.0m/s v2i = 0m/s Find v 1f and v 2f . This is an elastic collision so energy and momentum are conserved. From the conservation of momentum, m1 v1i + m2 v2i = m1 v1f + m2 v2f 2m2 v1i = 2m2 v1f + m2 v2f 2v1i = 2v1f + v2f v 2 v1i = v1f + f 2 11
(109) (110) (111) (112)
v1f = v1i −
v2f 2
(113)
1 v12f = v12i − v1i v2f + v22f 4
(114)
From the conservation of energy, 1 1 1 1 m1 v12i + m2 v22i = m1 v12f + m2 v22f 2 2 2 2 2 2 m1 v1i = m1 v1f + m2 v22f
(115) (116)
2m2 v12i = 2m2 v12f + m2 v22f 1 v12i = v12f + v22f 2 1 1 v12i = v12i − v1i v2f + v22f + v22f 4 2 3 v12i − v12i = −v1i v2f + v22f 4 3 2 v1i v2f = v 4 2f 3 v1i = v2 4 f 4 v2f = v1 = 5.33m/s 3 i
(117) (118) (119) (120) (121) (122) (123)
and, v1f = v1i
− v2
2f
= 1.3m/s
(124)
7.6.60, 6.6.84 a county fair, two childern ram each other head on while riding on the bumper cars attraction. Jill and her car travling left at 3.50m/s, have a total mass of 325kg. Jack and his car, traveling to the right at 2.00m/s, have a total mass of 290kg. Given: Assuming the collision to be elastic, determine their velocities after the collision.conservation of moment gives, m1 v1 − m2 v2 =
−m v f + m v f 1 1
2 2
(125)
conservation of energy gives, 1 1 1 1 m1 v12 + m2 v22 = m1 v12f + m2 v22f 2 2 2 2 12
(126)
rewrite equation ( ??) as, m1 v12 + m2 v22 = m1 v12f + m2 v22f
(127)
m1 (v12 − v12f ) = m2 (v22f − v22 ) m1 (v1 − v1f )(v1 + v1f ) = m2 (v2f − v2 )(v2 + v2f )
(128) (129)
rewrite equation ( ??) as, m1 (v1 + v1f ) = m2 (v2 + v2f )
(130)
Divide equation (??) by equation ( ??), m1 (v1 − v1f )(v1 + v1f ) m2 (v2f − v2 )(v2 + v2f ) = m1 (v1 + v1f ) m2 (v2 + v2f ) v1 − v1f = v2f − v2 v1 + v2 = v1f + v2f
(131) (132) (133)
Solve for v 1f , v1f = v 1 + v2 − v2f
(134)
Now we can solve for v 2f from the conservation of momentum, m1 (v1 + v1 + v2 − v2f ) = m2 (v2 + v2f ) 2m1 v1 + m1 v2 − m1 v2f = m2 v2 + m2 v2f 2m1 v1 + m1 v2 − m2 v2 = m1 v2f + m2 v2f 2m1 m1 − m2 v2f = v1 + v2 m1 + m2 m1 + m2 v2f = −3.81m/s
(135) (136) (137)
(138) (139)
Similarly we can solve for v 1f ,
m2 − m1 2m2 v1f = v2 + v1 = 1.69m/s m1 + m2 m1 + m2
(140)
note: the positive sign means that the initial assignment of a sign to the momentum of mass one final is correct.
13
7.6.63, 6.6.87 A 1.0kg object moving at 10m/s collides perfectly inelastically with a stationary 2.0kg object. After the collision the blocks slide up a plane inclined at 37 . How far up the plane will the combined system travel? Given: m1 = 1.0kg m2 = 2.0kg v0 = 10m/s θ = 37 ◦
◦
use conservation of momentum to find the velocity at the bottom of the ramp and then the conservation of energy to find the distance up the ramp. conservation of momentum, m1 v1 = m1 v2 + m2 v2 m1 v2 = v1 m1 + m2
(141)
(142) (143)
conservation of energy, 1 (m1 + m2 )v22 = (m1 + m2 )gh 2
(144)
h is the change in vertical height of the blocks, and is related to the distance along the incline, l, by, (145) h = l sin θ this leads to, 1 (m1 + m2 )v22 = (m1 + m2 )gh 2 1 (m1 + m2 )v22 = (m1 + m2 )gl sin θ 2 v22 l = 2g sin θ m21 v12 = = 0.94m (m1 + m2 )2 2g sin θ
(146) (147) (148)
(149)
—-, 6.6.89 Two balls with masses of 2.0kg and 6.0kg travel toward each other at speeds of 12m/s and 4.0m/s, respectively. If the balls have a head-on, inelastic collision and the 2.0kg ball recoils with a speed of 8.0m/s, how much kinetic energy is lost in the collision? 14
Given:
m1 = 2.0kg m2 = 6.0kg v3 = −8.0m/s
v1 = 12m/s v2 =
−4.0m/s
∆ p = 0 = m 1 v1 + m2 v2 − m1 v3 − m2 v4 m1 v1 + m2 v2 − m1 v3 m1 = (v − v3 ) + v2 = 2.67m/s v4 = m2 m2 1 1 1 ∆K = ( m1 v32 + m2 v42 ) − ( m1 v12 + frac12m2 v22 ) 2 2 2 1 K i = m1 v12 + f rac12m2 v22 = 192J 2 1 1 K f = m1 v32 + m2 v42 = 85J 2 2 ∆K = 1.1 × 102 J
(150)
(151) (152) (153) (154) (155)
7.6.65, 6.6.90 Two balls approach each other as shown in Fig 6.36, where m = 2.0kg, v = 3.0m/s, M = 4.0kg, and V = 5.0m/s. If the balls collide sand stick together at the origin, (a) what are the components of the velocity v of the balls after the collision, and (b) what x M = 4.0kg V = 5.0m/s is the angle θ? Given: m = 2.0kg v = 3.0m/sˆ This is an inelastic collision, momentum is conserved, energy is not. Write down momentum conservation in the x and y directions,
x : mv = (m + M )vx y : M V = (m + M )vy mv = 1.0m/s vx = (m + M ) M V = 3.3m/s vx = (m + M ) θ = tan 1 (MV/mv) = 73
(156) (157)
(158)
(159)
−
◦
(160)
7.6.73 Show that the fraction of kinetic energy lost in a ballistic pendulum collision is equal to M /(m + M ). To find the fraction of energy lost we need to know the energy before and after the collison. The energy will all be kinetic and as such we need to know the velocity
15
before v and after V the collision. 1/2mv2 − 1/2(m + M )V 2 Fraction of energy lost = 1/2mv2 mv2 − (m + M )V 2 = mv 2
(161) (162)
The conservation of momentum gives. pi = pf mv = (m + M )V m V = v m + M
(163) (164)
(165) (166)
2 Substituting this into the energy fraction gives, mv 2 − (m + M )V 2 Fraction of energy lost = (167) mv2 m mv 2 − (m + M )( m+ v)2 M = (168) mv2 mv 2 m m + M − m M − = (1 ) = = (169) mv 2 m + M m + M m + M
7.6.76, 6.6.106 Find the center of mass of a system composed of thress shperical objects with masses of 3.0kg, 2.0kg and 4.0kg and centers locaed at (-6.0m,0), (1.0m,0) and (3.0m,0),respectively. m1 = 3.0kg m2 = 2.0kg m3 = 4.0kg Given: r1 = (−6.0cm, 0) r2 = (1.0cm, 0) r3 = (3.0cm, 0) by inspection y cm = 0, xcm =
m x i mi i i
(170)
i
m1 x1 + m2 x2 + m3 x3 = −0.44m m1 + m2 + m3 = (−0.44m, 0)
xcm = rcm
16
(171) (172)
7.6.77, 6.6.107 Rework exercise 69, using the concept of the center of mass, and compute ths distance the other fragment lands from the gun. Given: v = 90.0km/h θ = 60.0 At the maximum height the bullet explodes the center of mass travels on the initial parabolic path with an x-compoent of the velocity of v0x . For the bullet fragement to fall straight down it must have a speed v1x = 0 or − vx relative to the center of mass. This means the other fragment must have a velocity of v2x = 2v0x relative to the ground (its speed is +vx relative to the center of mass). The time required for the bullet fragments to fall the ground are the same, and are the same as the time required to reach the maximum height, t max (assuming the ground is flat). This means the total distance traveled is, ◦
x = v0 xtmax + v2x tmax = v 0 xtmax + 2v0xtmax = 3v0x tmax = 3vo cos θtmax
(173) (174)
To find tmax, vytop = v0y − gt max v0y v0 sin θ = tmax = = g g
(175)
(176)
where v ytop = 0, now we write, vo2 cos θ sin θ = 82.8m x = 3 g 3 vo2 = sin2θ 2 g
(177) (178)
or 3/2 times the range equation.
7.6.79, 6.6.109 A piece of uniform sheet metal measures 25cm by 25cm. If a circular piece with a radius of 5.0cm is cut from the center of the sheet, where is the sheet’s center of mass now? The distribution of the mass as far as its symmetry doesn’t change so the center of mass is still in the center.
17
7.6.80, 6.6.110 Locate the center of mass of the system shown in Fig 6.38 (a) if all the masses are equal; (b) if m4 = m2 = 2m3 = 2m1 ; (c) if m1 = 1.0kg, m2 = 2.0kg, m3 = 3.0kg, and m 4 = 4.0kg. Given: r1 = (0m, 0m) r2 = (0m, 4m) r3 = (4m, 4m) r4 = (4m, 0m) xcm =
m x i mi i i
xcm =
(179)
i
m1 x1 + m2 x2 + m3 x3 + m4 x4 m1 + m2 + m3 + m4
(180)
m1 y1 + m2 y2 + m3 y3 + m4 y4 m1 + m2 + m3 + m4
(181)
similarly in the y-direction, ycm = a) m 1 = m 2 = m3 = m4 rcm = (2.0m, 2.0cm) b) 2m1 = m 2 = 2m3 = m4 rcm = (2.0m, 2.0cm) c)m1 = 1.0kg, m 2 = 2.0kg, m3 = 3.0kg, m 4 = 4.0kg rcm = (2.8m, 2.0cm)
18
