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Unformatted text preview: k = µ k FN , where µk is the coefficient of kinetic friction and FN
is the magnitude of the normal force that acts on the twocar system. There are only two
vertical forces that act on the system; they are the upward normal force FN and the weight
(m1 + m2)g of the cars. Taking upward as the positive direction and applying Newton's
second law in the vertical direction, we have FN – (m1 + m2 ) g = (m1 + m2 )a y = 0 , or
FN = (m1 + m2 ) g . Therefore, f k = µk (m1 + m2 ) g , and we have x= (m1 + m2 )vf2 2 µ k (m1 + m2 ) g = vf2 2µk g = (8.9 m/s) 2
= 5.9 m
2(0.68)(9.80 m/s 2 ) 40. REASONING Considering the boat and the stone as a single system, there is no net
external horizontal force acting on the stoneboat system, and thus the horizontal component
of the system’s linear momentum is conserved: m1vf 1x + m2 vf2x = m1v01x + m2 v02x (Equation
7.9a). We have taken east as the positive direction. SOLUTION We will use the following symbols in solving the problem:
m1 = mass of the stone = 0.072 kg
vf1 = final speed of the stone = 11 m/s
v01 = initial speed of the stone = 13 m/s
m2 = mass of the boat
vf2 = final speed of the boat = 2.1 m/s
v02 = initial speed of the boat = 0 m/s
Because the boat is initially at rest, v02x = 0 m/s, and Equation 7.9a reduces to
m1vf 1x + m2 vf2x = m1v01x . Solving for the mass m2 of the boat, we obtain m2vf2x = m1v01x − m1vf1x or m2 = m1 ( v01x − vf1x )
vf2x 372 IMPULSE AND MOMENTUM As noted above, the boat’s final velocity is
horizontal, so vf2x = vf2. The horizontal
components of the stone’s initial and final
velocities are, respectively, v01x = v01 cos 15°
and vf1x = vf1 cos 12° (see the drawing). Thus,
the mass of the boat must be m2 = m1 ( v01x − vf1x )
vf2 = vf1
v01x 12° vf1x 15° v01
Initial and final velocities of the stone ( 0.072 kg ) (13 m/s ) cos15o − (11 m/s ) cos12o 2.1 m/s = 0.062 kg 41. SSM REASONING The two skaters constitute the system. Since the net external force
acting on the system is zero, the total linear momentum of the system is conserved. In the x
direction (the east/west direction), conservation of linear momentum gives Pf x = P0 x , or
( m1 + m2 ) v f cosθ = m1 v 01
y Note that since the skaters hold onto each other, they move away
with a common velocity vf. In the y direction, Pf y = P0 y , or p ( m1 + m2 ) v f sin θ = m2 v 02 These equations can be solved simultaneously to obtain both the
angle θ and the velocity vf. p 1 SOLUTION
a. Division of the equations above gives θ = tan −1 F v I = tan L kg)(7.00 m / s) O 73.0°
m
(70.0
=
G v J M kg)(3.00 m / s) P
m
(50.0
HK N
Q
2 −1 02 1 01 b. Solution of the first of the momentum equations gives
vf = m1 v 01
( m1 + m2 ) cosθ = (50.0 kg)(3.00 m / s)
= 4.28 m / s
(50.0 kg + 70.0 kg)(cos 73.0 ° ) 2 x θ P f Chapter 7 Problems 373 42. REASONING
+355 m/s
a. The conservation of linear
Block 1
Block 2
momentum can be applied to this
threeobject system (the bullet and
(a) Before collisions
the two blocks), even though the
second collision occurs later than
the first one. Since there is no
vblock 2
+0.550 m/s
friction between the blocks and the
horizontal surface, air resistance is
negligible, and the weight of each
mblock 2 = 1530 g
mblock 1 = 1150 g
block is balanced by a normal
mbullet = 4.00 g
force, the net external force acting
on this system is zero, and the
(b) After collisions
conservation of linear momentum
applies. This principle will allow us to determine the velocity of the second block after the
bullet imbeds itself.
b. The total kinetic energy of the threebody system is not conserved. Both collisions are
inelastic, and the collision with block 2 is completely inelastic since the bullet comes to rest
within the block. As with any inelastic collision, the total kinetic energy after the collisions
is less than that before the collisions.
SOLUTION
a. The conservation of linear momentum states that the total momentum of the system after
the collisions [see part (b) of the drawing] is equal to that before the collisions [part (a) of
the drawing]: mblock 1vblock 1 + ( mblock 2 + mbullet ) vblock 2 = mbullet vbullet
14243
14444444 244444444
4
3
Total momentum
Total momentum after collisions
before collisions
Solving for the velocity vblock 2 of block 2 after the collisions gives vblock 2 = mbullet vbullet − mblock 1vblock 1
mblock 2 + mbullet ( 4.00 × 10−3 kg ) ( +355 m/s)
= − (1.150 kg )( +0.550 m/s ) 1.530 kg + 4.00 × 10−3 kg = +0.513 m/s b. The ratio of the total kinetic energy (KE) after the collision to that before the collision is 374 IMPULSE AND MOMENTUM KEafter
=
KE before = 1m
v2
2 block 1 block 1 + 1
2 2
( mblock 2 + mbullet ) vblock 2 1m
v2
2 bullet bullet 1
2 (1.530 kg + 4.00 × 10−3 kg ) ( 0.513 m/s)2 = 1.49 × 10−3
1 4.00 × 10−3 kg 355 m/s 2
)
)(
2( (1.150 kg )( 0.550 m/s )2 + 1
2 43. REASONING The ratio of the kinetic energy of the hydrogen atom after the collision to
that of the electron before the collision is
KE hydrogen, after collision
KE electron, before collison = 1
2 2
mH vf,H 1
2 2
me v0,e where vf,H is the final speed of the hydrogen atom, and v0,e is the initial speed of the
electron. The ratio mH/me of the masses is known. Since the...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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