Unformatted text preview: collision.
m
v
(144+ m )3
4
2444
bullet block f Momentum after collision = mbullet v0,bullet + mblock v0,block
1444 24444
4
3
Momentum before collision 384 IMPULSE AND MOMENTUM Solving this equation for the speed vf of the bullet/block system just after the collision gives
vf =
= mbullet v0,bullet + mblock v0,block
mbullet + mblock ( 0.00250 kg )( 425 m/s ) + ( 0.215 kg )( 0 m/s ) =
0.00250 kg + 0.215 kg 4.89 m/s b. Just after the collision, the total mechanical energy of the system is all kinetic energy,
since we take the zerolevel for the gravitational potential energy to be at the initial height of
the block. As the bullet/block system rises, kinetic energy is converted into potential energy.
At the highest point, the total mechanical energy is all gravitational potential energy. Since
the total mechanical energy is conserved, we have m
m ) gh
v
(144+2444 = 144 2444
( m 4+ m )3
4
3
bullet block f Total mechanical energy at
the top of the swing,
all potential 1
2 bullet block 2
f Total mechanical energy at
the bottom of the swing,
all kinetic Solving this expression for the height hf gives
hf = 1
2 vf2
g = 1
2 ( 4.89 m/s ) 2 9.80 m/s 2 = 1.22 m 56. REASONING AND SOLUTION According to the impulsemomentum theorem, Equation 7.4, ( ΣF ) ∆t = mv f − mv 0 , where ΣF is the net average force acting on the person. Taking
the direction of motion (downward) as the negative direction and solving for the net average
force ΣF , we obtain ΣF = m ( v f − v0 )
∆t = ( 62.0 kg ) [ −1.10 m/s – ( − 5.50 m/s)]
1.65 s = +165 N The plus sign indicates that the force acts upward . 57. REASONING AND SOLUTION The momentum is zero before the beat. Conservation of
momentum requires that it is also zero after the beat; thus
0 = mpvp + mbvb
so that
vp = –(mb/mp)vb = –(0.050 kg/85 kg)(0.25 m/s) = −1.5 ×10−4 m/s Chapter 7 Problems 385 58. REASONING The impulse that the wall exerts on the skater can be found from the
impulsemomentum theorem, Equation 7.4. The average force F exerted on the skater by
the wall is the only force exerted on her in the horizontal direction, so it is the net force;
ΣF = F .
SOLUTION From Equation 7.4, the average force exerted on the skater by the wall is F= mvf − mv 0
∆t = ( 46 kg ) ( −1.2 m/s ) − ( 46 kg ) ( 0 m/s )
0.80 s = −69 N From Newton's third law, the average force exerted on the wall by the skater is equal in
magnitude and opposite in direction to this force. Therefore,
Force exerted on wall = +69 N
The plus sign indicates that this force points opposite to the velocity of the skater. 59. REASONING Let the total amount of gas in the completely filled propulsion unit be mtotal,
and the amount of gas ejected during the spacewalk be mejected. The total mass of the gas is
the sum of the mass mejected of the ejected gas and the mass mleft of the gas left over at the
end of the spacewalk: mtotal = mejected + mleft. The mass of the leftover gas, in turn, is the
difference between the mass m1 = 165 kg of the astronaut with the partially full propulsion
unit and the mass mempty = 146 kg of the astronaut with the completely empty propulsion
unit: mleft = m1 − mempty = 165 kg − 146 kg = 19 kg. Thus, the percentage of gas propellant
in the completely filled propulsion unit that is depleted during the space walk is Percent depleted = mejected
mtotal ×100% = mejected
mejected + mleft × 100% (1) We will find mejected by considering the spacewalk. During the spacewalk, the astronaut,
the propulsion unit, and the gas form an isolated system. Although the propulsion unit and
the ejected gas exert forces on one another, as do the astronaut and the propulsion unit, these
are internal forces. The parts of the system interact only with each other, so the net external
force on the system is zero. Hence, the total momentum of the system is conserved. We will
apply the momentum conservation principle m1vf1 + m2 vf2 = m1v01 + m2 v02 (Equation 7.7b),
with the following identifications:
m1 = 165 kg (mass of astronaut and partially empty propulsion unit)
m2 = mejected = ? (mass of ejected gas) 386 IMPULSE AND MOMENTUM vf1 = −0.39 m/s (final velocity of astronaut and partially empty propulsion unit)
vf2 = +32 m/s (final velocity of ejected gas)
No part of the system is moving before the gas is ejected, so we also have v01 = v02 = 0 m/s.
We will determine the unknown mass m2 = mejected of the ejected gas from Equation 7.7b.
SOLUTION The momentum conservation principle yields the mass m2 = mejected of the
ejected gas: 0
m1vf1 + m2vf2 =
14 244
4
3 14 244
4
3
Total momentum
after gas is ejected or m2 = Total momentum
before gas is ejected − m1vf1 − (165 kg ) ( −0.39 m/s )
=
= 2.0 kg
vf2
( +32 m/s ) Equation (1) then yields the percent of propellant gas depleted during the spacewalk: Percent depleted = mejected
mejected + mleft × 100% = 2.0 kg
×100% = 9.5%
2.0 kg + 19 kg 60. REASONING AND SOLUTION Equation 7.10 gives the center of mass of this twoatom
system as
m x + mo xo
xcm = c c
mc + mo
If we take the origin at the center of the carbon atom, then xc =...
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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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