The Medical College - Page 65
into work done by friction in stopping the
block.
(The energy is dissipated as frictional
heat.)
1/2 mv
2 = 1/2
4 kg
(5 m/sec)
2 = 50
joules
91.
B.
A lens with a negative focal length is
a
diverging lens. The power in Diopters is the
reci-
procal of the focal length in meters;
thus
P = –1/0.5 = –2.0 D.
92.
B.
The product of force and time, Ft, is
the
impulse and equals the change in momentum
of
an object, pf
– pi. The initial momentum of
both
masses is zero. Since F and t are the same
for
both masses, their final momenta must be
the
same, so 2 N
4 sec = 8 kg-m/sec. (Note that the
3 kg mass is moving faster but has the
same
momentum as the 6 kg mass.
93.
C.
Gamma (
) emission is the emission of a
photon from an excited nucleus and will
change
neither the charge nor mass number. The
resulting
nuclide is the (unexcited) Cu
64. Emission of a
neg-
ative electron,
–, from the nucleus
leaves it with
one additional unit of positive charge so that
the
atomic number, z, increases by 1. The result
is
30Zn
64. Emission of a
positron,
+, leaves
the
nucleus with one less positive charge and
z
decreases by 1. The result is
28Ni
64. Note that
the
mass changes are so small for beta decays that
the
superscript mass number (64) remains the same
in
all three cases.
94.
A.
The index of refraction is n = c/v, where c
is
the speed of light in a vacuum (3
10
8 m/sec)
and v is the speed of light in the material.
Here, vs
= 0.83 vw. Or vs = 0.83 c/nw = c/ns. Cancel c and
solve for ns
= 1.6.
95.
B.
The Coulomb force law between
charged
particles depends inversely on the square of
the
distance between the particles. F =
kq
1q2/r2. Here
the force increases by a factor of 4 because
r2 is
one-half of r1.
96.
A.
Newton's second law states that the net
force
equals the product of mass and acceleration:
Fnet =
ma. In this case the net force is the
difference
between the applied force (18 N) and the force
of
friction (9 N): Fnet = Fa – Ff = 18 – 9 = (90)a.
97.
B.
The total momentum of the two-cart
system
is zero before the students push off, and
remains
zero after they separate. Then: 0 =
m1v1 +
m2v2.
Thus, v2
= 60 kg(1.2 m/sec)/(0.8 m/sec).
98.
B.
The initial kinetic energy is used up in
doing
work against the frictional force as
the
student-cart is brought to rest. Using the
conser-
vation of energy principle:
Ff x = 1/2
mv
2, where
x = 8 m. Then : Ff = 20 N.
99.
C.
We know that the total angular momentum
of
the student is conserved (constant). Because he
is
capable of changing his moment of inertia about
the
vertical axis of spin by extending his arms or
pulling
them in, we can find the ratio of the moments
of
inertia by using the conservation law in the
form: I1
1 =
I2
2. Then
I2/I1 =
1/
2 = 1.5.
100.
D.
Linear momentum is conserved in this
prob-
lem. The addition of mass causes a decrease in
the
velocity as given by: m1v1
= (m1
+ msand)V2. V2 =
1.6 m/sec.
101.
B.
The center-of-mass will not move because
the
forces exerted by pulling on the rope are
internal
forces to the two-cart system. It is nearer the
larger
mass, and is found by using the formula for the
cm:
Xcm
= (m1x1 +
m2x2)/(m1 +
m2). We are
allowed to
place the origin of the coordinate system
anywhere
we choose. Placing the origin at the original
posi-
tion of the larger mass, x1 = 0 and x2 = 2 m. Xcm =
(0 + (80)(2))/(200 kg) = 0.8 m.
102.
C.
Conservation of momentum applies to
this
completely inelastic collision as follows:
m1v1 =
(m1
+ m2)V. Then V = (80 kg)(4 m/sec)(80 + 80) =
2 m/sec.
103.
B.
The net force accelerates the 100 kg
mass
according to Newton's Third Law. The net
force
here is the difference between the applied 90
N
force and the opposing friction force.
Fnet = 90 –
f
= ma = 100 kg
0.8 m/sec
2. Solving for f we
find
that f = 10 N.
104.
B.
The falling weight can be isolated so
that
we can apply Newton's third law to it alone.
The
weight force due to the mass acts downward
and
the 90 N tension upward. The third law
then
shows that: mg – 90 = ma, which we solve
for
the unknown mass, m. m(g – a) = 90 N. m(9.8
–
0.8) = 90. Then m = 10 kg.
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