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CIRCLE
THE SINGLE BEST ANSWER FOR ALL
MULTIPLE CHOICE QUESTIONS. IN MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION
SHOW WORK FOR PARTIAL CREDIT.
- In a
contest, two tractors pull two identical blocks of stone the same distance
over identical surfaces. However, block A is moving twice as fast as block
B when it crosses the finish line. Which statement is correct?
a. Block A has twice as much energy as block
B.
(8)
b. Block B has lost twice
as much energy as block A.
c. Both blocks have had equal
losses of energy to friction.
d. No energy is lost to friction because the
ground has no displacement.
2. Two eggs of equal mass are thrown at a
blanket with equal velocity. Egg A hits the wall instead but egg B hits the
blanket. Compare the work done on the eggs in stopping them:
a. More work was done on A than on B.
b. More work was done on B than on A.
(8)
c. The amount of work is the same
for both.
d. It is meaningless to compare the amount
of work because the forces were so different.
3.
Cubical blocks of mass m and side L are piled up in a vertical column. The
total gravitational potential energy of a column of three blocks relative to
the ground is.
a. 5/2 mgL.
b. 3 mgL.
(8)
c. 9/2 mgL. U = ˝ mgL + 3/2 mgL + 5/2 mgL
d. 6 mgL.
4.
A ball falls
to the ground from height h and
bounces to height h’. Momentum is
conserved in the ball-earth system:
a. No matter what height h’ it reaches.
b. Only if h’ < h.
(8)
c. Only if h’ = h.
d. Only if h’ > h.
5. Take a look at the graph of gravitational
potential energy of a point mass versus the separation distance between the
object and the Earth’s center. Point A
is the point of:
a. Stable equilibrium.
b. Unstable equilibrium.
(8)
c. Neutral equilibrium.
d. None of the above.
6. A plate drops onto a smooth floor and
shatters into three pieces of equal mass. Two of the pieces go off with equal speeds
v at right angles to one another.
Find the speed and direction of the third piece.
pix = 0 piy
= 0 hence pfx
= 0 pfy = 0
pfx = p1fx + p2fx + p3fx
pfy = p1fy + p2fy + p3fy
p1fx
= mv p2fx = 0 p3fx = mVx (my choice here is: piece one goes in
x direction, piece two goes in y direction)
p1fy
= 0 p2fy = mv p3fy = mVy
mv + 0
+ mVx = 0 Vx
= - v
0
+ mv + mVx =
0 Vy
= - v
V = sgrt(2) v tan(q) =Vy/Vx
q = 225 o
7.The two masses in the Atwood’s machine shown
below are initially at rest at the same height. After they are released, the
large mass, m2, falls through a height h and hits the floor, and the
small mass, m2, rises through a height h.
a. Find the speed of the masses just before
m2 lands. Give your answer in terms of m1, m2,
g, and h.
Ei = Ef
Ei = 0 J, since V1i = V2i
= 0 m/s y1i = y2i = 0 m (this is my choice of reference
level)
Ef = ˝ m1V1f 2
+ ˝ m2V2f
2 + m1 g y1f + m2 g y2f where V1f = V2f
= V y1f = + h y2f = - h
Ef
= ˝ m1V 2 + ˝ m2V2 + m1
g h - m2 g h = ˝ (m1 + m2)V 2
+ g h (m1 - m2) = 0
(20) V 2 = 2
g h (m2 – m1)/(m1 + m2)
V = sqrt( 2 g h (m2 – m1)/(m1
+ m2))
b.
Now suppose that m2 has an initial upward velocity of the
magnitude V. How high does m2 rise above its initial position before
momentarily coming to rest?
Ei = Ef
Ei
= ˝ m1V 2 + ˝ m2V2 , since V1i = V2i = V
and y1i = y2i =
0 m
Ef = - m1 g x + m2 g x , since V1f
= V2f = 0 m/s and y1f = -x y2f = + x
˝ m1V 2 + ˝ m2V2
= - m1 g x + m2 g
x
˝ (m1 + m2)
V2 = (- m1 + m2 ) g x
x
= (m1
+ m2) V2/2 g ( m2 - m1 )
9.
Dr. K is trying once again to move a box full of physics textbooks (total mass
of 30 kg) up the incline by exerting a varying force parallel to the surface of
incline. The magnitude of the force she
exerts on the box varies as F(x) = (300 – 6.00 x2) N, where x is the
distance along the incline. The
surface is rough and a constant friction force of 40 N magnitude acts on the
box. Nevertheless, she manages to move the box over 3.00 m distance.
- Calculate
the work done by Dr. K on the box.
WDr.K = int 03
(300 – 6.00 x2) dx = 300 x – 6.00 x3/3
|03 = 846 J
- Calculate
the work done on the box by friction.
Wf = fk
d cos(q)
= - fk d = (-40.0 N) (3.00 m) = -120 J
- If the
box starts from rest and has a speed of 1.00m/s at the end of its
displacement,
what is the angle of the incline?
Wnet = WDr.K
+ Wf = 846 J –120 J = 726 J
Wnet = Ef
–Ei
Ei = 0 J (box starts from rest and I assume
that yi = 0 m)
Ef = ˝ mV2 + m g yf
Wnet = ˝ mV2 + m g yf yf
= (Wnet
- ˝ mV2 )/m g = 2.42 m
tan(q) = yf
/d = 0.807
9.
A block of mass m is released from rest on a frictionless inclined plane as
shown below. The horizontal floor is rough (the coefficient of kinetic friction
for m on the floor is mk = 0.250). The mass m slides 0.20 m and makes a completely
inelastic collision with another mass M that is initially at rest.
- What
was the speed of mass m at the bottom of the incline?
(30)
Ei = Ef
Ei = m g y1i = m g L sin(q)
Ef = ˝ mVf
2
m g L sin(q) = ˝ mVf 2 Vf
2 = 2 g L sin(q)
Vf = 4.43 m/s
- The
composite particle has a speed of 0.250 m/s immediately after the
collision.
Find the ratio M/m of the two masses.
Wf = - fk
d = - mk m g d = Ef
– Ei = ˝ mVbefore2
- ˝ mVf 2
-
2 mk g d = Vbefore2 - Vf
2
Vbefore2 =
Vf 2- 2 mk g d Vbefore = 4.37 m/s
m Vbefore= (m + M) Vafter
1
+ M/m = Vbefore /Vafter
= (4.37 m/s) / (0.250 m/s) = 17.48
M/m = 16.48