Download Mid Term S05 My Solutions PDF with thumbnails 05/26/05

NAME __Me___Hulan E. Jack Jr.________My Solutions_______________March 27, 2005
Borough of Manhattan Community College
Course Physics 110
Instructor: Dr. Hulan E. Jack Jr.
Date March 22, 2005
Mid Term Exam - My Solutions
1.
a.
Suppose that critters are discovered on Mars who measure distance in glups and time in clicks.
What would the units of velocity be ? Explain. (5 pts)
Velocity = displacement/time = ∆x/∆t
b.
What would the units of acceleration be? Explain. (5 pts)
Acceleration = velocity/time = ∆v/∆t
c.
units = (glups/clicks)/clicks
What would the units of speed be in this system. Explain. (5 pts)
Speed has same dimensions as velocity
2.
units = glups/clicks
units = glups/clicks
Given the two vectors A and B .
a.
On the diagram, add the two vectors to get the vector A+B (10pts)
b.
Explain what you did. (5 pts).
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= glups/clicks2
3.
A car travels along a straight road as pictured by the x vs t
graph shown. x is the displacement and t is time.
a.
During which of the time intervals, A, B, C or D,
does the velocity have the largest magnitude? ( 2 pts)
Explain. (5 pts)
x
B
C
D
A
v = ∆x/∆t = slope of x vs t curve . D has the
t
steepest slope, the largest magnitude. Note relative
sizes of the ∆t ‘s and ∆x’s in the picture.
b.
During any of these time intervals does the car have
an acceleration? (3 pts)
Explain (5 pts)
v = ∆x/∆t = slope of x vs t curve . The slope of is
continuously change in interval B. Hence v is continuously
changing . But, a = ∆v/∆t = slope of v vs t curve . So
the car accelerates in interval B. Note in the picture how
the slope (the dashed blue lines) changes in the B
interval.
4.
You push a box across a rough (friction) flat horizontal floor.
a.
b.
On the picture, sketch all of the action-reaction pairs acting
between the three ; you-box, you-floor, and box-floor. (10
pts)
myou
mbox
rough floor
On the picture, add whatever is needed make the free body diagrams (FBD) of you and
the box. (5pts)
Notes: 1. Always have action-reaction forces touch body on which they act.
2. Weight always acts from the center of the body.
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NAME __Me___Hulan E. Jack Jr.________My Solutions_______________March 27, 2005
Borough of Manhattan Community College
Course Physics 110
Instructor: Dr. Hulan E. Jack Jr.
Date October 29, 2004
Mid Term Exam - My Solutions
5.
An elevator of mass m is pulled upward by a cable causing it to have an upward acceleration a
a.
Sketch the Free Body Diagram
(FDB) of the elevator (5pts)
b.
What net external force F is required to for
the acceleration to be a=3g up?
Explain (5 pts).
FBD
Newton’s 2nd Law of Motion Fnet
c.
Express this force in terms of the elevator’s weight W.
Explain. (5 pts)
W =mg,
6.
= ma
= m(3g) = 3mg.
So Fnet
= 3mg = 3W.
A body having mass m is on the surface of the earth, a distance RE from the center of the earth.
The acceleration due to gravity on the surface of the earth is gE .
a.
Write the formula for weight of the body W in terms of its mass m. (5pts)
W = mg .
b.
From Newton’s Law of Gravitation FG = G
M
show that the g E = G E2
RE
1.
2.
3.
M m
R2
(10 pts)
W equals the force of gravity on mass m. So FG = W = mg .
On earth M = ME , R = RE and g = gE
FG = G
(G
ME m
= W = mg E
R E2
ME
M
) m = mg E . Cancelling m gives g E =G E2
2
RE
RE
Notes: b1 and b2 are part of my explanation - NOT required from students.
b3 is required from students.
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ME
-FG RE FG
O
m
7.
A spring gun is loaded with a rubber dart, the gun is cocked, and then fired at a target on the
ceiling. Describe the energy transformations that take place in this process.
Explain what these
pictures have to do with
providing the asked for
description. Include any
physical laws and
formulas that apply. (15
pts)
This is an example
of Conservation of Energy
PE + KE = const (constant)
.
At 1 - all PE spring
At 2 - half the PEspring converted to KE.
At 3 - All PEspring convert to KE,
At 4 - Some KE converted to PEgravity .
Note that the size of the energy box does not change !!! That is the meaning of the
Conservation of Energy
8.
Initially, Car A has mass mA = 1000 kg travels eastward with a velocity vA = 20 m/s east, and
Car B with mass mB = 800 kg travels westward with a velocity vB = 15 m/s west. They collide
in a head-on collision and stick together.
a.
Calculate the linear momentum of Car A, pA . (3 pts)
pA = mAvA = 1000 kg 20 m/s east = 20000 kg m/s east
b.
Calculate the linear momentum of Car B, pB . (3 pts)
pB = mBvB = 800 kg 15 m/s west = 12000 kg m/s west = - 12000 kg m/s east
c.
Calculate the initial linear momentum, pinit . (3 pts)
pinit = pA + pB = (20000 - 12000) kg m/s east = 8000 kg m/s east .
d.
What is the final momentum ? Explain - a formula helps. (6 pts)
Conservation of Linear Momentum
So,
pfinal = pinit .
pfinal = 8000 kg m/s east.
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