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Massachusetts Institute of Technology Physics Department
Physics Fall
Solutions for Assignment
by Dru Renner
Problem
a We need to mathematically describ e the motion of the rst stone This is one
dimensional motion in the vertical direction with a constant acceleration due to the
Earths gravity Lab el the vertical direction x with the x axis p ointing up from the surface
of Earth and with the origin x at the surface of Earth If we set v 0 m s and
g m s
2 and cho ose the time t to coincide with the throwing of the stone then
the motion is
x v 0
t �
2
g t
Now we can ask when is the rst stone at the height h m This question
leads to the equation
h v 0 t �
2
g t
which is a quadratic equation in t with the two solutions
2 v 0 �
q
v 0
� g h
t �
) t �
s and t
s
g
0 0
If we lab el the quantities for the second stone x
0
t and v 0
then similarly
0 0 0
02 x v 0
t � g t
But you must b e careful here t
0 is the amount of time elapsed since the second stone was
thrown and the second stone is thrown s after the rst stone is thrown So there
is a delay of t s b e tween the throwing of the stones and thus we can write the
relationship b etween t and t
0
as
0
t t � t
Now at the times t t �
or t
0 t �
� t t he two stones are required to hit implying
that x
0
h This leads to the equation
0
h v 0
t� � t � g t� � t
2
with the solution
h
1
2
g t �
� t
2
0
v 0
t �
� t
So for the two p ossible values t �
there are two p ossible velo cities v
0
0
0 t �
s with v 0
m s mph
and
0
t+ s with v 0
m s m ph
The sp eed of mph would b e a very go o d fast ball and almost as go o d for a tennis
serve which could b e nearly mph therefore the plausible answer is m s
mph
b Now you must wait s b efore throwing the second stone The equation ab ove
0
v h
1
2
g t �
� t
2 t �
� t still works but now t s The rst thing you 0
should notice is that t �
s is less than t s So the rst ball reaches the
height h for the rst time b efore you even throw the second ball So the only p ossibility
is to strike the rst ball at the second time
0
t
s with v 0
m s mph
Problem
If an ob ject exp eriences free fall for a length of time t then the distance it falls is
2
given by d
1
g t Thus measuring b oth t and d provides a value for g 2
d
g
t
2
If the error for the measurement of d is d and the error for the measurement of t is t
then by our simple metho d the error for g is
d d
t � t
2
� g
where the rst quantity represents the largest value of g consistent with d and t
The data from lectures the resulting values for g with the exp erimental error and
the consistency with the value of g m s
2 for Boston are given b el ow Both
lectures make the identical measurement for the m drop
which gives
A
� 1 � 1
cos
x
cos
�
jAj
p
x
Similarly
A
� 1 � 1
cos
y
cos
�
jAj
p
y
and
A
� 1 � 1
cos
z
cos
�
jAj
p
z
Problem
You are given the vector
v x y z
The magnitude is jv j
p
so the vector
^0
v v x y z
jv j
has the same direction as v but magnitude jv
0
j Finally the vector
^00 ^0
v v x y z
has the same direction as v
0 which has the same direction as v but now has magnitude
v
00
j j
Problem
You are given the distances to Venlo km and Eindhoven km but you are
not given any information concerning the directions to these cities The greatest distance
b etween the two cities is km km km This o ccurs when the two cities are
in opp osite directions The shortest distance b e tween the two cities is km km
km This o ccurs when the two cities are in the same direction The actual distance is
km which is greater than k m a nd less than km
Problem
You are given the vectors
� y
A x and B � x ay � z
Now you must cho ose a value of a that makes A and B p erp endicular Mathematically
you must satisfy
A � B
) A B A B A B
) � � a �
�
2
x x y y z z
a 3
Problem
You are given the vectors
A �x � y z
and B x y � z
The calculations for parts a b and c are all done comp onent wise
a A B � x � y � z
�x � y � z
b A � B � � x � � y � � z
�x � y z
c A � B � � x � � y � � z
� x � y z
d A � B � � �
B � A � � �
In fact it is true that A � B B � A for any two vectors
e A � B � � � x � � � y � � � z
x � y z
B � A � � � x � � � y � � � z
�x y � z
� x � y z
In fact it is true that A � B �B � A for any two vectors
Problem
The p osition as a function of time is given as
2
r t � t
2
� tx y � t � t z
a The velo city vector at any time t is given by
d r
v
dt
d d d
� tx t � t
2
y
dt dt
� � ty
� t � t
2
x
dt
x � � tz
�x � ty � � tz
So in particular at t
v �x � y � � z
� x � y z
b The sp eed at t i s given by the magnitude of v at t which is
jv j
q
2
2
2
p
So the sp eed is m s
c The acceleration vector at any time t is given by
d v
a
dt
d d d
x t y tx
dt dt dt
x y z
y z
So in particular at t
a y z
The magnitude of a at t is
2 jaj
q
2
2
p
So the magnitude of acceleration at t is m s
2
Problem
Cho ose the z axis p ointing vertically up the x axis p ointing horizontally along the
direction of the car and cho ose the x z origin to b e the corner of the ramp Also cho ose
the t origin to b e the moment when the car is at the corner of the ramp and let v 0 x
b e the
unknown sp eed of the car at that moment The subsequent motion of the car is pro jectile
motion
The vertical motion of the car is given by
2
z � g t
So we can ask the question when do es the car fall the distance h m This gives the
equation
�h � g t
2
with solution
v
s
u
h u^ m
t
t
s
g ms
2
The horizontal motion of the car is given by
x v 0 x
t
so at t t
the horizontal p osition is
x v 0 x
t
For the stunt driver to avoid crashing x
must b e larger than the distance d m
So
x d
v
0 x
t
d
v
0 x
d
t
v
0 x
q
d using t
q
2 h
2 h
g
g
Therefore to clear all the cars
24 0
v 0 x
p^
2 � 2 0
9 8
ms
2 cos sin
d m
The ball will travel m � m m farther
c Supp ose the angle was greater by
� We want to determine d when
�
�
�
and v m s
cos sin
d m
The ball will travel by m � m m farther
Problem
The motion of the skier is identical to the motion of the car in problem if you
m
cho ose a similar co ordinate system with v
km h
�
3
m s So the s
motion is given by
x v
t and z � g t
a The equations ab ove descib e the motion of the skier but the equations do not
know where the ground is We must supply that information Normally we cho ose the
co ordinates so that the ground corresp onds to z but that is not true in this case The
gound is really a hill It starts at x and z but then it slop es down at an angle
of
� i e it is a line through the origin with slop e The mathematical description of
the line is
z g r ound
�x g r ound
The skier landing on the slop e is mathematically indicated by the i ntersection of the curve
of the skier and the curve of the ground Supp ose this intersection o ccurs at the horizontal
p osition x b ecause this is a p oint on the ground we k now that the corresp ondingly z �x
So if the skier lands at time t t
then
x �z
v
t
x
z �
g t
v t
)
g
t
) t
or
t
v 0
g
of which the non zero solution is the one that corresp onds to the skier landing At that
time
2
ms
2
x v 0
t
v 0
m
g
and z �x � ms
So the distance down the slop e is
p
d x
2 z
2 m
b The skier attains large sp eeds which make considerations of air resistance necessary
Air resistance makes the actual distance shorter than our calculated distance This will
b e explored in more detail in lecture
Problem
If the radius of the orbit is R
11 m then the circumference of the orbit is
C R
11
m
11
m
The Earth travels this distance once in the time year s
7 s So its sp eed is
C R
11 m
v
4
ms
7 s
So the centrip etal acceleration is given by
� � 2
2
2 R
v
2
R
4 ms
2
a
R R
2
2
� 3
ms
11 m
Problem
In problem we actually calculated the centrip etal acceleration in terms of the
radius of the orbit and the p er io d of the orbit That result a
2
R
2
is true for all
the planets This result do es assume that all planets have circular orbits but most orbits
are elliptical so the radius we use is the average distance of the planet to the sun This
was referred to as R in the lecture supplement for (^) Elliptical orbits will b e
discussed later in the course
a Here is a table of information found by asking what is the distance to out planets
on the web site httpaskcom and by using the formula ab ove for a
You should notice that all the planets indep endent of their mass lie on this curve
This indicates a relationship b etween radius and acceleration log a � log R C
which implies that a K R
2
where C and K are constants indep endent of the mass of
the planet As was shown in lecture Notice that the expression from problem
related a and R but this expression is indep endent of indicating that there must b e
additional relationships b etween and R
Here are a few lines of co de that will pro duce graphs similar to the one ab ove First
you need to nd Mathlab Log onto an Athena workstation and either cho ose the
Numerical Math option from the top then cho ose Analysis and Plotting and MAT
LAB or at the command line typ e add matlab matlab Then at the prompt which
should lo ok like typ e the following lines
radii
acc
plot logradii logacc o logradii logacc
Problem
Let w b e the sp eed of the wind blowing from A to B v b e the sp eed of the plane
measured relative to th e air and d b e the distance from A to B
For the trip from A to B the ground sp eed sp eed of the plane measured relative t o
the ground would b e
v w
and the time it would take to travel that distance would b e
d
t A!B
v w
For the trip from B to A the ground sp eed would b e
v � w
and the time of ight would b e
d
t B !A
v � w
At this p oint you should notice that we could make t B !A
b e very large by making w � v
which already indicates that the round trip with wind will b e longer
The total time of ight would b e
d
tw tA!B tB !A d
v w v � w v � w
2 v
2
The travel time without the wind w is
t 0
d v
so we can write tw as
t w
t 0
� w
2 v
2
For jw j v
w
2 v
2
so t w
t 0
and the round trip with wind is longer But for jw j v
w
2 v
2
which would make t w
So clearly we should b e careful here
The sp ecial p oints w v cause problems the ab ove expression for t w
b ecomes
in nite In fact for w v the ground sp eed for the B to A p ortion is v w indicating
that the plane will never b e able to leave p oi nt B This means that someone standing
on the ground at B will see the plane oating still in the air Where as a bird in the
wind will see the plane as moving forward So all considerations of the trip from B
to A are irrelevant the plane never makes that p ortion of the trip This is revealed in
the expression for t B !A
which b ecomes in nite for w v So the analysis ab ove breaks
down we can not talk ab out the round trip b ecause it do esn t o ccur In fact for w v
the ground sp eed is v w and the plane continues to b e blown further in the A to
B direction
If the wind were to blow from B to A instead the ab ove results would work by
considering w And when w v when the wind blows from B to A with sp eed v
the ground sp eed from A to B would b e v w The plane would never even b egin
its trip since it couldn t leave A And for w v the plane immediately is blown further
away from B So for this case the plane never even completes the rst half of the round
trip
