QUESTION:
A ball dropped from the top of the mast of a moving ship falls straight down parallel to the mast. But with the movement of the ship, it also falls down on an inclined LONGER path. In other words, it travels over two different distances at the same time. Which is the actual 'real' distance traveled?

ANSWER:
There is no such thing as "the
actual 'real' distance traveled ". The distance traveled in one
frame of reference will be different than for another frame. For your
question, if the ball is dropped from a mast of height H , it
will travel a distance H in a time t where t = √(2H /g )
where g =9.8 m/s^{2} is the acceleration due to gravity.
Now, look at the path of the ball from a dock the boat is passing; if
the boat moves with a speed v , the distance the boat travels
before the ball hits the deck is vt=v √(2H /g ).
The distance between the starting and ending points of its path (blue
line) is √[(2v ^{2} H /g )+H ^{2} ].
For example, if H =4 m and v =5 m/s, then t =0.9
s and the distance is 6.03 m. The path followed by the ball in the dock
frame is a parabola (red line), so its length is the distance the ball
has traveled; it is rather complicated mathematically to compute this
distance.

ADDED
THOUGHT:
I got curious and calculated the distance the ball actually traveled
in the dock frame. You need to integrate ds = √[(dx )^{2} +(dy )^{2} ]=[√(1+(dy /dx )^{2} )]dx= [√(1+ 4x ^{2} )]dx
from x =0 to x=vt . The result is s =¼[2vt √(1+4v ^{2} t ^{2} )+sinh^{-1} (2vt )];
for the example above, s =9.19 m.

FOLLOWUP QUESTION:
Still confused, I appreciate your reply. The ball in the first frame of reference hits the deck at the same time as in the second, where it travels the longer parabolic path or distance. These are two observable facts.
To travel the longer distance in the same time in the second frame of reference, the ball must either accelerate faster than 9.8 m/sec^{2} - or time itself must slow down, which is both unlikely.
Your 'complicated mathematical computation' supposedly addresses the problem. But can a mathematical computation correct two conflicting observable facts?
I am thinking of Albert Einstein's skeptical quote, "So far as the theories of mathematics are about reality, they are not certain; so far as they are certain, they are not about reality."

ANSWER:
The photograph above is a time-lapse photo of two balls falling. The red
ball is simply dropped. The white ball is given a horizontal velocity.
The only force on each ball is its own weight which points vertically
down and is responsible, of course, for the downward acceleration of the
balls. Since there is no force in the horizontal direction, the white
ball's velocity in the horizontal direction remains constant. The thing
to note is that both balls, having equal vertical accelerations, fall at
exactly equal distances in equal times. This is exactly what is
happening for your question. Your error was the false assumption that
"…the ball must…accelerate
faster than 9.8 m/sec^{2} …"

QUESTION:
My dad told me about your website, very interesting reading. My question deals with molecules. When a molecule emits a photon, the mass of the molecule decreases to account for the energy in the photon. So, the mass of the molecule as a whole decreases, but this mass does not come from the "parts" of the molecule. In other words, the mass of the constituent electrons does not decrease, the mass of the protons does not decrease, so the energy must come from the electric field between the electrons and protons.
But the electric field has energy, not mass. Now mass is a form of energy, but I don't think you can say that the field has mass? But yet, it is said that the mass of the molecule decreases. The electric field contributes to the mass of the molecule, but yet it is incorrect to say that the field has mass?

ANSWER:
Actually, this is not as complicated as you are trying to make it. It
all boils down to the fact that mass is a form of energy and must be factored into
any energy conservation that occurs in an isolated system. You say that
the masses of the protons and electrons do not change, but that is not
right. Look at the simplest case, a hydrogen atom. If you measure the
mass of this atom it will be less than if you measure the masses of a
free electron and a free proton. Here is how you can see that: if you
pull the electron away from the proton, that is you ionize the atom, do
you have to do any work? Of course you do because the electron and
proton are bound together. So, you have added energy to the system (p+e)
and that energy shows up as mass. In a system as complicated as a
molecule you cannot say which particle or particles changed their
masses, but you can say for sure that the total mass of the molecule
changed by exactly the energy of the emitted photon divided by c ^{2} .

QUESTION:
A wheel rolls without slipping with angilar velocity ω and radius r what is the angular velocity of a point in the rim at the same level as the centre ?

ANSWER:
I believe that you are asking the wrong question; you must be asking
what the velocity of the point is, not its angular velocity. It is
important to understand how the wheel is rotating. Since the point of
contact with the ground is at rest at any instant, the whole wheel is
rotating about that point at that instant. The angular velocity ω
about this point is v /R where R is the radius
of the wheel and v is the speed which the center of the wheel
is moving forward with speed v . Now, the point on the rim also
has the same angular velocity ω , but its distance from
the axis of rotation is R' . Because the angle which R'
makes with the ground is 45 º, it is easy to show that
R'=R √2. Therefore,
ω=v /R=v' /R'=v' /(R √2),
and so v'=v √2; the direction of the velocity
v' is 45º below the horizontal, as shown above.

QUESTION:
I am trying to understand how to predict stopping distances at other speeds if I know a stopping distance at one speed.
Let's say a car going 25 mph with studded tires on packed powder stops in 20 meters. (I got that from a study on various types of tires on various surfaces- it's realistic). What would be a good approach to finding the the stopping distance of the same car with the same tires on the same surface, if it is going 20 mph?
We do not know the mass of the car, but perhaps it is accurate to create an acceleration constant that will apply for other speeds?
a=v/t and v=d/t so therefore by combining these we get a=v^2/d
25mph is 11.176 m/s
deceleration constant: a = (11.176m/s)^2 / 20m = 6.245 m/s^2
then can we apply the deceleration constant to the other question? What will be the stopping distance at 20 mph?
20mph = 8.94 m/s; stopping distance is 12.8m
Is this valid?
My applications include predicting future stopping distances at other speeds if one has experienced a stopping distance at a particular speed, and also, finding a reasonable speed for a traffic situation near where I live, where there is limited visibility.

ANSWER:
Your final answer of 12.8 m is right, but your method is wrong. Your answer
for the acceleration (and stopping time if you had calculated it) is wrong.
Your principle error was that d /t is the average velocity,
not the starting velocity; the velocity is changing the whole time. In this
case, uniform acceleration, the average velocity is ½v _{0
} so your acceleration is wrong by a factor of 2; a =-3.12 m/s^{2} .
The correct expression for the stopping distance is d =½v _{0} ^{2} /a
where v _{0} is the initial velocity and a is the
magnitude of the acceleration. The force which stops the car depends only on
the nature of the rubbing surfaces and the weight of the car. But the weight
of the car cancels out when you calculate the acceleration, so the
acceleration is independent of both the weight and the initial speed.
Therefore your guess that you can use one datum to determine the
acceleration and then use that for all other speeds is correct; the added
bonus is that it is also independent of how heavy the car is as long as it
is the same road surface with the same tires.

QUESTION:
If you were able to drive your car with forward acceleration in outer space and you rolled the window down and stuck out your arm, would it be pushed back like it would here on earth? This isn't a homework or test question, it is a question that came up in a fun discussion at work the other day...

ANSWER:
You had better have your space suit on because outside the car there is no
air in outer space. And the fact is that the main thing which causes your
arm to be pushed back here on earth, particularly for high speeds, is the
air drag, the apparent wind you perceive to be blowing from in front of you.
That is totally absent in space. However, you have specified that the car is
accelerating. When you accelerate forward, it feels as if there is a force
pushing you (and your extended arm) backward. There is no such force (which
is why we call it a fictitious force), but it feels like it because your
accelerating arm needs a forward force (your shoulder provides it) to
accelerate it and your brain interprets this as there is something pushing
your arm backward which your shoulder counters. Here on earth, unless your acceleration is
very large and/or your speed is very small, this force is pretty small
compared to the air drag. So the answer to your question is that your arm
will be pushed back but not " …like it would here on earth…"

QUESTION:
If a train is traveling at 60 m/s and a person runs and jumps off the rear of said train (opposite direction of travel) at 3 m/s. Which direction would the person travel?
Would they continue to move in the opposite direction of the train, move with the train at relative speed, or drop straight downward?

ANSWER:
Variations on this question are probably the most frequent
question I answer. I always say that you always have to specify velocity
relative to what . In your question, I take it that the velocity of the
train relative to the ground is 60 m/s and the velocity of the person
relative to the train is 3 m/s in the opposite direction as the train's
velocity. The velocity of the person relative to the ground is 57 m/s in the
direction of the train's velocity. An observer on the ground would see the
person with a horizontal velocity of 57 m/s regardless of whether he was on
the train or had jumped off the back. Once he is off the train he will begin
falling and hit the ground with both horizontal and vertical components of
his velocity. For example, if he starts at a height of 2 m above the ground,
his vertical component when he hits the ground will be about 6.3 m/s so his
total speed will be √(57^{2} +6.3^{2} )=57.3 m/s.

QUESTION:
I had an argument with my roommate I was hoping to resolve. Basically, can a dart bounce off of the metal rims of a dart board past the point from where it was thrown, say 8 feet. My point was that if the force is strong enough (enough force to go 50 feet w/o obstructions) when it hit the board there would be 42 feet (I dont know if this is correct) of force left over to push back. This would be enough to push it back further than the 8 feet. He seemed to believe that due to the angle of the collision, it would bounce off and not make it.
I gave an example that from 1 feet it would surely bounce back far enough, but he didn't seem to agree. Is there a physics law or example that can explain.
Note: Dart bars are curved, so a dart could come down on an angle and hit the top curve of the metal bar and get a different angle. But we were also arguing that even if the wall was flat, it could still happen.

ANSWER:
A very light object (dart) can bounce off a very heavy object (board
attached to wall) with a speed about equal to the speed it came in with if
the collision is elastic. You may know that the maximum range for a
projectile is achieved (neglecting air friction) if it starts out at a 45^{0}
angle relative to the horizontal. Suppose the dart is thrown with an angle
of say 10^{0} , so that it comes back to the same level it was thrown
from when it gets to the dart board. Then, if it was thrown at that same
speed but at 45^{0} , it would clearly go farther (assuming the board
and wall weren't there), right? So, if it bounces elastically off the
"bar" at an angle of 45^{0} it would go back farther than it came in
from.

QUESTION:
If a motorcyclist is riding at 80 mph, is he facing or creating 80 mph winds behind him? My boyfriend insists he's facing 80 mph winds because his bike is moving at and 'hitting' air mass at 80 mph. I dont understand how his motion of mass would contribute to air mass in creating 'wind' in such precise and equal measurement?

ANSWER:
The concept here is called Galilean velocity addition. If you are in
a car going 60 mph north and there is another car going 60 mph south, you
see that other car approaching you with a speed of 120 mph. In this case,
think of the air as the other car and it is at rest relative to the ground,
that is it is a calm day; then if your velocity is 80 mph north, you see the
air approaching you with a speed of 80 mph going south. Or, try this: if you
look down at the ground, you see it going backwards, that is south, with a
velocity of 80 mph, and since the air and ground are indisputably at rest
with respect to each other, the air is also going 80 mph south to you.

QUESTION:
My girlfriend and I were wondering whether an
(1) airplane travels faster over the ground as it flies east to west due to the earth's rotating underneath it, leaving out tailwind or headwind. She thinks it does not travel fast because it is in the atmosphere and she equates an example she recalled where if someone in a bus going 80 mph
(2) throws a ball straight up in the air, the ball doesn't go backwards. That just sparked an argument which I'm losing. I said it does go backward, you just can't notice because it's not in the air long enough to see a difference. It seems to me that if the bus was
tall enough and you could throw it straight up high enough, the ball
would come down behind where it was thrown because it is slowing down
while in the air.
(3) After a bit of searching, it seems that
the consensus is that the ball will come down in the "exact same" place
due to what some call conservation, others preservation of momentum. I
understand that the objects inside the bus are going 80 mph just as the
bus is - and the bus is a container. I don't understand why the ball
thrown up on the bus does not start to decelerate AT ALL after leaving
the person's hand if it has not contact with anything else beside the
air in the bus. It seems to me that it must decelerate some. I know it
would not rapidly decelerate as it would if thrown out of the window
into oncoming air; and as compared to a ball thrown 80 mph in a
ballpark, I could see if the deceleration on the bus is less rapid
because the air on the bus is going 80 and maybe it pushes the ball
forward some. But it seems to me that the air on the bus shouldn't have
much more effect than an 80 mph wind would have on a ball thrown up by a
person standing outdoors. And maybe the plane does not get an increase
in ground speed exactly equal to the speed of the earth's opposite
rotation due to whatever reasons, but it seems it should get some
increase.
(4) The atmosphere is not as fixed to the
earth as a tree, is it? Doc, (5) I'd really
rather you say something that can be used against my girlfriend here,
but any help will be appreciated. I've thoroughly enjoyed the site and
now wish I had taken physics.

ANSWER:
Whoa! Too many things going on in this question�airplanes,
balls, buses, girlfriends�I have taken the liberty to try to organize your
question by numbering parts of it. There are two issues here. First, you
always have to specify a velocity with respect to something else. What
someone on the ground measures and someone on the bus measures are
different. Second, you have to decide whether air friction is important or
not to discuss the velocity of something which moves through the air.

The airplane
flies relative to the air and if the air is at rest relative to the
ground, it makes no difference which direction the plane flies, it
travels the same ground distance in a given time.

If the air in the
bus is at rest relative to the bus and if the bus is traveling with
constant speed in a straight line, the ball will go straight up and
straight back down as seen by somebody on the bus. Somebody on the
ground will see the ball travel in a parabolic path. These statements
are if air friction is negligibly small.

The reason the
ball comes down to exactly the same place in the bus is determined by
what forces act on it. If you neglect air friction, the only force is
gravity which acts straight down causing it to slow down going up and
speed up going down. If air friction is not negligible, it is a force
which always acts in the direction opposite the velocity vector, that is
it never has a component which is horizontal which is what you need to
make it start moving (relative to the bus) horizontally. If air friction
is important, like if you throw it really fast and really high, it will
still move vertically but it will take different times to go up and
down. Unless there is a horizontal force (a wind, for example),
something moving in the vertical direction will move only in the
vertical direction.

Since you
stipulated no head wind or tail wind effects, yes, the atmosphere is
just as rigidly attached as a tree.

Sorry, your
girlfriend has a much better grasp on this than you have!

One final proviso:
remember where I said "traveling with constant speed in a straight line"?
Well, this is never rigorously true because the earth is rotating and
therefore Newton's laws are only approximately true. The effects of this
(centrifugal and coriolis forces) are so tiny that any ball throwing will be
unmeasurably small. However, for very long-range problems they can be
important. These "forces" are what are result in the circulation of weather
systems.

QUESTION:
What would the transit time between Earth and Neptune be (pick your own planetary positions) assuming an acceleration/deceleration curve that maintained 1 g of force on the spacecraft for the entire journey? In other words if our hypothetical spaceship wanted to simulate gravity by means of constant acceleration as well as deceleration (for orbital capture) how long would it take to get to Neptune. I realize that one would have to take into account the trajectory of the spacecraft since getting from one planet to another is never a straight line, but any rough answer would be much appreciated.

ANSWER:
A constant acceleration of g ≈10
m/s^{2} to the halfway point will take the same time as the
deceleration the rest of the way, so we can just calculate the time for the
first half of the trip and double it. The distance from earth to Neptune is
on the order of 4x10^{12} m. Starting from rest, the position is
x= 5t ^{2} =2x10^{12} , so t =6.3x10^{5}
s. This calculation is correct only if the maximum speed is much smaller
than the speed of light; the biggest speed is v =10t =6.3x10^{6}
m/s which is still about 50 times smaller than the speed of light, 3x10^{8}
m/s. So, the time would be about 1.3x10^{6} s≈15 days. Pretty
speedy!

QUESTION:
My car travels at 55 mph. Are the outer portions of my tires moving at a faster speed than my car? Im thinking it does, but have no formula for it.

ANSWER:
The point of your tire which is touching the ground is at rest. The
axel of that wheel is going forward with a speed of 55 mph. The top of the
tire is going forward with a speed of 110 mph.

QUESTION:
I'm trying to program a robot to shoot a basketball into a
basket. I've been trying to simplify doing this with only projectile motions
equations (The ball weighs 11.2 oz and probably is effected by drag). The
robot would be able to find the height and the distance. I was wondering if
it is possible to find the initial velocity and the angle I would need, just
from the height and distance?

ANSWER:
Possible, yes. Easy, no. It is a pretty complicated problem which does not
have a simple formula you can use. I would try the simple model first and
see how it works.

FOLLOWUP QUESTION:
So are you telling me to just use the projectile motion formulas? When I do them I end up with 3 variabled, time, initial velocity and theta

ANSWER:
I thought your concern was the air drag. So, when you do the
kinematic equations, there are two:

One is the horizontal position which must end
up at x=R where R is how far horizontally the basket is
from the launch point; R=v _{0} t cosθ
where v _{0} is the initial velocity and t is
the time of flight.

The other is the
horizontal position which must end up at y=h where h
is how far vertically the basket is above the launch point; h =v _{0} t sinθ- �gt ^{2
} where g is the acceleration due to gravity (9.8 m/s^{2} ).

So, you have two
equations with three unknowns,
v _{0} , θ , and
t . The first thing to do is reduce it to one equation with two
unknowns by solving the R equation for t and substituting that
into the h equation. The result is h=R tanθ- �gR ^{2} /( v _{0} cosθ )^{2} .
Now, think about it�you cannot mathematically know both θ and
v _{0} because you have only one equation; but you
shouldn't expect to from a physical point of view either. If you change v _{0
} you will have to change
θ too to hit the target. So, there are an infinite number of
right answers here, not one unique answer. Nevertheless, if you program your
robot to always shoot with the same velocity, you can solve for the angle.
So, you need to solve the above equation for θ and that will tell you
how to aim. It is not trivial algebra, but not real hard either and I do not
intend to present the general solution because it is complicated. Be aware
that there will be two solutions for every situation which you can easily
understand, I think. Think of a rifle shooting at a target where you have to
aim a little above the target to hit it; if you also shot way up in the air,
there would be some large angle where it would come back down through the
target. I will leave the general solution to you; I will do one simple
special case to illustrate, R =4 m, h =2 m, v _{0} =10
m/s, and I will approximate g as 10 m/s^{2} : 2=4sinθ /cosθ- 0.8/cos^{2} θ.
Rearranging and simplifying (and using sinθ= √(1-cos^{2} θ )),
0.4+cos^{2} θ= 2cosθ √(1-cos^{2} θ ). Now
square this equation and rearrange: 5cos^{4} θ- 3.2cos^{2} θ+ 0.16=0.
This is a quadratic equation in cos^{2} θ. Thus, cos^{2} θ= (0.585
or 0.055) and cosθ= (0.765 or 0.234) and θ=( 40.1^{0} or
76.5^{0} ). Of course, you could also fix θ and solve for v _{0} .
This is easier. For example, if you fixed θ at 40.1^{0} , 2=4tan40.1^{0} -80/( v _{0} cos40.1^{0} )^{2}
which leads to, guess what, v _{0} =10 m/s. However, you will
run into trouble here because since you are playing basketball, the ball
must be coming down when it gets to the basket, and that will not always be
the case for a randomly chosen set of R , h , and θ. For
example, in the example I did, 40.1^{0} would not work because the
ball would be going up when it got to the basket (v _{y} =1.2
m/s); for the other solution, 76.5^{0} , the ball would be going down
(v _{y} =-7.4 m/s) like you need to make the basket. I got
these by solving for the times from the x equation and then using the
velocity equation (v _{y} =v _{0} sinθ-gt ).
You have your work cut out for you!

QUESTION:
Can you settle an agrugment here Please!!!!! If you have two cars and they are driving towards each other can a accurrate speed be determinded. My son thinks that physics would apply here.

ANSWER:
It is not really clear what you are asking. If you want to know if
one car can deduce how fast the other is going (relative to the road) then
the answer is yes if the the first car knows what his speed is (relative to
the road). For example, if the first car is going 50 mph north and measures
the second car to be approaching him from the south at a speed of 120 mph,
the speed of the second car (relative to the road) is 70 mph. Did you get a
speeding ticket from a cop in a car approaching you and with a radar gun? He
can do it!

QUESTION:
if i have a set distance, an acceleration rate and a deceleration rate is there a way of calculating the maximum attainable speed in which to still stop within that distance? ie distance of 52.15m, acc 2.19m/s/s and deccelareation rate of 6.67m/s/s. what would the theoretical maximum speed be?
[ Not
home work just work! I'm an accident investigator for lancashire
police.]

ANSWER:
The equations of motion for uniform acceleration a are given
by x (t )=v _{0} t +�at ^{2}
and v (t )=v _{0} +at where
x (t ) is the distance traveled in time t ,
v (t ) is the velocity at time t , and v _{0} =v (t =0).
I will adopt the notation that

x _{1}
is the distance traveled by the accelerating object in a time t ,
v _{1} its velocity at time t , t _{1}
is the time after starting that acceleration stops, and a _{1}
is the magnitude of its acceleration; and

x _{2}
is the distance traveled by the decelerating object in a time t ,
v _{2} its velocity at time t , t _{2}
is the time after starting deceleration that it stops, and a _{2}
is the magnitude of its deceleration.

The object starts at
rest and so the distance it travels in time t is x _{1} (t )=�a _{1} t ^{2
} and its velocity is v _{1} (t )=a _{1} t.
Now the object starts decelerating (physicists do not like that word!)
beginning with a speed of v _{0} =v _{1} (t _{1} )=a _{1} t _{1
} so the distance it travels in time t is x _{2} (t )=a _{1} t _{1} t -�a _{2} t ^{2
} and its velocity is v _{2} (t )=a _{1} t _{1} -a _{2} t
(note the minus sign because it is slowing down). Now, v _{2} (t _{2} )=0=a _{1} t _{1} -a _{2} t _{2
} (it comes to rest at the end), so t _{2} =a _{1} t _{1} /a _{2} .
The total distance traveled is s =x _{1} (t _{1} )+x _{2} (t _{2} )=�a _{1} t _{1} ^{2} +a _{1} t _{1} t _{2} -�a _{2} t _{2} ^{2} =�a _{1} t _{1} ^{2} +a _{1} ^{2} t _{1} ^{2} /a _{2} -�a _{2} (a _{1} t _{1} /a _{2} )^{2} =�a _{1} t _{1} ^{2} (1+(a _{1} /a _{2} )).
Final results, using your numbers:

t _{1} =√[2s /(a _{1} (1+(a _{1} /a _{2} )))]=5.99
s,

t _{2} =a _{1} t _{1} /a _{2} =1.97
s, and

v _{max} =v _{1} (t _{1} )=a _{1} t _{1} =13.1
m/s==29.3 mph.

QUESTION:
i got a good question if threw a rock downward at the same speed i threw it upward... wouldnt they meet at one point because they all go the same speed?

ANSWER:
The equation for motion is y=v _{0} t- �gt ^{2}
where v _{0} is the initial velocity, g is the acceleration due to
gravity, y
the height above (or below) where you threw it from, and t is the
time since it was thrown. I assume you are throwing the two at the same
time, so one has an equation of motion
y _{1} =v _{0} t- �gt ^{2}
and the other
y _{2} =-v _{0} t- �gt ^{2} .
When
y _{1} =y _{2} , the two are passing, so v _{0} t- �gt ^{2} = -v _{0} t- �gt ^{2} ,
and the two will pass only when
v _{0} t=-v _{0} t, only when t= 0;
this is when you threw them �they will
never pass again. They do not go the same speed, they have the same
acceleration. Their speeds are always different (except at the instant they
were thrown).

QUESTION:
If two cars are traveling at 50 mph towards each other what is there
speed when they pass? My boyfriend said it doubles to 100 but that makes
no sense! They do not increase speed nor does their speed differ! How
can the action passing another car moving at the same spew miraculously
double your speed?

ANSWER:
Velocity is one of those things which you must specify what it is
relative to. In your example, the speeds 50 mph are the speeds of cars
with respect to the road . As they pass, their speeds remain the same 50
mph. What your boyfriend is talking about is the speed of one car with
respect to the other. Each car sees the other one approaching at a rate of
50+50=100 mph; there are no sudden changes in speed, each sees the other
going 100 before, during, and after passing. This is called the velocity
addition theorem. If the two cars collide head-on, it is the same as if one
car were parked and the other car going 100 mph. If the two cars were
traveling in the same direction, each sees the other as being at rest,
50-50=0 mph; the minus sign signifies moving in the opposite direction.