Conceptual Physics: Solving Problems in Math & Science

[Dsoto489]

Hello friends!
I have just started a conceptual physics class and have not taken math or science in 6 yrs! Therefore, even some of the most fundamental physics questions are troubling me. I am not looking for answers, just logical steps to proceed with each problem. Below I will post some ideas that are giving me a headache.

1.Given a constant speed and time to stop, what is the acceleration?
I know that slowing down will cause a negative acceleration, so my answer should be negative.

2.Given two intervals of speed and the time is took to go from speed interval 1 to 2, what is the required acceleration? What I did was subtract the 2nd interval from the first and divided it by the time- I still do not know if it is correct.

3.Ball rolling off table with horizontal velocity, time to fall- how high is the table? I know that gravity will effect the ball by 9.81m/s/s- the book I have tells me to round the acceleration of gravity to 10m/s/s- which I do not like; I have no idea where to start though.

4. Same as 3 but given table height and horizontal velocity, what is the distance from base of the table where the ball will strike? Both 3/4 deal with gravity and this damned ball. Having gravity in my problem seems to throw me off. Is horizontal velocity dependent on vertical velocity? Does time to hit the floor depend on the balls horizontal velocity? Should I be worrying about Initial velocity? or can I assume that the ball is initially still, thus showing an initial velocity of 0?

Thanks again to anyone that can assist me. Again I am not looking for final solutions, just ways to think about these problems when I see them and which formulas to use appropriately. These are the 4 questions out of 20 that are giving me trouble.

~Dsoto489

 

[Dsoto489]

Am I posting incorrectly? I saw the format while trying to post, but do not understand which formulas to use, thus I cannot work out these problems. If anyone could help, please do!

Thanks!

 

[proculation]

Well we don't see any formulas

 

[Dsoto489]

ok, the equations I have in my notes that may work are

Acceleration=Vf-Vi/t

d=Vot +1/2at*t

v=Vo+at

 

[proculation]

But what are the questions ? You are giving basics formulas of dynamics.

 

[Dsoto489]

here is one real problem.

A ball rolls off a shelf which is 2m high, if it leaves the shelf with a horizontal velocity of 8m/s, calculate the distance from the base of the shelf at which the ball will strike the ground. I know that I need to find how long it will take the ball to strike the ground and I can use that to find the horizontal distance.

So,

Height of table/shelf: 2meters
Horizontal velocity: 8m/s
Horizontal distance: ?
Time to fall: ?
Vertical Velocity(Acceleration of Gravity): 9.81m/s/s

 

[proculation]

I don't want to answer your problem for you but I will give you some clues:

- You always need to separate the vectors in 2 equations in those case: x (horizontal) and y (vertical). You use trigonometry to do that.

- I know that basic problem and a lot of people does the same mistake. What is the acceleration of the ball after it quits the edge ? Both in X and Y.

You need to solve that problem with the formulas you gave here but in two times: horizontally and vertically.

Try that.

(look at you trigonometric circle for sinus and cosines)

 

[Dsoto489]

Ok, so I have a horizontal velocity of 8m/s and the height of the shelf is 2m, so cancelling and dividing, I get the time to fall is 4 seconds. Assuming the ball was at rest then at 8m/s the intial V is 0 making the acceleration is 8m/s.

A=8m/s
t=4s
d=Vot+1/2at*t

d=1/2(8)(4*4)

d=64m

 

[proculation]

You mixed velocity (speed) with acceleration.

 

[Dsoto489]

hrm, sorry I am sooo lost.

 

[proculation]

Another clue:
(horizontally)

Vo = 8 m/s
Vf = 8 m/s
There is NO horizontal acceleration since F=MA and there's is no horizontal force when the ball is in the air.

 

[Dsoto489]

but if the problem states that it is traveling at a horizontal velocity of 8m/s how do i assume that that the acceleration is 0?

a=V-Vo/t

so a=8-8/4? a=0?

F=ma, i think i know this formula, but the problem does not give me the mass of the ball...

 

[Dsoto489]

So horizontal velocity only stays the same on a flat surface, once the surface is gone, i.e. the table and the fall of the ball, gravity is the only velocity?

 

[Dsoto489]

I think i am getting me velocity and acceleration totally mixed up. Velocity is the change in direction over time, acceleration is the rate of change over time> am i correct for that assumption?

 

[proculation]

You problem doesn't ask you for the mass. The mass doesn't change anything. 1 ton of steel will it the ground at the same speed than one golf ball (leaving away the resistance of the air).

You resolved the first equation, the X one. You cannot get the time (t) from it.

Now, go for the Y equation.

Hint: there is an acceleration.

 

[Dsoto489]

for the Y the acceleration must be gravity, 9.81m/s/s

 

[proculation]

I think i am getting me velocity and acceleration totally mixed up. Velocity is the change in direction over time, acceleration is the rate of change over time> am i correct for that assumption?

Velocity is the change of DISTANCE over time. Acceleration is the change of velocity (speed) over time. All that in a plan frame in your problem. X and Y.

 

[Dsoto489]

how can i easily tell which equation/formula works for x and for y?

 

[proculation]

They are the same. Newton said so.

 

[Dsoto489]

a=f/m and f=ma

1st law states an object in motion will remain at a constant velocity unless acted on by a net force.
2nd Law states the acceleration of an object is always directly prop. to the net force and inversely prop. to the mass.

so, the ball will continue to travel at a horizontal v of 8m/s until it hits the floor, it hits the floor because of the acceleration of gravity...

 

[proculation]

a=f/m and f=ma

1st law states an object in motion will remain at a constant velocity unless acted on by a net force.
2nd Law states the acceleration of an object is always directly prop. to the net force and inversely prop. to the mass.

so, the ball will continue to travel at a horizontal v of 8m/s until it hits the floor, it hits the floor because of the acceleration of gravity...

you got it right buddy.

(by leaving out the friction of the air of course but that's not in your problem)

 

[Dsoto489]

So what really makes me frustrated is which do i compute first, x or y? Must I always start with the x? or start with whichever I have more data for? Also, I am not even sure which equation is for x and for y...eggaaad. physics is destroying me...thanks for the help so far...

 

[proculation]

You always separate the two to get the much information you can (variables result). You tried X. You got nothing. Now, go for Y.

Hint: you could have begun with Y and you would have got the answer right away.

 

[Dsoto489]

but which of these solves for the horiz(x) and the vert(y)

d=Vot+1/2at*t

V=Vo+at

 

[Dsoto489]

I think horizontal and vertical requires this:

a=
Vo=
V=
d=
t=

I know
Vo=0
V=8m/s
d=?
t=?

 

[proculation]

you solved X !

Now, go with Y !

What are the info you have ?

aY=9.81
VoY= 0
VfY= ?
dY = 2
t(x,y) = ?

You have two unknown variables. You need two equations to resolve it.

 

[Dsoto489]

ok so for vertical

I know

a=-9.81m/s/s
Vo=0
V=?
d=2m
t=?

so, 2=0(t)+1/2(-9.81)t*t


somehow i have 2=1/2(-9.81)t*t

2=-4.905t*t
2/-4.905=t*t

 

[proculation]

The minus only give you the direction of the vector (up or down). You can remove it.

 

[Dsoto489]

somehow i am getting a negative number for time

 

[Dsoto489]

is t= the square root of .407747? lol omg

 

[proculation]

actually, the "d" should have been "-2", so the it should have been:

-2/-4.905=t*t
2/4.905=t²

 

[Dsoto489]

is the d (-) because it is falling?

 

[proculation]

Because you are starting from the shelf (y=0) to the ground (y=-2) and you use Vf-Vo for the difference.

 

[proculation]

So now you have "t", put it in your X equation to get the distance from the edge of the shelf where it falls. And Voila !

 

[proculation]

Physics is not that difficult you see :-)

 

[proculation]

be back in 10 minutes, corner store time for some beers.

 

[Dsoto489]

ok, now i have

X
a=0
Vo=8m/s
V=8m/s
d=?
t=.64s

y

a=9.81
Vo=0
V=6.26
d=2m
t=.64s

 

[Dsoto489]

I keep trying to plug the equation only to cancel out most of the numbers and get D=T

When and if you get back on here, could you just run me through the steps when approached with these types of problems? maybe give me another example I could run through? I really appreciate the help here-

 

[proculation]

d=Vo(or Vf it's the same) X t

remember, distance = velocity X time

 

[Dsoto489]

ok so my D=5.1 m/s/s

 

[proculation]

You run a car at 100 mph for 2 hours. How much distance will you travel ?

d = 100 x 2 = 200 miles

 

[Dsoto489]

agreed, i understand the d=v/t formula, its the bigger one, that deals with this entire problem that i keep tripping on.

 

[proculation]

d = 8 m/s X 0.64s

meters/seconds X seconds

seconds/seconds = nil

only meters stay in unit.

 

[Dsoto489]

yea, after i reread the problem i saw that i kept the s. So 5.1m is the distance the ball will end up at when i reaches the floor.

 

[proculation]

Was it so difficult ? :-)

 

[Dsoto489]

totally! ahah, ok so i have a similar problem which has the same horiz velocity of 8m/s but instead height of the table they give me time(.24s). Do i work the problem the same way?

 

[Dsoto489]

or can i just use the D=v/t

 

[proculation]

I think you have all the information you need.

Switch the variable in Y and resolve with the X equation like you did.

 

[Dsoto489]

d=8(.24)+1/2(o)(.24*.24)

 

[Dsoto489]

so the height of the table is 1.92m tall