Simple Physics Problem (need help)

[Freak'n Big Panda]

Here's the problem:

Luke, sitting in a car traveling north @ 30m/s throws a ball directly across the car @ 3 m/s. What is the velocity of the ball with respect to the road?

First off doesn't the direction that the ball is thrown matter? My teacher said it didn't but 3 m/s East != 3m/s West... they are the opposites of each other. Anyways I assumed the ball was thrown west to east and got an answer of v = 30.153 [5.71° E of N]. Apparently the answer is 30.153 [5.71° E of S].... this makes absolutely no sense to me whatsoever. Is my teacher just on drugs or am I wrong?

Here's the logic I used to solve the problem:

-Ball initially has a velocity of 30m/s north and is not accelerating
-Luke applies a force to move the ball 3 m/s west to east within the car
-The velocity of the ball from the reference point of the car is 3 m/s [E]
-The velocity of the ball from the reference point of the road is 3m/s[E] + 30m/s[N] = 30.153 [5.71° E of N]


Anyways if somebody could please point me in the right direction I'd apperciate it. Thanks!

 

[Neeyik]

I must be dense too (which would be the first time) because I'm getting the same answer as you - regardless as to which direction the ball is thrown (E to W, W to E), an observer on the road will still view the ball travelling North. Basically your teacher seems to be saying the if one stands by a roadside and watches a car travel past, any object thrown at you from the car will shoot backwards down the road.

I'm actually having doubts over my own thinking here because I've tried it every which way I can (making it into a boat and current question or ball in the wind question); I really don't see how it can be going South.

 

[Basic]

The direction doesn't matter for the absolute value of the velocity. Maybe that was what your teacher meant.

If he says that the velocity is 30.153 [5.71° E of S] when the car is travelling north, it's just a typo. Teachers are humans too, and do those once in a while.

You seem to have understood how to calculate it, so no need to point you anywhere.

PS
One thing though, for some strange reason you answer is 30.153 instead of the 30.150 I get. Did you round off some intermediate values?

 

[Fred]

Your teacher is on drugs, that is all. The only way it could point south is if you had air resistance and were in the frame of the car. You can tell him an assistant physics professor said that =)

 

[Freak'n Big Panda]

awesome thanks guys. I'll talk to my teacher tomorrow. And no it wasn't a typo... she actually told us (in front of the whole class) the velocity of the ball relative to the road would be 30.153 [5.71° E of S]. So yeah she's on drugs well w/e. Are you guys absolutely 100% sure I'm right? I'd hate to start to get into a argument with her only to find out I'm in error x_x And since the value she wanted us to calculate was indeed a vector quantity the direction the ball was thrown does in fact matter, right? If the ball was thrown East to west the calculated velocity vector would be 30.153 [5.71° W of N], no?

One thing though, for some strange reason you answer is 30.153 instead of the 30.150 I get. Did you round off some intermediate values?

Yeah I must have, I just redid the calculation and got the same 30.150 you got.

 

[Basic]

Well, then it was the verbal equivalent of a typo.

There's no need to escalate anything into an argument of any sort. Just show some social skills and ask her if she meant that the direction is 5.71° E of N when the car is going north, and you're throwing the ball towards east. She'll say yes, or maybe "no I meant when the car was going south", or some other explanation.
And you'll say: "OK", since it isn't much to argue about.

The important part of the task was likely to get the figures 30.153 m/s and 5.71º, and then she let you decide if the car goes N or S and if you throw the ball E or W. Doesn't matter much if you just compensate for it in the answer.

 

[Freak'n Big Panda]

Well the problem is I think she actually doesn't know how to approach the problem from the correct angle... the equation she was telling us to use in class was the completely wrong equation to use. It could not be applied to the question above. This is directly off my hand out she gave us yesterday:

The velocity of an object A relative to an object B is:

Vab = Va - Vb

she plugged the car's velocity in as Vb and the ball's as Va and came up with that answer. This was all done in front of the class btw. I'm sure the majority of students belived it to be true. Obviously this wont give the correct result because the equation asumes both Va and Vb are mesured with the same frame of reference. Anyways it's not a big deal. I really couldn't care less, I'm not even going to bring it up unless she marks me wrong on the test.

 

[Xenus]

You should do a simple vector equation.

but it been two years so I do not remeber the equation to use.

It involves drawing two sides of a triangle and finding the hypotinus but how you figure it out for there I forget.

Ah that was easy then you use pythagoras's theorem to get 30.14962686

The problem now is I don't remember how to find the angle other than it requires the use of sin or cos.

I see that you already figured it out.

 

[Jew]

What if the ball was thrown out the other side of the car?

 

[StefanS]

Here's the problem:

Luke, sitting in a car traveling north @ 30m/s throws a ball directly across the car @ 3 m/s. What is the velocity of the ball with respect to the road?

First off doesn't the direction that the ball is thrown matter? My teacher said it didn't but 3 m/s East != 3m/s West... they are the opposites of each other. Anyways I assumed the ball was thrown west to east and got an answer of v = 30.153 [5.71° E of N]. Apparently the answer is 30.153 [5.71° E of S].... this makes absolutely no sense to me whatsoever. Is my teacher just on drugs or am I wrong?

Here's the logic I used to solve the problem:

-Ball initially has a velocity of 30m/s north and is not accelerating
-Luke applies a force to move the ball 3 m/s west to east within the car
-The velocity of the ball from the reference point of the car is 3 m/s [E]
-The velocity of the ball from the reference point of the road is 3m/s[E] + 30m/s[N] = 30.153 [5.71° E of N]

Anyways if somebody could please point me in the right direction I'd apperciate it. Thanks!

The direction in which the ball is thrown does, indeed matter, since the earth is rotating and your ball is influenced by the Coriolis force, twice actually. A) from the velocity vector due to gravitation b) from the movement of the thrown ball. The intial condition of the car going north has no influence on the ball, whatsoever, (the X product is zero, as the angular velocity of the Earth and the velocity vector of the car going north is zero). However, this problem requires dynamical treatment and require at least a time of measurement, for the complete treatment a starting height would be necessary as well...

 

[DemoCoder]

The initial condition of the ball is relevant to velocity of the ball with respect to the road. It's not irrelevent.

 

[StefanS]

The initial condition of the ball is relevant to velocity of the ball with respect to the road. It's not irrelevent.

Read the text again, before replying, it has not effect whatsoever on the coriolis force. This is implied by the text in the brackets, since the x-Product is only relevant in the context of the coriolis force

 

[Freak'n Big Panda]

I think you're making this question a little more complex than needed. Air ristance is being ignored, gravity is begin ignored, aceleration is being ignored. You assume luke somehow instanly transfered enough kenetic energy to the ball to make it move @ 3m/s. To solve it it's a simple addition of vecotrs which creates a triangle that can be solved using the trig ratios (SOH CAH TOA) and Pythagorean theorem.

 

[[maven]]

= SQRT(30^2 + 3+2) = 30.150 m/s
No sines or cosines needed (unless you want the angle but that'd be a bit stupid as we're dealing with vectors). What's the fuss all about?

 

[Basic]

What's the fuss all about?

The angle.

It's because it's a vector that the angle is interesting, not the other way. The vector is given in polar coordinates.

 

[StefanS]

I think you're making this question a little more complex than needed. Air ristance is being ignored, gravity is begin ignored, aceleration is being ignored. You assume luke somehow instanly transfered enough kenetic energy to the ball to make it move @ 3m/s. To solve it it's a simple addition of vecotrs which creates a triangle that can be solved using the trig ratios (SOH CAH TOA) and Pythagorean theorem.

Well, Coriolis force is the only reason why a movement from west to east is different from east to west except for the definition of the angle...

 

[Simon F]

Here's the problem:

Luke, sitting in a car traveling north @ 30m/s throws a ball directly across the car @ 3 m/s. What is the velocity of the ball with respect to the road?

First off doesn't the direction that the ball is thrown matter?

Yes it certainly does matter since velocity includes direction. I can only assume that your teacher is assuming Luke is sitting on one particular side of the car, but how you guess what that is is beyond me! Perhaps the teacher is assuming Luke is in the (front) passenger seat... but then is the car left- or righthand drive?

 

[Basic]

Perhaps the teacher is assuming Luke is in the (front) passenger seat...

But then he's throwing stuff at the driver, and that's not a smart thing to do at 108 km/h.

 

[Bolloxoid]

You assume luke somehow instanly transfered enough kenetic energy to the ball to make it move @ 3m/s.

Luke used the Force.

 

[Simon F]

But then he's throwing stuff at the driver, and that's not a smart thing to do at 108 km/h.

Good point... but he's probably he's probably not very smart anyway