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Inertia - throwing a ball off of a moving train

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Just wondered...

 

If a train is moving alogn at 70mph, and you throw a ball off the back at 70mph...does the ball effectivly 'stop'?

 

Dan

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the velocity of the ball, relative to the train, will be zero (as they cancel out) but its speed is still 70 mph... so no, it wouldn't 'stop'.... atleast, thats how I interpret it :)

 

-mak10

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Well the ball stops in the same way the train would stop, if it hit something. When you throw the ball from the train the ball is still moving at 70mph compared to the ground, so it will hit the ground moving at that speed.

 

edit: I neglected to read the part where he threw it off the back at the same speed.

 

Yes it would go to zero and drop, sorry about that.

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no it won`t.

 

+70 mph and -70mph leaves 0mph.

 

the ball will just hit the ground as if dropped from a stationary position, to YOU the ball thrower all will apear as normal as the ball stays still and you accelerate away from it at 70 mph.

to the guy standing at a the train station watching it, he will just see the ball hit the floor as if dropped from a stationary position.

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Yeah, the ball should just drop down in the reference frame of the ground

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I think not.Newton's first law of motion states that an object in motion will remain in motion whereas and object at remain at rest unless acted by an unbalanced force.Therefore,the ball will still move at 70mph when you throw it out.However,the ball will deaccelerate beacause there are no forces generated to counteract both vectors of gravity and friction.Owing to the fact that gravity exerts an acceleration of 9.8m/s squared,the ball will hit the ground first while moving forward and deaccelerate rapidly when it hits the ground.In contrast,the train remain at a steady motion(ie: it neither accelerate nor deacceralate) because the engine generate enough force to counteract gravity(its weight) and friction(air friction,ground friction....)It will remain in motion because all forces all balanced (as well as possible).The force exerted opon the ball, however is unbalanced.

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Oh,i forgot, the ball's inertia is its capability to withstand changes to its state of motion.The higher its inertia(ie:mass) the greater its ability to maintain its speed for a longer period of time.It will deaccerate at a slower rate,(2nd law of motion F=MA) assuming that the ball is smooth and the vectors are about the same.(friction will increase nonetheless)

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I think not.Newton's first law of motion states that an object in motion will remain in motion whereas and object at remain at rest unless acted by an unbalanced force.Therefore,the ball will still move at 70mph when you throw it out.However,the ball will deaccelerate beacause there are no forces generated to counteract both vectors of gravity and friction.Owing to the fact that gravity exerts an acceleration of 9.8m/s squared,the ball will hit the ground first while moving forward and deaccelerate rapidly when it hits the ground.In contrast,the train remain at a steady motion(ie: it neither accelerate nor deacceralate) because the engine generate enough force to counteract gravity(its weight) and friction(air friction,ground friction....)It will remain in motion because all forces all balanced (as well as possible).The force exerted opon the ball, however is unbalanced.

 

Throwing the ball is an unbalanced force. YT's post summed it, as it were, rather succinctly: +70 - 70 = 0

 

The ball will drop to the ground.

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Newton's first law of motion states that an object in motion will remain in motion whereas and object at remain at rest unless acted by an unbalanced force.Therefore,the ball will still move at 70mph when you throw it out.
..unless acted upon by an external force. The act of throwing the ball backwards is the external force. YT's explanation applies.

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I think we are arguing semantics here, or in other words reading the question differently.

 

The question was:

 

If a train is moving alogn at 70mph, and you throw a ball off the back at 70mph...does the ball effectivly 'stop'?

 

I read this the first time as meaning that you throw the bal off the back into the air while travelling at 70mph. I assume TWJian read it this way too. In this case the ball would continue at 70mph, with a little bit of a reduction from air resistance as TJWian said. However, I think the rest of you are reading it as throwing the ball backwards with a velocity relative to the thrower of -70mph. (Rereading the question I think this is what was intended.) In this case, as you all say, the ball would just drop straight down.

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it would just fall, nothing more.

 

as YT said the resulting vectors = zero.

 

still look like it is traveling -70 to the thrower but then the speed of the train is taken from ground, so to the ground it would be zero.

 

madness.

its like jumping off a moving buss, the fun never stops.

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it`s also why I want to know, when in the movies you see a car and a truck traveling at say 50 mph, the truck lowers a ramp and the car behind drives up it into the truck!???

 

wft is wrong with this picture? surely the car will then be traveling at 50mph inside the truck! with VERY limited stopping distance!

 

Knight Rider has alot to answer for! :)

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it`s also why I want to know' date=' when in the movies you see a car and a truck traveling at say 50 mph, the truck lowers a ramp and the car behind drives up it into the truck!???

 

wft is wrong with this picture? surely the car will then be traveling at 50mph inside the truck! with VERY limited stopping distance!

 

Knight Rider has alot to answer for! :)[/quote']

 

Think about conservation of energy, keeping in mind you have to view everything from one frame of reference. The car will not leap forward at some higher speed , as long as you take your foot off the gas at the right time. The excess energy that must be dissipated is in the rotation of the tires, but not in the body of the car.

 

Think of it this way - if the car were instead some kind of hovercraft landing on top of a moving platform, would there be a problem?

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it wouldn't just 'stop' and fall.... it stops relative to the train, because the velocities cancel out... but thats not say that its not travelling at 70 mph. it will still have the same speed and thus move some distance, from the point where it is thrown.... and then fall at some point.

 

-mak10

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I have yet to read any of the posts (Im in the middle of school, and am kinda ruished for time), but to answer your question, essentially, it would just fall to the ground.

It would have to be exact, but if you were to get everything perfect (70 mph train - 70 mph ball) then it would equal out to nothing...just falling to the ground

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it wouldn't just 'stop' and fall.... it stops relative to the train' date=' because the velocities cancel out... but thats not say that its not travelling at 70 mph. it will still have the same speed and thus move some distance, from the point where it is thrown.... and then fall at some point.

 

-mak10[/quote']yes it would, and no it wouldn`t be traveling at any speed, it would be kept stationary in space with the throw to cancel the trains forwards velocity.

the ball would be still, the train would move away at 70mph.

 

just the same as the balls velocity would be 140mph if thrown towards the front of the train

anyone IN the train will get hit with it at 70mph, anyone outside and stationary would get a good 140mph`s worth!

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yes it would, and no it wouldn`t be traveling at any speed, it would be kept stationary in space with the throw to cancel the trains forwards velocity.

 

velocity and speed arent the same thing... well, not almost. the velocities simply cancel off, as the ball and the train are travelling in opposite directions. but their respective speeds remain the same. if i were to imagine your secario and use your line of reasoning, that the ball would just fall without travelling any distance because its travelling at the opposite direction to the train, then so will the train... coz its travelling at the opposite direction from the ball... !!

 

-mak10

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wtf???

 

I have no other way to explain it any easier than I already have done.

 

the ball will have no velocity/speed relative to the ground frame, and thus will simply drop.

the thrower of said ball will see it leave his/her arm at 70mph, relative to the train only.

 

what`s the hard part????

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the ball will have no velocity/speed relative to the ground frame, and thus will simply drop

 

let me put in basic terms... velocity is a vector, it has both magnitude and direction. speed is scalar... it has only magnitude. so whichever direction you throw the ball, it is NOT going to affect the speed at which you have thrown. the ball will still travel some distance from the point from it is thrown... but it doesn't possess a velocity relative to the train, since velocities are direction-dependent, they get cancel off. how dont know how to explain this more simply... or maybe i am getting something wrong here... but thats how i see it

 

-mak10

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When you discuss velocity, it always has to be relative to a reference. The ball, after it is thrown, is moving 70 mph relative to the train, because that's what was stated. The train is moving 70 mph relative to the groundm which was also given. If the ball was thrown backward, it is now stationary (in the horizontal axis) relative to the ground. It isn't any more complicated than that.

 

And there's really nothing here that YT hasn't already said.

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if the ball's stationary, it must have zero speed. but since it has a speed, according to the question (70 mph), how is it stationary? (note: its velocity=0 relative to the train, that i agree wholeheartedly).

 

-mak10

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As I said before, I think this is a semantic problem. Mak10: go back and reread the question again. I think you are misinterpreting it. The ball is thrown out of the train with a speed of 70mph with respect to the train, or in other words at 0mph with respect to the ground...

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ok, maybe I mindunderstood since most of the intelligent people on board here disagree with me.... how i understand this is very simple, a ball is thrown off from a train at 70 mph opposite to the direction of the train which is also travelling at 70 mph, with respect to the ground. both their velocity vectors would cancel off, so they have a zero resultant velocity (as YT said +71-70 = 0). but i dont understand how this affects the magnitude of the speed, with the respect to the ground... maybe the poster of the question could be more clearer or something or I definitely need some work-out on motion principles.

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ok, maybe I mindunderstood since most of the intelligent people on board here disagree with me.... how i understand this is very simple, a ball is thrown off from a train at 70 mph opposite to the direction of the train which is also travelling at 70 mph, with respect to the ground. both their velocity vectors would cancel off, so they have a zero resultant velocity (as YT said +71-70 = 0). but i dont understand how this affects the magnitude of the speed[/i'], with the respect to the ground... maybe the poster of the question could be more clearer or something or I definitely need some work-out on motion principles.

 

The speed is the magnitude of the velocity. If the (instantaneous) velocity is zero, so is the speed.

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