1D Kinematics Problem
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1D Kinematics Problem
I was working on some problems with a teacher and he gave me this one and refuses to tell me the answer. There is a ramp and one ball is starting at the top and the other at the bottom. The ball at the top can travel the 8m long ramp in 4 seconds. The ball at the bottom is being hit by a spring or kicked or whatever with an unknown initial velocity and hits the other ball when its instantaneous velocity is equal to 0.
So, I know that for the first ball from the top, it's velocity is 2m/s and acceleration is 1m/s^2. For ball two, we know that at the specific time interval when they hit, its velocity is 0. Other than that, I have no idea.
Oh, I also know that the first ball's distance down the ramp can be measured by .5t^2, but that isn't too helpful as of yet.
Any hints/ideas?
So, I know that for the first ball from the top, it's velocity is 2m/s and acceleration is 1m/s^2. For ball two, we know that at the specific time interval when they hit, its velocity is 0. Other than that, I have no idea.
Oh, I also know that the first ball's distance down the ramp can be measured by .5t^2, but that isn't too helpful as of yet.
Any hints/ideas?
Re: 1D Kinematics Problem
I'm not quite sure your 2m/s velocity is correct, considering how the ball on the ramp is accelerated by forces from both gravity and friction and thereby not at constant speed. Other than that it would be helpful to know what exactly you are looking for, since you sort of forgot to mention that?
Re: 1D Kinematics Problem
Oh, I didn't state that we are assuming that the ramp is frictionless and it is free from gravity so that there is uniform acceleration. Also, what I'm looking for is the initial velocity of the ball thats being hit from the bottom of the ramp.
Re: 1D Kinematics Problem
If there is no gravity there is probably no acceleration at all, having a ramp isn't really relevant anymore and the only way the second ball can ever have a velocity of zero is if that equals it's initial velocity.
Re: 1D Kinematics Problem
I didn't mean no gravity but that we're assuming that it'll be a "perfect" situation without air resistance and everything
Re: 1D Kinematics Problem
Assume x to be 0 at the top of the ramp, 8m at the bottom. Position of the bottom ball is then x1 = 8m  v_in1 * t + ½at². Position of the top ball is x2 = 0m + v_in2 * t + ½at². The collision takes place when x1=x2 obviously. Does that help?
Re: 1D Kinematics Problem
Yeah. That's kind of where I'm stuck. I know the equations and where to find their intersection points its just past that I'm a bit lost. I can assume that they have the same acceleration(other than the one at the bottom is negative), so theres that. That simplifies the equations down to 8Vit.5t^{2} = .5t^{2}. Or 8  Vit = .5t^{2}. But once I get to that point I'm stuck with nowhere to go. I have the initial velocity in terms of time but I can't figure out how to solve for just the initial velocity...
Re: 1D Kinematics Problem
Hang on, is the first ball travelling under uniform acceleration or uniform velocity (i.e. no acceleration)? I can't believe it's the second, since no gravity implies that the second ball will *never* reach zero velocity. So we have uniform acceleration, which we assume is due to gravity, but are you sure about its magnitude?
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Re: 1D Kinematics Problem
Uniform acceleration. Sorry about that, keep causing a mixup.
Pretend my first post says:
I was working on some problems with a teacher and he gave me this one and refuses to tell me the answer. There is a ramp and one ball is starting at the top and the other at the bottom. The ball at the top can travel the 8m long ramp in 4 seconds. The ball at the bottom is being hit by a spring or kicked or whatever with an unknown initial velocity and hits the other ball when its instantaneous velocity is equal to 0. We also know that there is a constant acceleration so its velocity is a linear graph and it's position is a squaring etc.
So, I know that for the first ball from the top, it's velocity is 2m/s and acceleration is 1m/s^2. For ball two, we know that at the specific time interval when they hit, its velocity is 0. Other than that, I have no idea.
Oh, I also know that the first ball's distance down the ramp can be measured by .5t^2, but that isn't too helpful as of yet.
Any hints/ideas?
Pretend my first post says:
I was working on some problems with a teacher and he gave me this one and refuses to tell me the answer. There is a ramp and one ball is starting at the top and the other at the bottom. The ball at the top can travel the 8m long ramp in 4 seconds. The ball at the bottom is being hit by a spring or kicked or whatever with an unknown initial velocity and hits the other ball when its instantaneous velocity is equal to 0. We also know that there is a constant acceleration so its velocity is a linear graph and it's position is a squaring etc.
So, I know that for the first ball from the top, it's velocity is 2m/s and acceleration is 1m/s^2. For ball two, we know that at the specific time interval when they hit, its velocity is 0. Other than that, I have no idea.
Oh, I also know that the first ball's distance down the ramp can be measured by .5t^2, but that isn't too helpful as of yet.
Any hints/ideas?
Re: 1D Kinematics Problem
Sepens wrote:So, I know that for the first ball from the top, it's velocity is 2m/s and acceleration is 1m/s^2.
Why? If the ball is accelerating, then while its average velocity must be 2m/s (8 metres divided by 4 seconds), its instantaneous velocity is constantly changing  and, in fact, must start at less than 2 m/s since by the time it's reached the bottom it will have accelerated.
Is the ball, in fact, released from rest from the top? That would make a lot more sense for this kind of question, and would probably clear up some of the confusion here (and would give you a = 1 m/s^2 as you've said).
pollywog wrote:I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.
Re: 1D Kinematics Problem
Did I not say that the ball is starting from rest at the top? If not, yes the ball at the top is starting from rest. However, the ball at the bottom is starting at an unknown velocity and that is what we are trying to solve for.
Re: 1D Kinematics Problem
In the bit I quoted, you said "for the first ball from the top, it's velocity is 2m/s" which I took to meaning its velocity when it left the top.
However, it now seems we're happy with the following:
Uniform acceleration, a = 1 m/s^2 (note the negative if we're defining up from the bottom of the ramp to the top)
Ball 1 initial position x1,0 = 8 m
Ball 2 initial position x2,0 = 0 m
Ball 1 initial velocity u1 = 0 m/s
Ball 2 initial velocity u2 = ? m/s
Ball 1 velocity at impact = ? m/s
Ball 2 velocity at impact = 0 m/2
Ball 1 position at impact = Ball 2 position at impact = x = ? m/2
Time at impact t = ? s
So now you can write out the equations of motion for both balls, and work out which ones are relevant. However, it looks like you've already done that but made a significant mistake.
Are you using s = ut + 1/2 at^2? I suspect you are, so there are a couple of problems with that. Firstly, as I mention above, you need to make sure you treat up/down and positive/negative values consistently, otherwise you'll wind up with strange results like objects flying out into space. Secondly, that equation only holds for motion that starts at the origin. Look at the information you have  with no collision, at 0s the ball is at 8m, and at 4s the ball is at 0m. Does that fit your equation? Can you adjust your equation to make it work? Because if so, you should be several steps closer to solving your problem.
However, it now seems we're happy with the following:
Uniform acceleration, a = 1 m/s^2 (note the negative if we're defining up from the bottom of the ramp to the top)
Ball 1 initial position x1,0 = 8 m
Ball 2 initial position x2,0 = 0 m
Ball 1 initial velocity u1 = 0 m/s
Ball 2 initial velocity u2 = ? m/s
Ball 1 velocity at impact = ? m/s
Ball 2 velocity at impact = 0 m/2
Ball 1 position at impact = Ball 2 position at impact = x = ? m/2
Time at impact t = ? s
So now you can write out the equations of motion for both balls, and work out which ones are relevant. However, it looks like you've already done that but made a significant mistake.
Sepens wrote:Oh, I also know that the first ball's distance down the ramp can be measured by .5t^2, but that isn't too helpful as of yet.
Are you using s = ut + 1/2 at^2? I suspect you are, so there are a couple of problems with that. Firstly, as I mention above, you need to make sure you treat up/down and positive/negative values consistently, otherwise you'll wind up with strange results like objects flying out into space. Secondly, that equation only holds for motion that starts at the origin. Look at the information you have  with no collision, at 0s the ball is at 8m, and at 4s the ball is at 0m. Does that fit your equation? Can you adjust your equation to make it work? Because if so, you should be several steps closer to solving your problem.
pollywog wrote:I want to learn this smile, perfect it, and then go around smiling at lesbians and freaking them out.Wikihow wrote:* Smile a lot! Give a gay girl a knowing "Hey, I'm a lesbian too!" smile.
Re: 1D Kinematics Problem
Okay, I feel like I'm getting steps but theres still something that I'm missing. The first ball's distance can be defined as .5x^2+8 in this situation. And Ball 2's distance can be measured with Vi*t + 1/2 a t ^2. Do we know that the second balls acceleration is simply the opposite of the first balls because they are accelerating uniformly? Otherwise do I use that a = Vf  Vi / t, and since Vf is 0(where the instantaneous velocity is 0), then the acceleration is Vi/t?
Then the equation is rewritten as Vi * t  (Vi * t)/2 or as (Vi * t) / t. But then when you set the two balls to be equal to each other for their distances your still stuck. There must be something else I can't see.
Then the equation is rewritten as Vi * t  (Vi * t)/2 or as (Vi * t) / t. But then when you set the two balls to be equal to each other for their distances your still stuck. There must be something else I can't see.

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Re: 1D Kinematics Problem
Assuming the ramp has constant slope, and both balls have the same moment of inertia (or, in the case of no friction, are really sliding rather than rolling), both balls should have the same acceleration  not opposite, since they're both accelerating in the same direction (towards the bottom of the ramp). This should simplify your x1 = x2 equation, and combined with the Vf = 0 equation, you should be able to solve for Vi.
Re: 1D Kinematics Problem
Both of the balls aren't coming from the same direction. One of them is starting from the top of the ramp with an initial velocity of 0, and the other is being kicked/punched/whatever from the bottom of the ramp with an unknown initial velocity.
Re: 1D Kinematics Problem
The acceleration doesn't care what direction the object is actually moving.
Re: 1D Kinematics Problem
There's a short cut, assuming the top ball starts at 0m/s:
Spoiler:

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Re: 1D Kinematics Problem
Sandor wrote:There's a short cut, assuming the top ball starts at 0m/s:Spoiler:
On the other hand, kinematics annoys me, and when possible, I like to fall back to a conservation of energy perspective. You can set mgh = 1/2 m v^2, you can set g = 1m/s^2, the masses cancel out, take h as the x position that Sandor intuited, and you're there.
Ten is approximately infinity (It's very large)
Ten is approximately zero (It's very small)
Ten is approximately zero (It's very small)
Re: 1D Kinematics Problem
Got it. Thanks so much for everyone's help. Everyone on XKCD is always very helpful and enthusiastic.
Re: 1D Kinematics Problem
For future reference, when you have a problem where you don't know the time and want distance, it's often better to use Vf^2 = Vi^2 + 2ad. This relates the initial and final velocities in terms of the position and acceleration. Since the final position is the same for both, you can use this equation to easily solve for the final velocity. Energy conservation will use pretty much this same equation, but if you haven't seen that before, this is the way to go.
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