Finding value of g given a motion map

[Kashmir]

Homework Statement

Finding value of g given a motion map

Relevant Equations

All the equations of motion for constant acceleration such as
v=u+at.
s=ut+1/2 at^2

I tried using all equations of motion but couldn't get the correct answer.
Any hint would greatly help. I've tried doing this for two days now!

 

[Steve4Physics]

You need to provide evidence of your own attempts before we can help. But note that the question is at fault – the numbers on the ruler have no unit (should be shown as cm).

This is a simple high school level question. It is completely out of character with your other questions, e.g. “Help in understanding this derivation of Lagrange Equations in Non-Holonomic case” (https://www.physicsforums.com/threa...ange-equations-in-non-holonomic-case.1055957/). What’s going on?

 

[Kashmir]

You need to provide evidence of your own attempts before we can help. But note that the question is at fault – the numbers on the ruler have no unit (should be shown as cm).

This is a simple high school level question. It is completely out of character with your other questions, e.g. “Help in understanding this derivation of Lagrange Equations in Non-Holonomic case” (https://www.physicsforums.com/threa...ange-equations-in-non-holonomic-case.1055957/). What’s going on?

Thank you for your kind response. I was helping my relative in his basic physics and he gave me this question. I tried solving it every way incase i was missing something. That's why I asked it here in case there was something else I missed.

 

[Steve4Physics]

Thank you for your kind response. I was helping my relative in his basic physics and he gave me this question. I tried solving it every way incase i was missing something. That's why I asked it here in case there was something else I missed.

The rules here require that (before we can help) you show some sort of evidence of your own effort, e.g. a calculation that didn't give the correct answer.

 

[Chestermiller]

The rules here require that (before we can help) you show some sort of evidence of your own effort, e.g. a calculation that didn't give the correct answer.

The data does not correspond to a body released from rest at time zero. It corresponds to a body released from z = 10 at time zero, and having an initial upward velocity of 100.

 

[Steve4Physics]

The data does not correspond to a body released from rest at time zero. It corresponds to a body released from z = 10 at time zero, and having an initial upward velocity of 100.

The original question (image in Post #1) specifically states "...released from rest at t=0s". I think the question is ok, however, the diagram appears to be inaccurately drawn.

 

[Chestermiller]

The original question (image in Post #1) specifically states "...released from rest at t=0s". I think the question is ok, however, the diagram appears to be inaccurately drawn.

No way. I had my graphics package fit a quadratic to s vs t, and the equation it delivered was $$s=10-100t+500 t^2$$This is an exact fit to the data provided.

 

[Steve4Physics]

No way. I had my graphics package fit a quadratic to s vs t, and the equation it delivered was $$s=10-100t+500 t^2$$This is an exact fit to the data provided.

I don't disagree with your analysis.

The problem is the inconsistent question/diagram. The question explicitly states that the ball is released from rest (and we assume acceleration is constant). But the diagram provided is inconsistent with this (as your analysis clearly shows).

I believe the intention of the author of the question was this:

The ball is released from rest at t=0 from the top of the ruler marked in cm.

The lowest position of the ball is to be used to find the acceleration (which gives the indicated answer of 6.8m/s²).

Unfortunately, the intermediate positions of the ball shown on the diagram do not then match their expected positions. (This could conceivably simply be due to careless drawing.)

Well, that’s my interpretation!

 

[Chestermiller]

I don't disagree with your analysis.

The problem is the inconsistent question/diagram. The question explicitly states that the ball is released from rest (and we assume acceleration is constant). But the diagram provided is inconsistent with this (as your analysis clearly shows).

I believe the intention of the author of the question was this:

The ball is released from rest at t=0 from the top of the ruler marked in cm.

The lowest position of the ball is to be used to find the acceleration (which gives the indicated answer of 6.8m/s²).

Unfortunately, the intermediate positions of the ball shown on the diagram do not then match their expected

Even if we use the 2nd order accurate finite difference formula for equal increments in time (which is exact for a quadratic variation), we get exactly the same value of 1000 cm/s^2 for all 5 equal interval combinations: $$\frac{d^2s}{dt^2}=\frac{(s_{j-1}-2s_j+s_{j+1})}{(\Delta t)^2}$$

 

[Steve4Physics]

Even if we use the 2nd order accurate finite difference formula for equal increments in time (which is exact for a quadratic variation), we get exactly the same value of 1000 cm/s^2 for all 5 equal interval combinations: $$\frac{d^2s}{dt^2}=\frac{(s_{j-1}-2s_j+s_{j+1})}{(\Delta t)^2}$$

No argument with your analysis. The issue is with the badly posed question. The alternatives seem to be:
a) the question is stated correctly but the diagram is inaccurate;
b) the question is stated incorrectly but the diagram is accurate.

Could this be the cause of the OP’s original difficulties? It would be good to hear from them!

 

[Chestermiller]

No argument with your analysis. The issue is with the badly posed question. The alternatives seem to be:
a) the question is stated correctly but the diagram is inaccurate;
b) the question is stated incorrectly but the diagram is accurate.

Could this be the cause of the OP’s original difficulties? It would be good to hear from them!

I pick option b.

 

[nasu]

It looks like the data satisfy very well a free fall with g=10 m/s, starting from rest at the position 5cm on the ruler, a shown in the figure.
The distance travelled during each 0.1 s intervals are:
5 cm during the first second
15 cm during the second second (3 x 5)
25 cm during the third second (5 x 5)
35 cm during the fourth secons (7 x 5).

A second order polynomial fit of the data set (positions of 5, 10, 25, 50, 85) versus time in 0.1 increments starting with zero gives a fit with 500 1000$cm/s^2$ for acceleration. The linear term is of the order of $10^{-12}$ and is just error from the fitting algorithm.

Edit. Corrected the value of 500 to 1000 after @Chestermiller's observation.

 

[Chestermiller]

It looks like the data satisfy very well a free fall with g=10 m/s, starting from rest at the position 5cm on the ruler, a shown in the figure.
The distance travelled during each 0.1 s intervals are:
5 cm during the first second
15 cm during the second second (3 x 5)
25 cm during the third second (5 x 5)
35 cm during the fourth secons (7 x 5).

A second order polynomial fit of the data set (positions of 5, 10, 25, 50, 85) versus time in 0.1 increments starting with zero gives a fit with 500 $cm/s^2$ for acceleration. The linear term is of the order of $10^{-12}$ and is just error from the fitting algorithm.
View attachment 334001

500 is wrong. The 2nd derivative of y with respect to x is 1000.

 

[Steve4Physics]

It looks like the data satisfy very well a free fall with g=10 m/s [edit - you mean 10m/s²], starting from rest at the position 5cm on the ruler, a shown in the figure.
The distance travelled during each 0.1 s intervals are:
5 cm during the first second
15 cm during the second second (3 x 5)
25 cm during the third second (5 x 5)
35 cm during the fourth secons (7 x 5).

A second order polynomial fit of the data set (positions of 5, 10, 25, 50, 85) versus time in 0.1 increments starting with zero gives a fit with 500 $cm/s^2$ [edit - you mean 1000 $cm/s^2$] for acceleration. The linear term is of the order of $10^{-12}$ and is just error from the fitting algorithm.

Aha! I took the start as t=0, s=0, v=0. But it looks like the intention is that it should be t=0, s=5cm, v=0.

That leads to a value for g which is consistent with the ball positions on the diagram.

So g=10m/s² though this is not on the list of available choices. Probably an error in the choice-list.

 

[PeroK]

Let's assume that the object is released at $t =0, s = 0, v = 0$. After $0.5s$ it has fallen $0.85m$. The average speed over this time is $1.7 m/s$, so the final speed is $3.4 m/s$. That gives an acceleration of $6.8 m/s^2$.

The penultimate data gives a fall of $0.5m$ after $0.4s$, giving an acceleration of $6.25 m/s^2$. Presumably this is to be taken as less accurate than the final data. As are the other previous data.

 

[Steve4Physics]

Let's assume that the object is released at $t =0, s = 0, v = 0$. After $0.5s$ it has fallen $0.85m$. The average speed over this time is $1.7 m/s$, so the final speed is $3.4 m/s$. That gives an acceleration of $6.8 m/s^2$.

The penultimate data gives a fall of $0.5m$ after $0.4s$, giving an acceleration of $6.25 m/s^2$. Presumably this is to be taken as less accurate than the final data. As are the other previous data.

Alternatively, take the initial conditions as $t=0, s=5cm, v=0$ and get an answer which is not on the answer list – but which gives an excellent fit to all the data.

It’s a badly posed question with one or more ambiguities/inconsistencies. IMO there can be no definitive correct answer.

Hopefully the OP has picked-up on all of this.

 

[nasu]

Let's assume that the object is released at $t =0, s = 0, v = 0$. After $0.5s$ it has fallen $0.85m$. The average speed over this time is $1.7 m/s$, so the final speed is $3.4 m/s$. That gives an acceleration of $6.8 m/s^2$.

The penultimate data gives a fall of $0.5m$ after $0.4s$, giving an acceleration of $6.25 m/s^2$. Presumably this is to be taken as less accurate than the final data. As are the other previous data.

How do you get the 0.5 s? You assume that the first position shown on the diagram is not the initial position but after the first 0.1 s interval?

 

[PeroK]

How do you get the 0.5 s? You assume that the first position shown on the diagram is not the initial position but after the first 0.1 s interval?

Yes, the initial position is not shown.

 

[nasu]

Aha! I took the start as t=0, s=0, v=0. But it looks like the intention is that it should be t=0, s=5cm, v=0.

That leads to a value for g which is consistent with the ball positions on the diagram.

So g=10m/s² though this is not on the list of available choices. Probably an error in the choice-list.

If it starts at t=0 at s=0 and it's at s=5 after 0.1 s and then it's at s=10 after another 0.1 s, the motion would have zero acceleration over this interval.

 

[nasu]

Yes, the initial position is not shown.

Then what is the first position shown? At what time from t=0?

 

[PeroK]

Then what is the first position shown? At what time from t=0?

The first position is at approx $0.05m$ at $t = 0.1s$.

 

[nasu]

So it moved 5 cm in 0.1 s and then again 5 cm (from 5 to 10) in another 0.1s. How can this be in a free fall?

 

[PeroK]

So it moved 5 cm in 0.1 s and then again 5 cm (from 5 to 10) in another 0.1s. How can this be in a free fall?

Experimental error!

PS note that the positions appear to be to the nearest $5cm$.

 

[nasu]

So you decide which data has more or less error based on the given answer? When you can find a consistent fit to the data as given and without any extra assumption.

 

[PeroK]

So you decide which data has more or less error based on the given answer? When you can find a consistent fit to the data as given and without any extra assumption.

We need the object to fall a significant distance, so that the $\pm 2.5cm$ measurement error is not significant. This is a practical experimental question and requires the ability to think beyond a few SUVAT equations.

 

[kuruman]

I too got $$s=5+500t^2$$ for a fit of position and time pairs. If I compare with the SUVAT equation $$s=s_0+v_0\,t+\frac{1}{2}a\,t^2,$$ I deduce that

1. $s_0=5$ units
2. $v_0=0$ which implies that the mass starts from rest at $s_0$
3. $\frac{1}{2}a=500~$units which implies that the acceleration is 1000 units/s².

That's what the numbers say and they don't lie. We don't know what the units on the scale are. However, if we accept that the correct answer is 6.8 m/s², then we can calibrate the scale and say that 1 unit =6.8 mm.

 

[kuruman]

It’s a badly posed question with one or more ambiguities/inconsistencies. IMO there can be no definitive correct answer.

I disagree. Assuming that the first position is reached as some unspecified clock time $\Delta t$, the other positions are reached at clock times $t_n=n\Delta t$ and are separated by equal time intervals. This brings to mind Galileo's experiment wherein he rolled a ball down an incline with small bumps separated so that the clicks heard when the ball went over bumps were at equal time intervals.

The spatial separation between events is easy to find. If the object starts from rest at the origin, the space intervals are at positions
$s_1=\frac{1}{2}a(\Delta t)^2$
$s_2=\frac{1}{2}a(2\Delta t)^2=4\frac{1}{2}a(\Delta t)^2=4s_1$
$\dots$
$s_n=\frac{1}{2}a(n\Delta t)^2=n^2 s_1.$
The spatial separation between adjacent events is an oddmultiple of $s_1$:
$\Delta s_n=s_{n+1}-s_n=\left[(n+1)^2-n^2\right]s_1=(2n+1)s_1.$
Here we have
$\Delta s_1=(0.10-0.05)~\rm{m}=0.05~\rm{m}=1\times(0.05)~\rm{m}.$
$\Delta s_2=(0.25-0.10)~\rm{m}=0.15~\rm{m}=3\times(0.05)~\rm{m}.$
$\Delta s_3=(0.50-0.25)~\rm{m}=0.25~\rm{m}=5\times(0.05)~\rm{m}.$
$\Delta s_4=(0.85-0.50)~\rm{m}=0.35~\rm{m}=7\times(0.05)~\rm{m}.$
It follows that $s_1=0.05~\rm{m}$ which allows us to find the acceleration
$$s_1=\frac{1}{2}a(\Delta t)^2\implies a=\frac{2s_1}{(\Delta t)^2}=\frac{2\times 0.05~\rm{m}}{(0.1~\rm{s})^2}=10~\rm{m}/{s}^2.$$ This result is in agreement with polynomial fits reported above. It is not in agreement with any of the choices offered in the formulation of the question. I also note that this method of analysis does not rely on knowing the initial position but only on the condition that the object start from rest which is given. It was started by @nasu in post #12, but not completed algebraically.

 

[Steve4Physics]

I disagree. Assuming that the first position is reached as some unspecified clock time $\Delta t$, the other positions are reached at clock times $t_n=n\Delta t$ and are separated by equal time intervals. This brings to mind Galileo's experiment wherein he rolled a ball down an incline with small bumps separated so that the clicks heard when the ball went over bumps were at equal time intervals.

The spatial separation between events is easy to find. If the object starts from rest at the origin, the space intervals are at positions
$s_1=\frac{1}{2}a(\Delta t)^2$
$s_2=\frac{1}{2}a(2\Delta t)^2=4\frac{1}{2}a(\Delta t)^2=4s_1$
$\dots$
$s_n=\frac{1}{2}a(n\Delta t)^2=n^2 s_1.$
The spatial separation between adjacent events is an oddmultiple of $s_1$:
$\Delta s_n=s_{n+1}-s_n=\left[(n+1)^2-n^2\right]s_1=(2n+1)s_1.$
Here we have
$\Delta s_1=(0.10-0.05)~\rm{m}=0.05~\rm{m}=1\times(0.05)~\rm{m}.$
$\Delta s_2=(0.25-0.10)~\rm{m}=0.15~\rm{m}=3\times(0.05)~\rm{m}.$
$\Delta s_3=(0.50-0.25)~\rm{m}=0.25~\rm{m}=5\times(0.05)~\rm{m}.$
$\Delta s_4=(0.85-0.50)~\rm{m}=0.35~\rm{m}=7\times(0.05)~\rm{m}.$
It follows that $s_1=0.05~\rm{m}$ which allows us to find the acceleration
$$s_1=\frac{1}{2}a(\Delta t)^2\implies a=\frac{2s_1}{(\Delta t)^2}=\frac{2\times 0.05~\rm{m}}{(0.1~\rm{s})^2}=10~\rm{m}/{s}^2.$$ This result is in agreement with polynomial fits reported above. It is not in agreement with any of the choices offered in the formulation of the question. I also note that this method of analysis does not rely on knowing the initial position but only on the condition that the object start from rest which is given. It was started by @nasu in post #12, but not completed algebraically.

I like your, solution – in particular the fact that it doesn’t require prior knowledge of the start position (thereby removing a potential ambiguity).

I still wouldn’t regard the question as completely unambiguous - there are missing units on the scale (so we must make an assumption). And if the official answer (6.8m/s²) is considered to be 'part of the whole', then there is a major flaw.

 

[PeroK]

There are two possible answers:

1) The wrong graph was included. And the graph shows an object falling from rest at $0.5 cm$ under the Earth's gravity for $0.4s$.

2) The graph shows an object falling from $0 cm$ for $0.5s$. The question required the answer which best fits the data (not necessarily perfectly fits the data). And, allowing for experimental error, $g = 6.8 m/s^2$ is a good fit for the data.

 

[nasu]

You don't need to assume what happened before the first point shown in the figure. As @kuruman shows, you can use the given points to find it from the fit. It could have been that the object had some vlocity at the first position shown but if you take this as a parameter, the best fit shows that it is zero.
But if you want to assume some error and that the motion started before the position indicated, there is no reason to assume that it started exactly 0.1 s before. If there is error, then why not in this value? Why to take the total time 0.5 s and not 0.45 or 0.55 or whatever? By taking different values of this pre-interval you can obtain all the suggested answers. Maybe the reaction time to turn on the stopwatch was so that the time before the first recorded position was 0.17 s and not 0.1 s. Then the average velocity for the 0.85 cm (assuming the scale is in cm) vill be 14.46 m/s and the acceleration 5.0 m/s ².

I think that all agree that the problem as posed has some "problems". It may be interesting to "guess" the author's mind (rather than assume that is a mistake in the answer key) but how is this helping a student who does know the answer in advance?
It may be that there was another answer below (c) (or more), not shown in the image cut, and there was bug in the software. What if there was a choice (d) 9.8 m/s ²?
We found this case (the wrong answer indicated as correct) several time in the online packages used for tests during the look-up. It is not even a mistale of the author of the tests but just some programming bug. It may be useful if the OP indicates the source. But I don't think he is interested any more.

 

[kuruman]

There are two possible answers:

1) The wrong graph was included. And the graph shows an object falling from rest at $0.5 cm$ under the Earth's gravity for $0.4s$.

I vote for Answer (1). I think that Answer (2) is untenable. Here is why.

Under constant acceleration, the instantaneous velocity is equal to the average velocity at the middle of the time interval over which the average is taken. Let $t_1$ be the clock time corresponding to event 1. The clock time corresponding to the subsequent events is $t_n=t_1+(n-1)\Delta t, ~n=2,3,\dots$
Now we can compute instantaneous velocities and times at which they occur.
Event 1: $s_1\rightarrow s_2$
$\bar v_1=\dfrac{s_2-s_1}{t_2-t_1}=\dfrac{s_2-s_1}{\Delta t}$
The instantaneous velocity $v_1=\dfrac{s_2-s_1}{\Delta t}$ occurs at clock time $\tau_1=t_1+\frac{1}{2}\Delta t.$

Event 2: $s_3\rightarrow s_2$
$\bar v_2=\dfrac{s_3-s_3}{t_3-t_2}=\dfrac{s_3-s_2}{\Delta t}$
The instantaneous velocity $v_2=\dfrac{s_3-s_2}{\Delta t}$ occurs at clock time $\tau_2=t_2+\frac{1}{2}\Delta t=(t_1+\Delta t)+\frac{1}{2}\Delta t=t_1+\frac{3}{2}\Delta t$

Now we can compute the constant acceleration $$\begin{align} & a_{\text{3-2-1}}=\frac{v_2-v_1}{\tau_2-\tau_1}=\frac{\frac{s_3-s_2}{\Delta t}-\frac{s_2-s_1}{\Delta t}}{t_1+\frac{3}{2}\Delta t-(t_1+\frac{1}{2}\Delta t)} =\frac{s_3-2s_2+s_1}{(\Delta t)^2} \nonumber \\& =\frac{(0.25-2\times 0.1+0.05)\rm{m}}{(0.1~\rm{s})^2}=10~\rm{m/s}^2.
\nonumber \end{align}$$ One needs at least three consecutive positions and times and here we have five. Using the rest of the triplets, $$\begin{align} & a_{\text{4-3-2}}=\frac{s_4-2s_3+s_2}{(\Delta t)^2}=\frac{(0.50-2\times 0.25+0.10)\rm{m}}{(0.1~\rm{s})^2}=10~\rm{m/s}^2 \nonumber \\
& a_{\text{5-4-3}}=\frac{s_5-2s_4+s_3}{(\Delta t)^2}=\frac{(0.85-2\times 0.50+0.25)\rm{m}}{(0.1~\rm{s})^2}=10~\text{m/s}^2 \nonumber \\
\end{align}$$This shows that one can calculate the constant acceleration from three position-time pairs without assuming anything about the initial conditions. Three points are sufficient to define a parabola when there is no experimental uncertainty as in this case.

In my opinion, all this is gorilla-glue strong evidence that the data support an acceleration $a=10~\text{m/s}^2$. In fact, the "triplet analysis" was (and maybe still is) how spark timer data were processed for finding $g$ experimentally before spreadsheets, photogates and lab analysis software came into being. Thus, I am forced to conclude that the graph is wrong. Furthermore, since all accelerations are exactly equal, it is highly likely that we have here a contrived set of data with zero uncertainty.

This concludes my justification of my vote for A. Now let's look at model B:

2) The graph shows an object falling from $0 cm$ for $0.5s$. The question required the answer which best fits the data (not necessarily perfectly fits the data). And, allowing for experimental error, $g = 6.8 m/s^2$ is a good fit for the data.

We can use the model to calculate the positions at the "measured" times. These are summarized in the Table below.

Time (s)​	Expt. Position (m)​	Calc. Position (m)​
0​	N.A.​	0​
0.1​	0.05​	0.03​
0.2​	0.10​	0.14​
0.3​	0.25​	0.30​
0.4​	0.50​	0.54​
0.5​	0.85​	0.85​

The discrepancy between the "experimental" values and the calculated values has been attributed to "experimental" error. I doubt that there ever was an experiment that generated these data, but I will play along. I will superimpose on OP's figure the predictions of the model A where the acceleration is $a_A=10~\text{m/s}^2$ and the object starts from rest at the 0.05 m mark (red circles) and model B where the position are calculated using an acceleration $a_B=6.8~\text{m/s}^2$ and the object starts from rest at the 0 m mark (blue circles.)

A more conventional presentation of model B is shown below. The "experimental" data are shown as discrete points connected with a smoothed line and the calculation is shown as a solid line. It looks like the assumption that the initial position of the object is zero throws the parabola out of kilter. This is no random variation due to experimental uncertainty.

Let the reader decide which is the best analysis of the data: one that yields a value for the constant acceleration that makes assumptions about the initial conditions or one that does not need such assumptions.