What is the ratio of the distances traveled by these two players?

[posto002]

Homework Statement

Two hockey players push off from a clinch, recoiling from rest. Jean-Claude has an inertia that is 70% greater than Pierre's inertia. After they push off and move across the ice in opposite directions, they lose kinetic energy at the same rate until they come to a stop.

What is the ratio of the distances which players traveled?

Relevant Equations

None that I can come up with at the moment!

I couldn't really understand how to approach this problem with just given the inertia with Pierre have a mass of 70% less than Jean-Claude. As I understand it, because Jean-Claude is heavier, he must not have traveled as far as Pierre, but I am unable to find exactly how far each of them went. Could someone please help walk me through this? Thank you!

 

[hutchphd]

You are asked to find the ratio of the distances traveled. This does not require you to know each distance explicitly.

 

[posto002]

You are asked to find the ratio of the distances traveled. This does not require you to know each distance explicitly.

Okay. Sorry, I still don't quite understand how to do that exactly with just the given inertia.

 

[hutchphd]

What does "inertia" mean algebraically?

 

[posto002]

What does "inertia" mean algebraically?

So from what I remember, F=ma. If two objects are acted on my force, (Let's say F-subz, for instance. F-subz=m1a1, and F-subz=m2a2 therefore m1a1=m2a2 and that also should mean that m1/m2=a1/a2 because F-subz is constant and doesn't change. This is really all I could recall from what we've talked about in class. Is this going in the right track?

 

[hutchphd]

m1a1=m2a2 and that also should mean that m1/m2=a1/a2

1. do the algebra correctly!
2. then you know that each acceleration lasted for the same time interval so what about the velocities?

 

[posto002]

1. do the algebra correctly!
2. then you know that each acceleration lasted for the same time interval so what about the velocities?

For the relation between acceleration and velocity, acceleration is the time rate of change of velocity. Am I correct? So the velocity should have taken the same time because acceleration did.

 

[jbriggs444]

For the relation between acceleration and velocity, acceleration is the time rate of change of velocity. Am I correct? So the velocity should have taken the same time because acceleration did.

Yes, acceleration is the time rate of change of velocity. However, one can obtain a useful result far more directly by using conservation of momentum. Can you write an equation relating the momentum of Jean-Claude to that of Pierre?

 

[posto002]

Yes, acceleration is the time rate of change of velocity. However, one can obtain a useful result far more directly by using conservation of momentum. Can you write an equation relating the momentum of Jean-Claude to that of Pierre?

(m1u1+m2u2=m1v1+m2v2 where m1 is the mass of Jean-Claude, m2 is the mass of Pierre, u1 is the initial velocity of Jean-Claude, u2 is the initial velocity of Pierre, v1 is the final velocity of Jean-Claude, and v2 is the final velocity of Pierre.)

That would give me 0 (because initial velocity of both is at zero before they push away from each other)= (m1v1)+(0.70(m1)+m1)v2 ------Because that's the relationship between the mass of Jean-Claude to Pierre who is 70% lighter than Jean-Claude------- would I use some sort of other equation to figure out the possible velocities?

 

[jbriggs444]

That would give me 0 (because initial velocity of both is at zero before they push away from each other)= (m1v1)+(0.70(m1)+m1)v2 ------Because that's the relationship between the mass of Jean-Claude to Pierre who is 70% lighter than Jean-Claude------- would I use some sort of other equation to figure out the possible velocities?

Good so far. You have the equation:$$m_1v_1+(0.70m_1+m_1)v_2=0$$
You should be able to simplify that and solve for $v1$. See what you get.

 

[PeroK]

Because that's the relationship between the mass of Jean-Claude to Pierre who is 70% lighter than Jean-Claude

Just an arithmetic point. If one person is 70% heavier than a second person, that doesn't mean the second person is 70% lighter than the first.

In other words, 170 is 70% more than 100, but 100 is not 70% less than 170.

 

[posto002]

Good so far. You have the equation:$$m_1v_1 + (0.70m_1+m_1)v2=0$$
You should be able to simplify that and solve for $v_1$. See what you get.

I got v1= -[(0.70m1+m1)v2]/m1 Please correct me if I'm wrong, but would I be able to further simplify that to v1=-[(1.70m1)v2]/m1 ?

 

[posto002]

Just an arithmetic point. If one person is 70% heavier than a second person, that doesn't mean the second person is 70% lighter than the first.

In other words, 170 is 70% more than 100, but 100 is not 70% less than 170.

That makes sense, thank you

 

[jbriggs444]

I got v1= -[(0.70m1+m1)v2]/m1 Please correct me if I'm wrong, but would I be able to further simplify that to v1=-[(1.70m1)v2]/m1 ?

You have $m_1$ in both the numerator and denominator. That offers an easy simplification.

 

[posto002]

You have $m_1$ in both the numerator and denominator.

So I can just pull it out to get -0.7v2=v1?

 

[jbriggs444]

So I can just pull it out to get -0.7v2=v1?

Be careful. You seem to have accidentally mangled the constant there.

And convention calls for the variable you are solving for to appear alone on the left hand side. [It is striking how much readability is improved by consistently following conventions like that].

 

[posto002]

Be careful. You seem to have accidentally mangled the constant there.

And convention calls for the variable you are solving for to appear alone on the left hand side. [It is striking how much readability is improved by consistently following conventions like that].

Oops! so taking v1=-[(1.70m1)v2]/m1, I would subtract 1 from -1.70m1 from the numerator to get rid of the m1 in the denominator to get v1=-0.7m1v2, right?

 

[jbriggs444]

Oops! so taking v1=-[(1.70m1)v2]/m1, I would subtract 1 from -1.70m1 from the numerator to get rid of the m1 in the denominator to get v1=-0.7m1v2, right?

No. That is not how it works.

Let's typeset that equation and remove extraneous braces:$$v_1=-\frac{1.70m_1v_2}{m_1}$$
Flip the term order to make things a bit more obvious:$$v_1=-\frac{1.70v_2m_1}{m_1}$$
We can rewrite that as $$v_1=-1.70v_2\frac{m_1}{m_1}$$
And we can rewrite that as $$v_1=-1.70v_2$$

 

[posto002]

No. That is not how it works.

Let's typeset that equation and remove extraneous braces:$$v_1=-\frac{1.70m_1v_2}{m_1}$$
Flip the term order to make things a bit more obvious:$$v_1=-\frac{1.70v_2m_1}{m_1}$$
We can rewrite that as $$v_1=-1.70v_2\frac{m_1}{m_1}$$
And we can rewrite that as $$v_1=-1.70v_2$$

Ohhh! okay! Thank you so much! So I can then use this equation to solve for v2 as well, I'm assuming

 

[jbriggs444]

Ohhh! okay! Thank you so much! So I can then use this equation to solve for v2 as well, I'm assuming

No. You can solve for neither $v_1$ nor $v_2$ independently. There is not enough information provided for that.

But you do not need to know either. You have enough information to know their ratio.

Refer back to the original problem. What is it that we are trying to determine?

they lose kinetic energy at the same rate until they come to a stop.

What is the ratio of the distances which players traveled?

There are a few ways to proceed from here. We could write down a ratio of velocities. But I am going to suggest that you write down a formula for the kinetic energy of each skater. Do not use numbers. Use the variables $m_1$, $m_2$, $v_1$ and $v_2$.

 

[posto002]

No. You can solve for neither $v_1$ nor $v_2$ independently. There is not enough information provided for that.

But you do not need to know either. You have enough information to know their ratio.

Refer back to the original problem. What is it that we are trying to determine?There are a few ways to proceed from here. We could write down a ratio of velocities. But I am going to suggest that you write down a formula for the kinetic energy of each skater. Do not use numbers. Use the variables $m_1$, $m_2$, $v_1$ and $v_2$.

KE (of Pierre)=0.5(m2)(v2^2)
KE (Of JC)= o.5(m1)(v1^2)

 

[jbriggs444]

KE (of Pierre)=0.5(m2)(v2^2)
KE (Of JC)= o.5(m1)(v1^2)

Good. Now, since we want to be able to compare the one against the other, it would help if both were expressed using the same variables. Can you write the KE of Pierre ($\frac{1}{2}m_2v_2^2$) using $m_1$ and $v_1$ instead?

 

[posto002]

Good. Now, since we want to be able to compare the one against the other, it would help if both were expressed using the same variables. Can you write the KE of Pierre ($\frac{1}{2}m_2v_2^2$) using $m_1$ and $v_1$ instead?

KE of Pierre= 1/2(0.70m1+m1)(v1/-1.70) From what we got from above (v1=-1.70v2), was it right to be able to solve for v2 to get v1/-1.70 by dividing both sides by -1.70? Then to use that and plug in v2 into the KE equation?

 

[jbriggs444]

KE of Pierre= 1/2(0.70m1+m1)(v1/-1.70)

This is the right idea, but the velocity term is supposed to be squared. Can you fix that problem and proceed to simplify?

 

[posto002]

Sorry, my mistake: KE of Pierre=0.29411m1v1^2 is what I got once I multiplied everything through including squaring the velocity term.

(KE of Pierre= (1/2)(1.70m1)([v1/-1.70]^2)
=(1.70m1/2)(v1^2/2.89)
=(0.85m1)(v1^2/2.89)
=([0.85m1v1^2]/2.89)
=(0.294117647m1v1^1)

 

[posto002]

Sorry, my mistake: KE of Pierre=0.29411m1v1^2 is what I got once I multiplied everything through including squaring the velocity term.

(KE of Pierre= (1/2)(1.70m1)([v1/-1.70]^2)
=(1.70m1/2)(v1^2/2.89)
=(0.85m1)(v1^2/2.89)
=([0.85m1v1^2]/2.89)
=(0.294117647m1v1^1)

Oh wait! I got it! Thank you for your time and help! I greatly appreciate it!

 

[jbriggs444]

Sorry, my mistake: KE of Pierre=0.29411m1v1^2 is what I got once I multiplied everything through including squaring the velocity term.

(KE of Pierre= (1/2)(1.70m1)([v1/-1.70]^2)
=(1.70m1/2)(v1^2/2.89)
=(0.85m1)(v1^2/2.89)
=([0.85m1v1^2]/2.89)
=(0.294117647m1v1^1)

The final result is obviously incorrect. The velocity term is raised only to the first power.

It is more difficult to tell at a glance whether the constant factor is correct. Folding the $\frac{1}{2}$ into the number and expressing the fractional result as a decimal actually reduces clarity rather than increasing it.

Let's do the calculation a bit differently and a bit more slowly. You would have started with $$KE=\frac{1}{2}m_2v_2^2$$
You would then have substituted in using $m_2=1.70m_1$ and $v_2=\frac{v_1}{1.70}$. This would yield:$$KE=\frac{1}{2}1.70m_1\frac{v_1}{1.70}^2$$
Re-arranging things, that could be written as $$KE=\frac{1}{2}\frac{1.70m_1v_1^2}{1.70^2}$$
We can cancel the one 1.70 on the top with one of the 1.70's on the bottom yielding:$$KE=\frac{1}{2}\frac{m_1v_1^2}{1.70}$$
While you could take the next step and calculate $\frac{1}{2 \cdot 1.70}$ to arrive at 0.294, you would be better served to express the result as:$$KE=\frac{1}{1.70}\ \frac{1}{2}m_1v_1^2$$
Compare this to the formula for the KE of Jean Claude.
You can read off the ratio of the kinetic energies very easily in this form.

Resist the urge to do arithmetic too early.

 

[PeroK]

What is the ratio of the distances which players traveled?

I thought the question was to find the ratio of distances travelled? That seems not very easy to me. Unless I'm missing something, all the calculations so far are just the preliminaries!

 

[posto002]

I thought the question was to find the ratio of distances travelled? That seems not very easy to me. Unless I'm missing something, all the calculations so far are just the preliminaries!

I understand where I went wrong. I was currently waiting for my Professor's Office hours to open up so I could talk to him about it. While I was waiting, I thought I'd try and solve it myself the night before and leading to this morning. Thank you once again for your help. With my Professor I was able to come up with the final solution for this practice problem I was doing.

 

[posto002]

The final result is obviously incorrect. The velocity term is raised only to the first power.

It is more difficult to tell at a glance whether the constant factor is correct. Folding the $\frac{1}{2}$ into the number and expressing the fractional result as a decimal actually reduces clarity rather than increasing it.

Let's do the calculation a bit differently and a bit more slowly. You would have started with $$KE=\frac{1}{2}m_2v_2^2$$
You would then have substituted in using $m_2=1.70m_1$ and $v_2=\frac{v_1}{1.70}$. This would yield:$$KE=\frac{1}{2}1.70m_1\frac{v_1}{1.70}^2$$
Re-arranging things, that could be written as $$KE=\frac{1}{2}\frac{1.70m_1v_1^2}{1.70^2}$$
We can cancel the one 1.70 on the top with one of the 1.70's on the bottom yielding:$$KE=\frac{1}{2}\frac{m_1v_1^2}{1.70}$$
While you could take the next step and calculate $\frac{1}{2 \cdot 1.70}$ to arrive at 0.294, you would be better served to express the result as:$$KE=\frac{1}{1.70}\ \frac{1}{2}m_1v_1^2$$
Compare this to the formula for the KE of Jean Claude.
You can read off the ratio of the kinetic energies very easily in this form.

Resist the urge to do arithmetic too early.

I understand where I went wrong. I was currently waiting for my Professor's Office hours to open up so I could talk to him about it. While I was waiting, I thought I'd try and solve it myself the night before and leading to this morning. Thank you once again for your help. With my Professor I was able to come up with the final solution for this practice problem I was doing.

 

[jbriggs444]

I thought the question was to find the ratio of distances travelled? That seems not very easy to me. Unless I'm missing something, all the calculations so far are just the preliminaries!

Yes. My aim was at a preliminary result which would be the ratio of the kinetic energies. Then one could try to figure out whether the kinetic energy loss was a rate over distance or a rate over time.

A rate over distance would make the rest of the problem easy using work and energy. A rate over time would either call for the evaluation of an integral or a clever way of comparing the two situations.