Force and Impulse on a Railway Locomotion

[ozni]

Homework Statement

A railway locomotive pushed a stationary wagon and accelerated to 4ms. Did the same to another wagon with same force and time and accelerated it to 6ms. The two now stationary wagons are bow locked together and the same impulse force is applied to the wagons to move them across the yard. What velocity could the two wagons achieve if the loco applies the same impulse force as before?

Homework Equations

Ft =m(delta)v

The Attempt at a Solution

Then a bit stuck...[/B]

 

[PhanthomJay]

Using your first 2 equations, What is the relation between M1 and M2?

 

[ozni]

Using your first 2 equations, What is the relation between M1 and M2?

They are subjected to the same pulse when joined together as they were when they were not joined together. I just can't seem to relate it mathematical

 

[quinoa19]

Spoiler

From first equation, $m_1=Ft/4$, from second, $m_2=Ft/6$. Plugging $m_1$ and $m_2$ into third, $1=\left(\frac{1}{4}+\frac{1}{6}\right)v$, then $v=12/5=2.4$ m/s

 

[ozni]

Spoiler

From first equation, $m_1=Ft/4$, from second, $m_2=Ft/6$. Plugging $m_1$ and $m_2$ into third, $1=\left(\frac{1}{4}+\frac{1}{6}\right)v$, then $v=12/5=2.4$ m/s

Thanks - getting there. Could you explain

I'm home schooled so any assistance is greatly appreciated.

 

[quinoa19]

Thanks - getting there. Could you explain View attachment 240693 I'm home schooled so any assistance is greatly appreciated.

Sure, try plugging first two equations into second, you can see that $Ft$ cancels on both sides, after that, 1 remains the left, and $\left(\frac{1}{4}+\frac{1}{6}\right)v$ on the right. Then, rearrange to get $v$.

 

[ozni]

Thankyou for that. So its fractioned ( a quarter and a sixth). You could use decimels?

 

[quinoa19]

Yes, one could use decimals, although it's better to convert to decimals at the very end not to lose precision, e.g. if I were to convert 1/6 as ≈0.1667 early I would lose some precision since 1/6 has infinite number of digits in decimal notation which would be lost.

 

[ozni]

Great - such useful extra info! I'm not 100 percent with plugging in as I'm trying to fill gaps in with my algebra - plugging into 3rd equation?

 

[quinoa19]

I mean 3rd equation in your original post, $Ft=\left(m_1+m_2\right)v$. Try going through the whole derivation from the beginning and see if you can get $v$

 

[PhanthomJay]

You have the right equations of Physics, but the algebra is bugging you.
One of the great axioms of Algebra is that
“Things equal to the same thing are equal to each other” (Credit: Euler). Since the right side of your first equation and the right side of your second equation are both equal to the same thing on the left side, then both right sides must be equal to each other. So set them equal to solve for m1 in terms of m2, then plug that result into equation 3.

 

[ozni]

You have the right equations of Physics, but the algebra is bugging you.
One of the great axioms of Algebra is that
“Things equal to the same thing are equal to each other” (Credit: Euler). Since the right side of your first equation and the right side of your second equation are both equal to the same thing on the left side, then both right sides must be equal to each other. So set them equal to solve for m1 in terms of m2, then plug that result into equation 3.

Set them equal?

 

[haruspex]

Set them equal?

@PhanthomJay is saying that you have $Ft=m_14(ms^{-1})$ and $Ft=m_26(ms^{-1})$. So what equation can you write relating $m_14$ and $m_26$?

 

[sojsail]

Set them equal?

haruspex may be unavailable, so I will try to jump in.
Go back to the 11th item in this thread and reread the great axiom of Algebra credited to Euler until the meaning is clear. haruspex was trying to encourage you to write a new equation using that great axiom of Algebra.

 

[ozni]

I see what I need to do. Ft is equal to both equations its the combining into one is the issue

 

[sojsail]

Let us know what you come up with.

 

[ozni]

*note the time for both impulses is constant as well as the force applied

Because we know that masses are irrelevant to the answer we can also substitute in values for them. I also made the time one second so the values would be easy

So assuming mass of cart one is 10kg we find the force to be 40N. F=ma where F=10*4 F=40

further more using this value we can find the assumed mass of cart 2. F=ma where 40= m*6 so m=6 and two thirds.

the final equation of the carts is the combination of the mass of both carts so. F=ma where 40={10+6+2/3}*a using inverse operations...

40=16.66667*a so 40/16.666667=a=2.4ms-1

 

[neilparker62]

$$ \frac{Ft}{m_1}=v_1⇒\frac{m_1}{Ft}=\frac{1}{v_1} $$ $$ \frac{Ft}{m_2}=v_2⇒\frac{m_2}{Ft}=\frac{1}{v_2} $$ Adding the right hand equations: $$\frac{m_1+m_2}{Ft}=\frac{v_1+v_2}{v_1v_2} $$ Hence: $$\frac{Ft}{m_1+m_2}=\frac{v_1v_2}{v_1+v_2} $$

 

[PhanthomJay]

*note the time for both impulses is constant as well as the force applied

Because we know that masses are irrelevant to the answer we can also substitute in values for them. I also made the time one second so the values would be easy

So assuming mass of cart one is 10kg we find the force to be 40N. F=ma where F=10*4 F=40

further more using this value we can find the assumed mass of cart 2. F=ma where 40= m*6 so m=6 and two thirds.

the final equation of the carts is the combination of the mass of both carts so. F=ma where 40={10+6+2/3}*a using inverse operations...

40=16.66667*a so 40/16.666667=a=2.4ms-1

Now be careful here the acceleration values are not given nor are they asked for; you can't find the acceleration during the impact because the time of the impact is not given. You can't find the force of the impact either. Instead of writing F = ma , you should be writing Ft = m(\Delta v), like you did in your original post. You are solving for v, not a. Your approach is OK.

 

[ozni]

Okay, then follow through on what neilparker62 suggested?

 

[PhanthomJay]

Okay, then follow through on what neilparker62 suggested?

I know you are having trouble with your algebra, and sometimes when plugging in assumed values to avoid dealing with letter variables, you plug in one too many, and get nonsense. For now, what you did in post 17 gave you the right answer for 'v' (as noted in the 'spoiler' some posts ago), not 'a', so make some adjustments in your workings, and refresh your knowledge of algebra, which is critical for Physics problem solutions.