What Speed Does the Railway Carriage Achieve After an Inelastic Collision?

[physicshelpppp]

Homework Statement

A railway carriage with a mass m = 2.05×104 kg is initially traveling to the left with a speed Vi = 3.24 m/s. It collides and couples with (get stuck to) two identical railway carriages move to the left with a speed of Ui = 2.36 m/s.

What is the speed of the railway carriage after the collision?

Homework Equations

P = m1v1 + m2v2

KE = 1/2mv2

KE (before collision) = 1/2m1v12 + 1/2m2v22

KE (after collision) = 1/2(m1+ m2)vf2

The Attempt at a Solution

1. P = (2.06×10^4 x 3.24) + 2(2.05×10^4 x 2.36)
= 1.6x10^5

∴ P = 1.6x10^5 = (m1 + 2m2)vf = 6.15x10^4
vf = (1.6x10^5) / (6.15x10^4)
vf = 2.60 m/s


But this answer is not correct... Can someone point out the parts I'm not doing correctly?

 

[Simon Bridge]

Welcome to PF;
What makes you think the answer is not correct?

1. P = (2.06×10^4 x 3.24) + 2(2.05×10^4 x 2.36)

... what is your reasoning behind this calculation?
Are you calculating the initial momentum of the carriages here?
What are the masses of the carriages?

It is usually best practice to complete the algebra symbolically before plugging numbers in.
A good discipline for these things is to go:
1. before $p_i=\cdots$ some equation
2. after $p_f=\cdots$ some other equation
3. conservation of momentum: $p_f=p_i \implies\cdots$ ... putting the RHSs above equal to each other.
... then some algebra.
Usually the equations simplify so you are less likely to make some arithmetic error.

 

[physicshelpppp]

I'm sorry if it isn't very logical what I'm saying. This is my first time ever learning physics. I'm trying to calculate the speed of the railway carriage after the collision. So I began by calculating the momentum by using P = m1v2 + m2v2 and then I used p = mxv and rearranged it to v = p/m to try and calculate the speed of the carriage after the collision.

The initial velocity of the carriage was 3.24 m/s and then it collides and couples to two identical carriages moves to the left at a speed of 2.36 m/s. And I need to calculate the speed of the carriage after this collision. I know my previous two answers have been incorrect because it says incorrect when I enter my value.

Thank you for your reply and in advance for any further clarification you can offer :)

 

[Simon Bridge]

I began by calculating the momentum by using P = m1v2 + m2v2

OK: so what is m1 and what is m2 ?
There is only one mass given in the problem statement.

I know my previous two answers have been incorrect because it says incorrect when I enter my value.

... from this I'm guessing you mean that a computer is telling you the answers are wrong?
One of the problems with these systems is that they can be very fussy about how the answer is entered ... get a digit out or something and it becomes "wrong".

 

[physicshelpppp]

The two carriages it collides with are identical so for the second mass I multiplied the mass of the singular carriage 2x(2.05x10^4)

 

[physicshelpppp]

yeh the program might be picky, but will let me continue trying until i get the correct answer :)

 

[Simon Bridge]

The two carriages it collides with are identical so for the second mass I multiplied the mass of the singular carriage 2x(2.05x10^4)

Are all three carriages the same mass?
If so - then why did you write a different mass for the 1st carriage?

 

[physicshelpppp]

oh my apologies, yes they are all identical, the same mass. That was meant to say 2.05x10^4, was a typo when posting on the forum. sorry!

 

[NascentOxygen]

1. P = (2.06x10^4 x 3.24) + 2(2.05×10^4 x 2.36)
= 1.6x10^5

How many sig fig to retain in intermediate calculations?
= 1.635x10^5

 

[physicshelpppp]

thank you so much it was sig fig :)