Mistake on official answer (conservation of linear momentum)

  • Thread starter Hernaner28
  • Start date
  • #1
263
0

Homework Statement


attachment.php?attachmentid=48759&stc=1&d=1341021527.gif

The problem says:

A mine provides a transport ystem constiting into carriages (of neglect mass) hanging on a riel free of friction.
In a specific moment two people of mass m=45kg are travelling when one of them says his breaks are not working. In that moment he was at V1=10m/s whilst the carriage ahead was at V2=8.0m/s.
Fortunately who travells in the second carriage can take a mass of 25Kg of sand and throw a fraction of it with a velocity V0=6.0m/s (respect to him).
Calculate the MINIMUM mass that the man will have to throw to avoid the colission.

** The "correct" answer is 23Kg

Homework Equations





The Attempt at a Solution



Well, obviously this is a conservation of linear momentum problem but I think it's not 23Kg. Here's what I did:

[tex] \displaystyle {{v}_{mass}}={{v}_{0}}+{{v}_{back}}[/tex]

That's the speed of the mass respect to the Earth. Vback is the additional speed the man should have when throwing the mass. If it asks the minimum then the man has to have a final speed of 10m/s, and therefore the "backward" speed should be 2m/s.(8+2=10):

[tex] \displaystyle {{v}_{mass}}=-6.0+2.0=-4.0[/tex]

[tex] \displaystyle {{p}_{i}}={{v}_{2i}}\left( m+{{m}_{x}} \right)[/tex]
[tex] \displaystyle {{p}_{f}}=\left( m+{{m}_{x}} \right){{v}_{2f}}+{{m}_{x}}\left( {{v}_{mass}} \right)[/tex]

m is 45Kg (man mass) and mx is the unknown thrown mass.
Notice that my initial mass is the unknown thrown mass plus the mass of the man, but solving for those values I get another answer BUT if I consider the initial mass as only the mass of the man I get 22.5 which is the correct answer. But should I consider both mass?

Thanks!
 

Attachments

  • carritos.gif
    carritos.gif
    4.6 KB · Views: 550
Last edited:

Answers and Replies

  • #2
77
0
Fortunately who travells in the second carriage can take a mass of 25Kg of sand and throw a fraction of it with a velocity V0=6.0m/s (respect to the second person).

Rereading this part of the problem might help.

If you have any more questions, ask!
 
  • #3
1,065
10
1. [tex] \displaystyle {{v}_{mass}}=-6.0+2.0=-4.0[/tex]

[tex] \displaystyle {{p}_{i}}={{v}_{2i}}\left( m+{{m}_{x}} \right)[/tex]
[tex] \displaystyle {{p}_{f}}=\left( m+{{m}_{x}} \right){{v}_{2f}}+{{m}_{x}}\left( {{v}_{mass}} \right)[/tex]

Thanks!


If second man drops a sand, what is the velocity of the sand with respect to him and what is velocity relative to ground ?

The final mass should be less than orginal mass since part has been thrown away.
 
Last edited:
  • #4
263
0
Sorry, it is a mminus sign, I typed it wrongly but I don't get the answer anyway. I get 23 Kg when I do what I told. I already know how to solve this problem, I just need you tell me what is correct, consider the whole man and sand or only the mass of the man".

Thanks!
 
  • #5
1,065
10
Sorry, it is a mminus sign, I typed it wrongly but I don't get the answer anyway. I get 23 Kg when I do what I told. I already know how to solve this problem, I just need you tell me what is correct, consider the whole man and sand or only the mass of the man".

Thanks!

You have to consider whole man and sand.
Your vmass is wrong.
 
  • #6
263
0
Why? The man throws the mass and he sees it at 6.0 m/s, so the Vmass respect to the ground is -6.0+ Velocity that the man gains which should be 2.0m/s so that he equals the velocity of the man without breaks and that would be the minimum.
 
  • #7
1,065
10
Take example for 1 seconds after they depart.

The man moves 10m to the right
Now relative to the man the sand moved 6m from him.

What is the distance moved according to a man on the ground?
 
  • #8
263
0
The man moved 10m to the right but saw from the ground. He actually has to gain a speed of 2m/s. Could you finally explain it? because I'm not getting it.

Thank you!
 
  • #9
1,065
10
So we are required to have minimum 10m/s to avoid collision.
10m/s is ground speed.
6m/s is relative to the man.
Every second the man moved 10m.
To the man, the sand moved 6m

Can you find what the distance travelled by the sand relative to ground in one second?
 
  • #11
1,065
10
16m/s ?

So what is the sand's ground velocity?
Wrong for 16m/s.
 
  • #12
263
0
4 m/s ? I don't know! Please tell me... if you explain it I will understand it.
 
Last edited:
  • #13
1,065
10
For 1 sec.

.------------------------ Man
0------------------------10m
|........|....................|
|........|<......6m.....>|
-------- Sand

You got the value and subtitute in your equation with man and sand.
 
  • #14
263
0
Seriously, I am not understanding those relative velocities, I need you to explain it to me.

If the man moved 10 m, and the sand moves 6 m respect to him, then the sand moved 4 m for a man in the ground. So it would be -4
 
  • #15
1,065
10
Seriously, I am not understanding those relative velocities, I need you to explain it to me.

If the man moved 10 m, and the sand moves 6 m respect to him, then the sand moved 16 m for a man in the ground.

At beginnig both at origin.
1 second later, man moves to the right 10m
Now the man check that the sand is 6m behind him.
If the sand speed 16m/s, it will overtake the man.

If -4 means, 14m behind him.
 
  • #16
263
0
At beginnig both at origin.
1 second later, man moves to the right 10m
Now the man check that the sand is 6m behind him.
If the sand speed 16m/s, it will overtake the man.

No, the sand moved 4 meters to the left so the velocity is -4.0 m/s. I don't know what's wrong with that
 
  • #17
263
0
[tex] \displaystyle {{v}_{2i}}\left( m+{{m}_{x}} \right)=\left( m-{{m}_{x}} \right){{v}_{2f}}+{{m}_{x}}\left( {{v}_{mass}} \right)[/tex]
[tex] \displaystyle 8\cdot 70=10\cdot 70-10{{m}_{x}}-{{m}_{x}}{{v}_{mass}}[/tex]

[tex] \displaystyle 8\cdot 70=10\cdot 70-10{{m}_{x}}-\left( -4{{m}_{x}} \right)[/tex]

[tex] \displaystyle {{m}_{x}}=23.33333[/tex]

Nothing was wrong with -4m/s!!!!

[tex]\displaystyle {{v}_{mass}}=-6.0+2.0=-4.0[/tex]

. I was mistaken with masses... not with relative speeds ¬¬
 
Last edited:
  • #18
1,065
10
Can you tell how the sign changes from +mxv to -mx in your 2nd. equation?
 
  • #19
263
0
Can you tell how the sign changes from + to - in your 2nd. equation?

Yeah, I have just realized that... then it should be

[tex] \displaystyle {{v}_{mass}}=-6.0+10.0=4.0[/tex]

right?

I thought I had to put the speed the man gained (2m/s), not the whole speed (10m/s).

But why is it positive? The mass has a velocity pointing at left!!
 
  • #20
1,065
10
Yeah, I have just realized that... then it should be

[tex] \displaystyle {{v}_{mass}}=-6.0+10.0=4.0[/tex]

right?

I thought I had to put the speed the man gained (2m/s), not the whole speed (10m/s).

But why is it positive? The mass has a velocity pointing at left!!

Yes sand's ground velocity is 4m/s to the right. .
 
  • #21
77
0
I think you did everything right except for what I was trying to point out. I haven't done the calculations myself but I noticed you did one small thing that would make your answer different on the order of 0.5 like you have.

The key is in this line:

"Fortunately who travells in the second carriage can take a mass of 25Kg of sand and throw a fraction of it with a velocity V0=6.0m/s (respect to him)."

In your calculations you assumed that the mass of sand with him was only the amount that he threw. This part of the question states that he can take 25 kg and throws a fraction of that total. I take this to mean that he takes the entire 25 kg with him so after he throws it, a non-zero amount is still with him.
 

Related Threads on Mistake on official answer (conservation of linear momentum)

Replies
6
Views
4K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
7
Views
2K
  • Last Post
Replies
1
Views
658
  • Last Post
Replies
2
Views
977
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
2
Views
1K
Top