How Do You Calculate the Mass of a Stationary Train After a Collision?

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In summary, the conversation focused on a problem involving a collision between a known mass train and a stationary train. The question asked for the mass of the stationary train, given that 27% of the initial kinetic energy was dissipated. The conversation referenced equations for conservation of momentum and energy, and the fact that the velocities are not unique. Ultimately, the solution was found by dividing the two equations and utilizing LaTeX code for future problems.
  • #1
EthanB
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Sorry that I don't have LaTeX, I'll do my best without it:

A train with known mass collides with a stationary train and the two link together. If 27% of the initial kinetic energy is dissipated, what is the mass of the stationary train? We are given no velocities.

M1 = mass of moving train (known)
M2 = mass of stationary train
Vi = initial velocity of moving train
Vf = final velocity of the 2-train system
Ki = initial kinetic energy
Kf = final kinetic energy
Wnc = work done by non-conservative forces
Pi = initial momentum
Pf = final momentum

Using conservation of momentum:
Pi = Pf
(M1)(Vi) = (M1+M2)(Vf)

Using conservation of energy:
Ki + Wnc = Kf
(1/2)(M1)(Vi)^2 - (.27)(1/2)(M1)(Vi)^2 = (1/2)(M1+M2)(Vf)^2
(.73)(M1)(Vi)^2 = (M1+M2)(Vf)^2

...2 equations, 3 unknowns. What am I forgetting?
 
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  • #2
EthanB said:
Sorry that I don't have LaTeX, I'll do my best without it:

A train with known mass collides with a stationary train and the two link together. If 27% of the initial kinetic energy is dissipated, what is the mass of the stationary train? We are given no velocities.

M1 = mass of moving train (known)
M2 = mass of stationary train
Vi = initial velocity of moving train
Vf = final velocity of the 2-train system
Ki = initial kinetic energy
Kf = final kinetic energy
Wnc = work done by non-conservative forces
Pi = initial momentum
Pf = final momentum

Using conservation of momentum:
Pi = Pf
(M1)(Vi) = (M1+M2)(Vf)

Using conservation of energy:
Ki + Wnc = Kf
(1/2)(M1)(Vi)^2 - (.27)(1/2)(M1)(Vi)^2 = (1/2)(M1+M2)(Vf)^2
(.73)(M1)(Vi)^2 = (M1+M2)(Vf)^2

...2 equations, 3 unknowns. What am I forgetting?
That you were not asked to find the velocities. Perhaps they are not unique. Does that give you any ideas?
 
  • #3
OlderDan said:
That you were not asked to find the velocities. Perhaps they are not unique. Does that give you any ideas?

I can't see that it does. I realize the velocity of the trains after impact is the same, that they become a system. I've already taken that into account, though.

Help?
 
  • #4
EthanB said:
I can't see that it does. I realize the velocity of the trains after impact is the same, that they become a system. I've already taken that into account, though.

Help?
Try dividing your two equations
 
  • #5
Righteous, dude. Can't believe that one slipped me by. I get so stuck on comparing numbers of equations and unknowns that I miss the little things. Thanks much.
 
  • #6
Sorry that I don't have LaTeX, I'll do my best without it:

But we do have LaTex. All you need next time is write the code between [ tex ] [ /tex ] tags and it comes out nicely.

Daniel.
 
  • #7
Oh, great. Where can I go to find out how to write the code?
 

What is a 'simple' collision?

A 'simple' collision is a type of collision where two objects come into contact with each other but do not rebound or stick together. This means that after the collision, the objects continue moving in their original directions with the same velocities as before the collision.

What are the factors that determine the outcome of a 'simple' collision?

The outcome of a 'simple' collision is determined by the masses and velocities of the objects involved. The angle and direction of the collision can also affect the outcome.

How is momentum conserved in a 'simple' collision?

In a 'simple' collision, the total momentum before the collision is equal to the total momentum after the collision. This is due to the law of conservation of momentum, which states that the total momentum of a closed system remains constant.

Can energy be lost in a 'simple' collision?

In theory, no energy is lost in a 'simple' collision. However, in real-world situations, some energy may be converted into other forms, such as sound or heat, due to factors like friction.

What are some real-life examples of 'simple' collisions?

Some examples of 'simple' collisions include billiard balls colliding on a pool table, cars colliding on a road, and a basketball bouncing off the ground.

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