Maximizing Train Speed with a Damped Buffer Stop: Impulse-Momentum Approach

  • Thread starter izzyosman
  • Start date
In summary, the maximum speed at which the train, with a mass of 100 Mg, can hit the buffer stop at the end of its 2.3m travel without exceeding 1.5m/s is xdot = 1.5 m/s. This is found by using the impulse momentum equations and the fact that the train and buffer stop move together after impact.
  • #1
izzyosman
6
0

Homework Statement



A buffer stop at the end of a railway track has a moving part of mass, 2 Mg, which can move 2.3m parallel to the track. The force resisting the motion of the moving part is given by cxdot where xdot is the velocity and c is 200kn/m s

What is the greatest speed with which a train, of mass 100 Mg, can hit the buffer stop if, at the end of its 2.3m travel is not to exceed 1.5m/s?

Assume that after impact, the train and the moving part of the buffer stop move together.

Homework Equations



The Attempt at a Solution



I started with the impulse momentum equations

Train: -F t = 100x10^3 (1.5-x) where x is the initial velocity of the train
Buffer: (F-cxdot) t = 2x10^3 (1.5-0)

I do not know how to proceed because I have not seen an impulse momentum equation which includes a damper. Please help!
 
Last edited:
Physics news on Phys.org
  • #2
I've come to the conclusion that the x and xdot are the same thing as the damping force does not apply until after impact. However I also notice that i only have 2 equations and 3 unknowns: xdot, t and F, and so I'm unable to continue, can anyone think of another equation I'm missing?
 
  • #3
Never mind people I got it!

Okay so what i did was start off with the integral form because the damping force changes over time:

Train: -Ft = 100 (1.5 - xdot)
Buffer: Int(F-cxdot) dt = 2(1.5-0) note that the two xdots are not necessarily the same as it changes over time.

The buffer eqn can reduce to Ft - c Int(xdot)dt = 2(1.5)
Since Int(xdot)dt = 2.3m, I now only have the initial velocity of the train, xdot, to deal with, and everything cancels out!

Yay me! :D
 

1. What is impulse and momentum?

Impulse and momentum are two important concepts in physics that describe the motion of objects. Impulse is defined as the change in an object's momentum, while momentum is the product of an object's mass and velocity.

2. How are impulse and momentum related?

Impulse and momentum are directly related, as impulse is equal to the change in an object's momentum. In other words, the more impulse an object experiences, the greater its change in momentum will be.

3. What is the difference between impulse and force?

While both impulse and force are related to the motion of objects, they are not the same thing. Impulse is a change in momentum, while force is a push or pull that causes an object to accelerate. Impulse can be calculated by multiplying force by the time interval over which it acts.

4. How is impulse calculated?

Impulse can be calculated by multiplying the average force applied to an object by the time interval over which it acts. This can be represented by the equation I = FΔt, where I is impulse, F is force, and Δt is the time interval.

5. What is the conservation of momentum?

The conservation of momentum is a fundamental law in physics that states that the total momentum of a closed system remains constant, regardless of any internal or external forces acting on it. This means that in a collision or interaction between objects, the total momentum before and after the interaction will be the same.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
24
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top