Momentum and springs - A2 Physics

[Vanagib]

Homework Statement

A train of total mass 1.02x10^5 kg strikes a buffer that behaves like a spring of stiffness 320kN/m with an initial velocity of 0.15m/s
Calculate the maximum compression of the spring.

Homework Equations

P=mv

The Attempt at a Solution

P=mv
= 15300kgm/s

Units of spring stiffness so assumed Stiffness = F/d
P=Ft
F=P/t

so S=P/dt, v=d/t
t=d/v

so, S=Pv/d^2

d^2=Pv/s
d=sqrt (pv/s)
final solution:

sqrt ((1.53x10^4 * 0.15) / 320x10^3)

= 0.0847m compression.


Comments:
Unsure of the real answer, and this method is totally invented. - Nor does it appear to take into account the changing force with respect to distance. - Is there a simpler method I have not taken into account? - This much assumption seems very off compared to the rest of the A2 syllabus

Thanks,
Vanagib.

 

[dx]

The answer is correct. (Didn't check your method btw.)

 

[saunderson]

Hi Vanagib,

i think you are right


another approach:

If the train is stopped by the spring, the hole kinetic energy of the train is saved in the spring, so

T = V​

\frac{1}{2} m v^2 = \frac{1}{2} k x^2​

x_{max} = \sqrt{\frac{m}{k}} \cdot v = 0.0847 m​

with best regards!

 

[SmashtheVan]

That's the correct answer.

a simpler way to do this would to consider the Kinetic/Potential energy systems. The train has initial Kinetic energy found by .5mv^2, and when the train stops the buffer has all of that energy transformed into potential energy, which for a spring is .5kx^2. set the two equal, and simple algebra to solve for x.

but the way you did it is fine too.

edit: it seems as in typing this I've been beat to my explanation, and with Latex too!

 

[Vanagib]

Many thanks all :)