Question

A runaway railroad car, with mass 30x10^4kg, coasts across a level track at 2m/s when it collides with a spring loaded bumper at the end of the track. If the spring constant of the bumper is 2x10^6N/m, what is the maximum compression of the spring during the collision?

Answer 1

This is a force problem with a kinematic twist involved. So we know that the bumper will stop the car after the spring is displaced, but we want to know this displacement. Now let's use our knowledge of Forces. We know that because the car is coasting the net forces on the car must be zero this tells us that \[\Sigma F _{net} = 0\], we will only concern ourselves with the X-direction. Knowing this we now have \[F _{net} = m*a = F _{spring} = k*\Delta s\] now we need to know the acceleration needed to stop the car, okay we only need it as intermediary step. So we have an initial velocity and final velocity and we want to find the final change in distance, we can use \[v ^{2} = v _{0}^{2} - 2*a*\Delta s\] ,this is the same change in s you are looking for, now solve for acceleration in this equation replace it into acceleration in the above equation and do some algebra to solve for your Delta s. Once you replace it back you should get \[\frac{m*v _{0}^{2} }{ 2*\Delta s } = k*\Delta s\] from here it's pretty straightforward.

Answer 2

@mikev does that make sense?

Answer 3

You can also look at in terms of conservation of energy

So you have the kinetic energy of the railway car
\[K = \frac{1}{2}mv^2\]

And you have the spring potential energy, where x is displacement away from its equilibrium position.
\[U_{spring} = \frac{1}{2} kx^2\]

Using conservation of energy, you know that that the initial energy is purely the kinetic energy of the car, and that the final energy is purely spring potential energy because the car has completely stopped when the spring is at its max compression.

Now you just equate the two
\[E_i=E_f\]
\[\frac{1}{2}mv^2 = \frac{1}{2}kx^2\]
\[x=v\sqrt{\frac{m}{k}}\]

and solve for x

Answer 4

that looks way more familiar

Answer 5

I'm trying to figure out why @murphyandhislaw 's answer is incorrect, because it seems sound but yields a false answer...

Answer 6

Conservation of energy is almost always easier methinks ^_^

Answer 7

thats what we are going over in class his method looked foreign

Answer 8

It should be possible to do it with kinematics like he did, just saying that the spring applies some acceleration over some distance, but it's not set up right for some reason...

Answer 9

There are more steps that way, however, and sticking with energy is almost always more pleasant

Answer 10

But this makes sense? ^_^

Answer 11

@murphyandhislaw It's because the force of the spring varies with distance, so it's not
\[ma=kx\]
it's
\[ma=\frac{1}{2}kx\]
You have to take the average force over the distance, where x_0 = 0
http://theory.uwinnipeg.ca/physics/work/node5.html

Answer 12

Anyways, happy physicsing! ^_^

Answer 13

ok thank you

Answer 14

what do i use to fill in the equation \[x=\v \sqrt\frac{ m }{ k}\]

Answer 15

v is the velocity of the rail car, m is its mass, and k is the spring constant of the spring.

Answer 16

i tried \[x=2 \sqrt{\frac{ 30x10^4}{ 1.8x10^11}} \]

Answer 17

your mass and velocity are correct, but the spring constant from above is given as
\[k=2\times 10^6\]

Where did you get 1.8x10^11 ?

Answer 18

youre right musta been lookin at a different problem thank you

Answer 19

do i square both of these? \[\sqrt{\frac{ 30x10^4 }{ 2x10^6 }}\]

Answer 20

or divide then square then multiply by 2?

Answer 21

You don't square either of them...
\[f^2≠\sqrt f\]
\[\sqrt f = f^{1/2}\]

So you take the value
\[\frac{30\times19^4 kg }{2\times 10^6 N/m}\]
and then take the square root of it, then multiply that value by 2 m/s

Answer 22

.78?

Answer 23

thank you for helping me work it out. I understand it now

Answer 24

.77 m yah

And welcome! Glad I could help ^_^