Question:
How do I answer this problem?
Ben D
2007-10-11 20:26:50 UTC
A car with a mass of 1480kg crashes in toa wall. The velocity of the car right before the collision is 14.9m/s. The bumper is compressed like a spring with the spring constant k = 1.14 x 10^7 N/m. What is the maximum deformation of the bumper? Answer in meters.
Three answers:
Gary B
2007-10-11 20:54:25 UTC
You have two equations here:



1. F=0.5kx (the equation of the spring)

But F=ma=mv/t

So, mv/t = 0.5kx

Or, x=mv/0.5tk



2. x=0.5 at^2 (the equation of the deceleration of the car)

Substitute a=v/t and rearrange this to: x=0.5(v/t) * t^2

Or, x = 0.5vt

Rearrange to: t=2x/v



Now, substitute for the "t" in the first equation with the 2x/v from the second equation:



x=mv / 0.5(2x/v) k

Or, x = mv^2/xk

Or, x^2 = mv^2/k



Now, this is good because we know everything on the right side of the equation. So,



x^2 = 1480(14.9)(14.9) / (1.14x10^7)

x^2 = 0.0288

Therefore, x = 0.17



You can substitute units in the equation to make sure the units work out right:



x^2 = mv^2/2k



Since m is in units of kg, v is in units of m/s, and k is in units of N/m of kg/s^2, substitute the units for m, v, and k:



units of x^2 = (kg)(m/s)^2 / kg/s^2



The kg's cancel, as do the s^2 units. This leaves:



units of x^2 = m^2



which means x is in meters, which is what we want. :-)
Jay B
2007-10-11 20:44:24 UTC
Approach this as a conservation of energy problem. I would make the assumption that at the instant of maximum deformation, all the kinetic energy of the car has gone into the bumper much like energy stored in a spring. (Recall that the energy in a spring 1/2*k*x^2).



So 1/2mv^2=1/2kx^2, solve for x.
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2017-01-03 17:13:13 UTC
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