Question

In a spring gun, a spring of mass 0.244kg and force constant 3500N/m is compressed 2.00cm from its unstretched length. When the trigger is pulled, the spring pushes horizontally on a 4.1×10−2kg ball. The work done by friction is negligible. (a) Calculate the ball's speed when the spring reaches its uncompressed length ignoring the mass of the spring.

Answer 1

ans is 5.8m/s ( i got this answer correct, but there is a second part of the qns which i cant solve)

part b: Calculate the ball's speed when the spring reaches its uncompressed length including, using the results of part A, the mass of the spring.

Answer 2

The part in this first bit is about deriving a term for the kinetic energy of a spring. The bottom part gives a try at the solution. If you have the answers, skip down there to see if it's worth reading through all this.

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Going from this guy (pdf, sorry), problem 4 eqns 37-39

http://www.google.com/url?q=http://www.stanford.edu/dept/physics/publications/oldquals/Qual2006/Qual2006Day1Solutions.pdf&sa=U&ei=Cy2cUrLJD8vwoATm-4CQCA&ved=0CB0QFjAAOAo&usg=AFQjCNHQPO15Wqa_CpbKMWvdc6UGqsod-g

you define a coordinate xi such that it exists at some bit of the spring, so runs from 0 to x, where x is the length of the spring. When the spring is moving, the length of spring from one instant to the next is then given by x+vdt, where v is the velocity of the spring. The density of the spring at position xi is given by M/x (so a smaller length of spring (compressed) yields a higher mass density, and vice versa), and the density is constant throughout.

Then the fraction xi over x is the fraction of the way down the spring our coordinate xi is. Since the spring moves linearly, *this must be constant* (as the spring decompresses, the coordinate xi is the same fraction of length of he way down the spring the whole time), so from one instant to the next

\[ \frac{ \xi + \mathrm{d}\xi}{x+v\mathrm{d}t} = \frac{\xi}{x}\]
\[(\xi + \mathrm{d}\xi ) x= ( x+ v \mathrm{d}t)(\xi)\]
\[x\mathrm{d}\xi=\xi v\mathrm{d}t\]
\[ \frac{\mathrm{d}\xi}{\mathrm{d}t}=\frac{\xi v}{x}\]
this term is the velocity of the coordinate xi, and therefore the velocity of the spring. This gives way to an expression for kinetic energy.

Then, looking at an expression for kinetic energy with variable mass, where d m is the mass of the spring, and u is the velocity of the spring

\[T= \int\frac{1}{2}u^2 \mathrm{d}m \]

we can see from above that dm is given by the density times the length differential d xi
\[d m = \frac{M}{x} \mathrm{d}\xi \]
and the velocity is given by
\[u = \frac{\mathrm{d}\xi}{\mathrm{d}t}\]
Plugging in those terms gives
\[ T=\int \frac{1}{2} \left(\frac{M}{x} \mathrm{d}\xi \right) \left(\frac{\mathrm{d}\xi}{\mathrm{d}t}\right)^2\]
And knowing from above that
\[ \frac{\mathrm{d}\xi}{\mathrm{d}t}=\frac{\xi v}{x}\]
It turns into
\[T = \int \frac{1}{2} \left(\frac{M}{x} \right) \left(\frac{\xi v}{x}\right)^2 \mathrm{d}\xi \]
which we can integrate xi=0 to xi=x
\[ T = \frac{Mv^2}{2x^3} \int_0^x \xi^2 \mathrm{d}\xi\]
So the kinetic energy of the spring is
\[T = \frac{1}{6}Mv^2\]
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Then, for applications to this problem, I think that since you know the total kinetic energy of the system when the spring is uncompressed, you equate that to a new system of kinetic energies that is comprised of the ball and the spring, knowing that they will both be traveling at the same velocity, so

\[T_{massless} = T_{ball} + T_{spring}\]
\[ \frac{1}{2}m(5.8m/s)^2 = v^2 \left(\frac{1}{2}m+\frac{1}{6}M\right) \]
and finally
\[v=(5.8m/s) \sqrt{ \frac{ \frac{1}{2}m}{\frac{1}{2}m+\frac{1}{6}M}}\]
\[v=(5.8m/s) \sqrt{\frac{m}{m+\frac{1}{3}M}}\]

Which gives an answer that is less than the velocity with a massless spring, so at least it's in the right ballpark! ^_^

Answer 3

Here they derive the expression using y instead of xi, so might be more readable

http://en.wikipedia.org/wiki/Effective_mass_%28spring%E2%80%93mass_system%29

The final solution still holds ^_^