Let, contraction of a horizontal spring, $x$, is $4$ cm and mass of a bullet, $m$, is $10 g$. Spring constant, $k$, is $200 N/m$.

When bullet leaves the gun, what will be the velocity of it?

Can I solve it in the following way?

$0.5mv^2=0.5kx^2$ Or, $mv^2=kx^2$ Or, $v=5.6569 m/s$.

Please tell me whether this method of solution is correct or not.

But if I want to solve it in an alternative way, I may write:

$$F=kx\\ \implies ma=kx\\ \implies 0.01a=(200)(0.04)\\ \implies a=800 m/s^2$$

Again we know $v^2=u^2+2as$ Or, $v=8 m/s$.

Which is totally different from the first method. Why is this occuring? I knew that, $v^2=u^2+2as$ can be used when accelaration, a, is constant over the distance, $s$. Since in case of spring, accelaration created due to expansion or contraction of spring is not constant with the displacement of edge from equilibrium position, we cannot use $v^2=u^2+2as$. But I want to solve it with alternatives way except the first one.

Please tell me accelaration of which point $a=800 m/s^2$ is indicating to, i.e at to which point accelaration will be $800 m/s^2$.

If I want to solve this mathematical problem in the second way, which steps I should follow?


closed as off-topic by CuriousOne, John Rennie, Kyle Kanos, ACuriousMind, Gert Apr 9 '16 at 0:53

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Your first method is using the conservation of energy for the spring bullet system.

You second method fails because the acceleration is not constant and so you cannot use a constant acceleration kinematic equation.

On seeing $ma = -kx$ you might reorganise that the acceleration is proportional to the displacement from a point (the position of the free end of the spring when it is not extended or compressed) and the acceleration is directed towards that point.
Thus the motion of the spring bullet system is simple harmonic with an amplitude $A$, the initial compression of the spring, and $\omega^2 = \frac k m$ where $k$ is the spring constant and $m$ is the mass of the bullet.

Without going through all the derivations the maximum speed of the bullet undergoing shm is when the spring is at tints uncompressed and this is where the bullet leaves the spring.

The maximum speed is $\omega A$ and so the maximum kinetic energy of the bullet is $\frac 1 2 m \omega^2 A^2 = \frac 1 2 k A^2$ exactly as you found using the law of conservation of energy.

At any position $x$ from the point where the bullet is released you can find the magnitude of the acceleration $(a = \sqrt{\frac k m} x$ and the speed $v^2 = \frac k m (A^2 - x^2 )$.

  • $\begingroup$ Again I want to ask the same question. Do you think that my first method is correct? A lot of thanks for your nice explanation. $\endgroup$ – Nazmul Hassan Apr 8 '16 at 12:23
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    $\begingroup$ Perhaps it was not clear enough but I tried to say that your first method using the conservation of energy is correct. $\endgroup$ – Farcher Apr 8 '16 at 12:34
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    $\begingroup$ A new questions needs to be started afresh. Those who set up this forum want there to be one question per thread. However in this case it is easier to write that you can again use the conservation of energy. $\endgroup$ – Farcher Apr 8 '16 at 12:35
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    $\begingroup$ I do not mind helping you. $\endgroup$ – Farcher Apr 8 '16 at 18:49
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    $\begingroup$ I do not mind helping you. What I have done is shown you that your conservation of energy method can be derived in a more long-winded way. The maximum kinetic energy of the bullet is equal to the energy stored in the spring. This occurs when $x=0$ so $v^2 = \omega^2 A^2 $so $\frac 1 2 m v^2 = \frac 12 k x^2$ just as you had at the start. $\endgroup$ – Farcher Apr 8 '16 at 18:54

As a supplement to Farcher's answer, pointing out that the constant acceleration equations fail here, we can still use $\vec{F} = m \vec{a}$ to solve for the speed of the bullet - by means of integration.

We have $- k \vec{x} = m \vec{a}$, but we can rewrite $\vec{a}$.

${a} = \frac{d^2 x}{d t^2} = \frac{ d v}{d t} = \frac{dv}{dx} \frac{dx}{dt} = v \frac{dv}{dx}$.

Putting this back in, we get $mvdv = -k x dx$. Integrating and adding in the limits we get $\int^v_0 mv dv= \int^0_{0.04}-kxdx$

$\frac{mv^2}{2} = \frac{kx^2}{2}|_{x = 0.04}$

Look familiar? We just end up at the energy equation again.

  • $\begingroup$ Again I want to ask the same question. Do you think that my first method is correct? A lot of thanks for your explanation. $\endgroup$ – Nazmul Hassan Apr 8 '16 at 12:27
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    $\begingroup$ Yup, working through the equations, $v = \sqrt{32} = 5.6569 ms^{-1}$ $\endgroup$ – Tweej Apr 8 '16 at 12:39

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