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- Homework Statement
- A particle of mass m moves under the action of a harmonic oscillator force with potential energy 1

2kr2. Initially, it is moving in a circle of radius a. Find the orbital speed v. It is then given a blow of impulse mv in a direction making an angle α with its original velocity. Use the conservation laws to determine the minimum and maximum distances from the origin during the subsequent motion. Explain your results physically for the two limiting cases α = 0 and α = π.

- Relevant Equations
- Conservation of angular momentum and Conservation of Energy

$$J= const$$ and $$E = const$$

I've already solved the orbital speed by equating the kinetic and potential energy in the circle orbit case.

$$\frac{1}{2}mv^2 = \frac{1}{2}ka^2.$$And so $$v^2 = \frac{k}{m}a^2$$Now when the impulse is added, the particle will obviously change course. If we set our reference point in time just after the impulse is added. We get back our Angular momentum and Energy Conservation.So,

$$Ei = Ef$$

$$\frac{1}{2}mv^2 + \frac{p^2}{2m} + \frac{1}{2}ka^2 = \frac{1}{2}mv'^2 + \frac{1}{2}kr^2$$where v' is the final velocity, and r is new extension of spring.

$$Ji = Jf$$ also gives

$$mva + mvacos(\alpha) = mv'r$$Now we can solve for v' then plug that in the Energy conservation. I will also substitute in the value for v.$$\frac{1}{2}ka^2 + \frac{1}{2}ka^2 + \frac{1}{2}ka^2 =\frac{ma^2v^2(1+cos(\alpha))^2}{2r^2} + \frac{1}{2}kr^2$$Dividing out the $\frac{1}{2}$s and Ks. Then multiplying both sides by $r^2$ we get something like:$$r^4 - 3a^2r^2 + a^4(1+cos(\alpha))^2=0$$which finally gives$$r^2 = \frac{3a^2\pm\sqrt{9a^4-4a^4(1+cos(\alpha))^2}}{2}$$

This is problematic since the term in the root becomes negative, at least for $(\alpha) = 0.$The answer in the book is:$$r^2 = \frac{1}{2}a^2(3+2cos(\alpha) \pm \sqrt{5+4cos(\alpha)}$$I tried solving the problem using the ##\frac{J^2}{2mr^2}## term instead of the ##\frac{p^2}{2m}## term in energy conservation but to no avail.Any help would be appreciated.

Edit. This is my first time using latex. Idk why the last two terms are weird here, they look fine on overleaf. Please excuse an strange looking things.

$$\frac{1}{2}mv^2 = \frac{1}{2}ka^2.$$And so $$v^2 = \frac{k}{m}a^2$$Now when the impulse is added, the particle will obviously change course. If we set our reference point in time just after the impulse is added. We get back our Angular momentum and Energy Conservation.So,

$$Ei = Ef$$

$$\frac{1}{2}mv^2 + \frac{p^2}{2m} + \frac{1}{2}ka^2 = \frac{1}{2}mv'^2 + \frac{1}{2}kr^2$$where v' is the final velocity, and r is new extension of spring.

$$Ji = Jf$$ also gives

$$mva + mvacos(\alpha) = mv'r$$Now we can solve for v' then plug that in the Energy conservation. I will also substitute in the value for v.$$\frac{1}{2}ka^2 + \frac{1}{2}ka^2 + \frac{1}{2}ka^2 =\frac{ma^2v^2(1+cos(\alpha))^2}{2r^2} + \frac{1}{2}kr^2$$Dividing out the $\frac{1}{2}$s and Ks. Then multiplying both sides by $r^2$ we get something like:$$r^4 - 3a^2r^2 + a^4(1+cos(\alpha))^2=0$$which finally gives$$r^2 = \frac{3a^2\pm\sqrt{9a^4-4a^4(1+cos(\alpha))^2}}{2}$$

This is problematic since the term in the root becomes negative, at least for $(\alpha) = 0.$The answer in the book is:$$r^2 = \frac{1}{2}a^2(3+2cos(\alpha) \pm \sqrt{5+4cos(\alpha)}$$I tried solving the problem using the ##\frac{J^2}{2mr^2}## term instead of the ##\frac{p^2}{2m}## term in energy conservation but to no avail.Any help would be appreciated.

Edit. This is my first time using latex. Idk why the last two terms are weird here, they look fine on overleaf. Please excuse an strange looking things.

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