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(C) Let the radius of the incoming disc be $r_2$. The impact parameter $b$ is given as $r_2$.For two discs of radii $r$ and $r_2$,the impact parameter

Vedclass · Accepted Answer

$\frac{r}{\sqrt{2}-1}$

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(C) Let the radius of the incoming disc be $r_2$. The impact parameter $b$ is given as $r_2$.For two discs of radii $r$ and $r_2$,the impact parameter $b$ is related to the scattering angle $	heta$ by the geometry of the collision.The line connecting the centers of the discs at the moment of impact makes an angle $\phi$ with the direction of the incoming disc,where $\sin \phi = \frac{b}{r+r_2}$.Given $b = r_2$,we have $\sin \phi = \frac{r_2}{r+r_2}$.For equal masses,the velocity vectors after collision are perpendicular if the collision is elastic. However,the scattering angle of the target disc is determined by the geometry of the impact.Given the target disc moves at $45^{\circ}$,the geometry implies $\phi = 45^{\circ}$.Thus,$\sin 45^{\circ} = \frac{r_2}{r+r_2}$.$\frac{1}{\sqrt{2}} = \frac{r_2}{r+r_2}$.$r+r_2 = \sqrt{2} r_2$.$r = r_2(\sqrt{2}-1)$.$r_2 = \frac{r}{\sqrt{2}-1}$.

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