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(C) In an elastic collision between two bodies of equal mass where one body is initially at rest,the conservation of linear momentum and kinetic energ

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$90$

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(C) In an elastic collision between two bodies of equal mass where one body is initially at rest,the conservation of linear momentum and kinetic energy leads to the condition: $\vec{p}_1 = \vec{p}_1' + \vec{p}_2'$ Since the collision is elastic,$K_1 = K_1' + K_2'$,which implies $\frac{p_1^2}{2m} = \frac{p_1'^2}{2m} + \frac{p_2'^2}{2m}$. This simplifies to $p_1^2 = p_1'^2 + p_2'^2$. Comparing this with the vector equation $\vec{p}_1 = \vec{p}_1' + \vec{p}_2'$,we can see that this follows the Pythagorean theorem $(a^2 + b^2 = c^2)$. Therefore,the vectors $\vec{p}_1'$ and $\vec{p}_2'$ must be perpendicular to each other. Thus,the angle between the two bodies after the collision is $90^{\circ}$.

When two identical masses,one of which is at rest,undergo an oblique elastic collision,what is the angle between the two bodies after the collision (in $^{\circ}$)?

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