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(C) For an elastic collision between two particles of equal mass where one is initially at rest,the conservation of linear momentum is given by: $m\ve

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$90$

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(C) For an elastic collision between two particles of equal mass where one is initially at rest,the conservation of linear momentum is given by: $m\vec{v_1} = m\vec{v_1}' + m\vec{v_2}'$,which simplifies to $\vec{v_1} = \vec{v_1}' + \vec{v_2}'$.Taking the dot product of both sides with themselves: $\vec{v_1} \cdot \vec{v_1} = (\vec{v_1}' + \vec{v_2}') \cdot (\vec{v_1}' + \vec{v_2}')$.This results in: $v_1^2 = v_1'^2 + v_2'^2 + 2(\vec{v_1}' \cdot \vec{v_2}')$ ... $(1)$.From the law of conservation of kinetic energy: $\frac{1}{2}mv_1^2 = \frac{1}{2}mv_1'^2 + \frac{1}{2}mv_2'^2$,which simplifies to $v_1^2 = v_1'^2 + v_2'^2$ ... $(2)$.Substituting equation $(2)$ into equation $(1)$: $v_1'^2 + v_2'^2 = v_1'^2 + v_2'^2 + 2(\vec{v_1}' \cdot \vec{v_2}')$.This implies $2(\vec{v_1}' \cdot \vec{v_2}') = 0$,so $\vec{v_1}' \cdot \vec{v_2}' = 0$.Since the dot product is $v_1' v_2' \cos 	heta = 0$,and the velocities are non-zero,we have $\cos 	heta = 0$,which means $	heta = 90^\circ$.

$A$ particle of mass $m$ moving with velocity $\vec{v_1}$ undergoes a two-dimensional elastic collision with another particle of mass $m$ at rest. If the particles move with velocities $\vec{v_1}'$ and $\vec{v_2}'$ after the collision,what is the angle between them in degrees?

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