A sphere P of mass m moving with velocity v u | vedclass

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(C) According to the law of conservation of linear momentum:$m \vec{v} = m \vec{v}_{P} + m \vec{v}_{Q}$$\vec{v} = \vec{v}_{P} + \vec{v}_{Q}$Taking the

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(C) According to the law of conservation of linear momentum:$m \vec{v} = m \vec{v}_{P} + m \vec{v}_{Q}$$\vec{v} = \vec{v}_{P} + \vec{v}_{Q}$Taking the dot product of both sides with themselves:$\vec{v} \cdot \vec{v} = (\vec{v}_{P} + \vec{v}_{Q}) \cdot (\vec{v}_{P} + \vec{v}_{Q})$$v^2 = v_{P}^2 + v_{Q}^2 + 2 \vec{v}_{P} \cdot \vec{v}_{Q}$$v^2 = v_{P}^2 + v_{Q}^2 + 2 v_{P} v_{Q} \cos 	heta \quad (i)$According to the law of conservation of kinetic energy for an elastic collision:$\frac{1}{2} m v^2 = \frac{1}{2} m v_{P}^2 + \frac{1}{2} m v_{Q}^2$$v^2 = v_{P}^2 + v_{Q}^2 \quad (ii)$Comparing equations $(i)$ and $(ii)$:$v_{P}^2 + v_{Q}^2 = v_{P}^2 + v_{Q}^2 + 2 v_{P} v_{Q} \cos 	heta$$0 = 2 v_{P} v_{Q} \cos 	heta$Since $v_{P} 
eq 0$ and $v_{Q} 
eq 0$,we must have $\cos 	heta = 0$,which implies $	heta = 90^{\circ}$.

$A$ sphere $P$ of mass $m$ moving with velocity $v$ undergoes an oblique and perfectly elastic collision with an identical sphere $Q$ initially at rest. The angle $\theta$ between the velocities of the spheres after the collision shall be ............... $^o$.

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