Question

A sphere $P$ of mass m and velocity $\vec{v_i}$- 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

• A. $0$
• B. $45^{\circ}$
• C. $90^{\circ}$
• D. $180^{\circ}$

Answer 1

Answer: (C)

According to law of conservation of linear momentum, we get
$m \vec{v}_{i}+m \times 0=m \vec{v}_{P f}+m \vec{v}_{Q f}$
where $\vec{v}_{p f}$ and $\vec{v}_{Q f}$ are the final velocities of spheres $P$ and $Q$ after collision respectively.
$\vec{v}_{i}=\vec{v}_{P f}+\vec{v}_{Q f}$
$\left(\vec{v}_{i} \cdot \vec{v}_{i}\right)=\left(\vec{v}_{P f}+\vec{v}_{Q f}\right) \cdot\left(\vec{v}_{P f}+\vec{v}_{Q f}\right)$
$=\vec{v}_{P f} \cdot \vec{v}_{P f}+\vec{v}_{Q f} \cdot \vec{v}_{Q f}+2 \vec{v}_{P f} \cdot \vec{v}_{Q f}$
or $v_{i}^{2}=v_{P f}^{2}+v_{Q f}^{2}+2 v_{P f} v_{Q f} \cos \theta$ ...(i)
According to conservation of kinetic energy, we get
$\frac{1}{2} m v_{i}^{2}=\frac{1}{2} m v_{p f}^{2}+\frac{1}{2} m v_{Q f}^{2}$
$\Rightarrow v_{i}^{2}=v_{P f}^{2}+v_{Q f}^{2}$ ...(ii)
Comparing $(i)$ and $(ii)$, we get
$2 v_{p f} v_{q f} \cos \theta=0$
$\Rightarrow \cos \theta=0$
or $\theta=90^{\circ}$