Particle A of mass $m_1$ moving with velocity $(\sqrt3\hat i + \hat j)\ \text{m s}^{-1}$ collides with another particle B of mass $m_2$ which is at rest initially. Let $\vec v_1$ and $\vec v_2$ be the velocities of particles A and B after the collision respectively. If $m_1 = 2m_2$ and after the collision $\vec v_1 = (\hat i + \sqrt3\hat j)\ \text{m s}^{-1}$, the angle between $\vec v_1$ and $\vec v_2$ is:
Answer: (D) $105^\circ$
Momentum conservation with $m_1 = 2m_2$:
$$2m_2(\sqrt3\hat i + \hat j) = 2m_2(\hat i + \sqrt3\hat j) + m_2\vec v_2$$
$$\vec v_2 = 2(\sqrt3 - 1)\hat i + 2(1 - \sqrt3)\hat j = 2(\sqrt3-1)(\hat i - \hat j)$$
$\vec v_1$ makes $60^\circ$ with the $x$-axis ($\tan^{-1}\sqrt3$), and $\vec v_2$ makes $-45^\circ$.
Angle between them $= 60^\circ + 45^\circ = 105^\circ$.
Solution by Sreeraj P, M.Sc Physics