# A particle of mass m and velocity V collides obliquely elastically with a stationary particle of mass 'm'. The angle between the velocity vectors of the two particles after the collision is

$\begin {array} {1 1} (a)\;45^{\circ} & \quad (b)\;30^{\circ} \\ (c)\;90^{\circ} & \quad (d)\;None\;of\;these \end {array}$

Now $\; \; \;mu_1 + mo = m\overrightarrow {v_1} + m\overrightarrow{v_2}$
$\overrightarrow {u_1} = \overrightarrow{v_1} + \overrightarrow{v_2}----(1)$ (vectorial addition)
Also since it is elastic collision
\begin{align*}\frac{1}{2} mu_1^2& = \frac{1}{2} mv_1^2 + \frac{1}{2} mv_2^2 \\ u_1^2 & = v_1^2 + v_2^2 - - - - (2)\end{align*}
From (1)
$\overrightarrow{u_1} . \overrightarrow{u_1} = (\overrightarrow {v_1} + \overrightarrow{v_2}) . (\overrightarrow{v_1} + \overrightarrow{v_2})$
$u_1^2 = v_1^2 + v_2^2 + 2(\overrightarrow{v_1}. \overrightarrow{v_2}) - - - - (3)$
From (2) an (3) we get
$2 \overrightarrow{v_1} . \overrightarrow{v_2} = 0$
$\implies \overrightarrow{v_1}. \overrightarrow{v_2} = 0$
$\implies v_1\perp v_2$
answered Nov 25, 2013 by
edited Mar 27