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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

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

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Collision is elastic.
Hence velocities along the line of impact will be exchanged.
Velocity of particle along line of impact $=V \cos \theta$
Velocity of particle perpendicular to line of impact
$ V \sin \theta= v \times \large\frac{1}{2}$
Velocity of disc $=\large\frac{ v \sqrt 3}{3}$
Particles will move perpendicular to disc.
The 2 bodies will move at right angles.
Hence c is the correct answer.
answered Nov 25, 2013 by meena.p
edited Apr 2, 2014 by meena.p
 

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