Question

Three particles A, B and C of equal mass are moving with the same velocity v along the medians of an equilateral triangle. These particle collide at the centre G of triangle. After collision A becomes stationary, B retraces its path with velocity v then the magnitude and direction of velocity of C will be

• A. v and opposite to B
• B. v and in the direction of A
• C. v and in the direction of C
• D. v and in the direction of B

Answer 1

Answer: (D)

v and in the direction of B

Answer 2

From the figure (I) it is clear that before collision initial

momentum of the system = 0

After the collision, A becomes stationary, B retraces its path with velocity v. Let C moves with velocity V making an angle θ from the horizontal. As the initial momentum of the system is zero, therefore horizontal and vertical momentum after the collision should also be equal to zero.

From figure (II) Horizontal momentum

$v\cos\theta + v\cos 30^{o} = 0$ …..(i)

Vertical momentum $v\sin\theta - v\sin 30^{o} = 0$ …..(ii)

By solving (i) and (ii) we get $\theta = - 30^{o}$ and V = v i.e. the C will move with velocity v in the direction of B.