question_answer

Three particles A, B and C of equal mass, move with equal speed v along the medians of an equilateral triangle as shown in the figure. They collide at centroid G of the triangle. After collision, A comes to rest and B retracts its path with speed v. What is the speed of C after collision?

A)  Zero                     

B)         \[\frac{\upsilon }{2}\]

C)  \[v\]                      

D)         \[2v\]

Correct Answer: C

Solution :

 The velocities of the three particles are directed as shown in the figure: It is clear that the three velocity vectors are inclined at an angle of\[{{120}^{o}}\]. We know that the resultant of three equal vectors inclined at an angle of \[{{120}^{o}}\] is zero. So, the resultant initial momentum of the system is zero. It means that the resultant final momentum of the system must also be zero. Since, it is given that the particle \[A\] comes to rest after the collision, the momentum of particle \[C\] must be equal and opposite to the momentum of particle\[B\], for the total final momentum to be zero. Thus, velocity of \[C\] should be equal to \[v\] and directed as shown in the figure.