Q16 / 105 · NTA Abhyas (JEE Main) 2020 · Physics · Center of Mass Momentum and Collision

Three identical balls are connected by light inextensible strings with each other as shown and placed over a smooth horizontal table. Length of each string is l .
At moment t=0 , ball B is imparted a velocity v0 perpendicular to the string and then the system is left on its own. The speed of A at the instant just before it collides with C is

 

Figure

  • A
    v03
  • Correct answer
    B
    2v03
  • C
    2v05
  • D
    2v09

Solution

Using conservation of linear momentum along the initial direction of motion of B we get,

mv0=3mv

v=v03

The total kinetic energy of the system will remain the same

12mv02=12mv2+212mv2+v12

where v1 is the component of velocity of A and C perpendicular to the initial direction of motion of B

v02=v032+2v032+2v12

v12=v023

Velocity of A=v2+v12=23v0

Hence, the correct answer is option: 'B'.

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