Question

Two particles A and B each of mass m are attached by a light inextensible string of length 2l .The whole system lies on a smooth horizontal table with B initially at a distance l from A.The particle at end B is projected across the table with speed u perpendicular to AB.Find the velocity of ball A.

Answer 1

Answer:

The velocity of ball A is √3u/4 unit

Explanation:

According to the problem the length of the string is 2l

Now let the initial velocity of the ball in AB axis is ucosθ

and the initial velocity of the ball in perpendicular of Ab is usinθ

Now from the law of momentum conservation we can say that,

2mv = mucosθ [ where v is the velocity of the ball A]

 => v = ucosθ/2

 Now from the diagram we can say that,

cosθ =√ (2l)^2-l^2/2l =√ 4l^2- l^2/2l

Therefore,

v= u/2 x√ 4l^2- l^2/2l

   = u/2 x √3l/2

 = √3u/4 unit