Question

Two masses a and/b connected with an inextensible string of length l lie on smooth horizontal plane. A is given velocity of v m/s along the perpendicular to line ab . Find the tension in string during subsequent motion.

Answer 1

Answer:

The answer  will be 2mv^2/2l

Explanation:

According to momentum conservation

v=  mx 2mv/ m+2m[ where m= mass vis the velocity]

v= 2v/3

centre of mass

y= 2m + m  xl/ 2m

y = l/3

velocity is va= v- 2v/3

va= v/3

Let the thension

T= 2mv^2/2l

Therefore the tension in string during subsequent motion is 2mv^2/2l