Two particles A and B of equal masses lie close together on a horizontal table and are connected by a light inextensible string of length `l`. A is projected vertically upwards with a velocity `sqrt(10gl)`. Find the velocity with which it reaches the table again.

Answer

B

Explanation

String becomes tight when A moves upwards by a distance 1. Let `v_1` be the velocity of A at this moment, then


`v_1^2=(sqrt(10gl))^2-2gl=8gl` or `v_1=sqrt(8gl)`
Let `v_2` be the common velocities of both A and B just after string becomes tight. Then from conservation of linear momentum.
`v_2=v_1/2=(sqrt(8gl))/(2)`
Both particles return to their original height with same speed `v_2`. String becomes loose after B strikes the ground and the speed v with which A strikes the ground is,
`v^2=v_2^2+2gl+(8gl)/(4)+2gl`
or `v^2=4gl` or `v=2sqrt(gl)`