The two particles of equal mass are joined by a rod of negli

The two particles of equal mass are joined by a rod of negligible mass. If they are released from rest in the position shown and slide on the smooth guide in the vertical plane, calculate their velocity v when A reaches B\'s position and B is at B\'. Answer. v=6.95 ft/s

Solution

>> Using Energy Conservation

As, at Initial Positon, both particles are at rest,

So, their Kinetic Energy = 0

and, taking reference as B level.

Potential Energy of System = mgh

wheren, h = 18 inch = 1.5 ft/s

=> Total Energy at initial position = 1.5*mg = 1.5*32.2*m = 48.3*m

>> Now, at final position,

Both are horizontal .

So, potential Energy = 0

>> As, both are moving with velocity, \"v\"

So, Kinetic Energy = 2*[(1/2)*m*V²] = m*V²

>> Total Energy at final position = m*V²

>> As, Energy is conserved

=> 48.3*m = m*V²

Solving,

V = Required velocity = 6.95 ft/s .......ANSWER......