Question

Two bodies of same mass tied with an inelastic string of length l lie together. One of them is projected vertically upwards with the velocity 6gl*1/2. Find the maximum height to which the centre of mass of system of two masses arises

Answer 1

Applying the law of conservation of energy on the given question to the initial moment and the moment when the first body reaches the height L
 we obtain = mv²2/mgL

Where V is the speed of the first body at height L
Thus
V = √Vi²-2gL = √6gL-2gL = 2√gL

At this moment both bodies start to move as one piece with its centre of mass at height L/2.
According to the law of conservation of energy the new speed of the system (mass is twice the mass of one body) is given by (the process is inelastic)V' = V/2 =√gL

After this moment  elevate the centre of mass of the system by the height ∆ℎ,
which is given by
∆ℎ =V'²/2g = gL/2g = L/2
Thus the maximum elevation of the centre of mass is
ℎ(max) = L/2 + Δh = L/2 + L/2 = L

Answer = L

Answer 2

Answer:

Hey mate

Answer :

Applying the law of conservation of energy on the given question to the initial moment and the moment when the first body reaches the height L

we obtain = mv²2/mgL

Where V is the speed of the first body at height L

Thus

V = √Vi²-2gL = √6gL-2gL = 2√gL

At this moment both bodies start to move as one piece with its centre of mass at height L/2.

According to the law of conservation of energy the new speed of the system (mass is twice the mass of one body) is given by (the process is inelastic)V' = V/2 =√gL

After this moment elevate the centre of mass of the system by the height ∆ℎ,

which is given by

∆ℎ =V'²/2g = gL/2g = L/2

Thus the maximum elevation of the centre of mass is

ℎ(max) = L/2 + Δh = L/2 + L/2 = L

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Final Correct Answer = L

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