Question

A small empty bucket of mass M is attached to a long inextensible cord of length l. The bucket is released from rest when the cord is in a horizontal position. In its lowest position the bucket scoops up a mass m of water, what is the height of the swing above the lowest position?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Let the bucket of mass reach the lowest postition with a speed 'V' . As it scoops up water of mass m, at the lowest postition , its speed decrease to V'. The speed of bucket just after scooping of water is given by conserving momentum of the system just before after scooping of water is given by conserving momentum of the system just before and after scooping of water we obtain.

MC = (M + M)V' $\Rightarrow$ v ' = $\frac{MV}{M=m}$ ... (a)

Coservation of energy of the bucket between A and B (Just scooping ) yields, $\frac{1}{2}$ m $v^2$ - mgl = 0

$\Rightarrow$ v = $\sqrt{2gl}$ ..(b)

Coservation of energy of the bucket + water between B and C (just after scooping) yields,

- $\frac{1}{2}$ m ${v^{\prime }}^2$ - mgh = 0 $\Rightarrow$ h $\frac{{v^{\prime }}^2}{2g}$ ......(c)

By using (a), (b), (c) , we obtain h = $\frac{M^2}{{(M+m)}^2}$ l