a) find the distance y from the top when the mass comes to rest for an instant, for the first time

b) what is the maximum velocity attained by the stone in this drop?

According to the figure, PE of stone = mechanical energy of string

So, we can write the following expression

mgy = ½ K(y-L)²

Where K represents the spring constant and L denotes the length of the string

Solving the above equation we get the following form:

Ky² – 2(KL + mg)y + KL² = 0

We may calculate the distance y from the top when the mass comes to a stop using the equation above.

D = b² – 4ac

Where a = K,

b = -2(KL +mg)

and c = KL²

Substituting the above values we get the following result:

$ \sqrt{D}=\sqrt{4mg(mg+2KL)}=2\sqrt{mg(mg+2KL)} $

Therefore, y can be solved as follows :

$ y=\frac{(KL+mg)\pm \sqrt{mg(2KL+mg)}}{K} $

(b)When the velocity is at its highest, the acceleration is zero, and the force is also zero. Let v be the stone’s maximum velocity. The greatest speed obtained by the stone is calculated using the law of conservation of energy. KE of the stone + PE obtained by the string = PE lost by the stone from p to Q’. Therefore, we can write :

mg (L + x) = 1/2 mv² + 1/2 Kx²

Solving the above equation, we get maximum velocity v as follows :

$v={{\left[ 2gL+\frac{m{{g}^{2}}}{K} \right]}^{\frac{1}{2}}}$