Question

An ideal spring supports a disc of mass $M$. A body of mass $m$ is related from a certain height from where it falls to his $M$. The two masses stick together at the moment they touch and move together from then on. The oscillations reach to a height a above the original level of the disc and depth $b$ below it. The constant of the force of the spring is_______

• A. $\frac{2mg}{b-a}$
• B. $\frac{mg}{b-a}$
• C. $\frac{2mg}{a-b}$
• D. $\frac{mg}{2(a-b)}$

Answer 1

The correct option is A $\frac{2mg}{b-a}$

When body fall to M the force spring experience is $(M+m)g$

The spring extension is $b-a$

The equlibrium between the force is $F_{spring}=T$

$kx=k(b-a)=(M+m)g$

$k=\frac{(m+M)g}{b-a}$

if M=m

$k=\frac{2mg}{b-a}$