Question

A rigid cord and an elastic cord support two small spheres of same mass $M$. The are deflected from the mean position through an angle of $90$${}^o$. When the spheres pass through the mean position the lengths of the two cords become same. If $v$${}_1$ is the velocity of the sphere attached to rigid cord and $v$${}_2$ is the velocity of the other sphere, then:

• A. $v_1=v_2$
• B. $v_1$ is more than $v_2$
• C. $v_1<v_2$
• D. $v_1=v_2=0$

Answer 1

Answer: (B)

$v_1$ is more than $v_2$

By applying the law of conservation of energy

for rigid cord there is no extension in cord thus no elastic energy is stored thus all the potential energy stored is converted into kinetic energy

$\frac{1}{2}m{v_1}^2=mgl\Rightarrow v_1=\sqrt{2gl}$

But in the case of the second rod, there will be some gain in elastic energy of cord.

As a result, total potential energy will be converted into kinetic energy and elastic energy

$\frac{1}{2}m{v_2}^2+\frac{1}{2}k{\delta l}^2=mgl$

$\Rightarrow v_2=\sqrt{2gl-\frac{k{\delta l}^2}{m}}$

this shows that

$v_2<v_1$