Question

A solid sphere of mass $2kg$ is rolling on a frictionless horizontal surface with velocity $6m/s$. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring $=36N/m)$

• A. $\sqrt{14}m$
• B. $\sqrt{2.8}m$
• C. $\sqrt{1.4}m$
• D. $\sqrt{0.7}m$

Answer 1

Answer: (B)

Given : $m=2kgv=6m/s$
Total kinetic energy of the sphere initially K.E
$=\frac{1}{2}m{v}^2+\frac{1}{2}I{w}^2$
Or K.E $=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{2}{5}m{r}^2{w}^2$
where $v=rw$ (pure rollinng)
$\therefore$ K.E $=\frac{1}{2}m{v}^2+\frac{1}{5}m{v}^2=\frac{7}{10}m{v}^2$
Let the compression in the spring be $x$.
Thus potential energy in the spring $P.E=\frac{1}{2}k{x}^2$
Potential energy stored in the spring is equal to the kinetic energy of the sphere i.e. $P.E=K.E$
$\therefore \frac{1}{2}k{x}^2=\frac{7}{10}m{v}^2$
Or $\frac{1}{2}(36)x^2=\frac{7}{10}(2)(6{)}^2$
$\Rightarrow x=\sqrt{2.8}m$