Question

A solid sphere of mass $2 \,kg$ is rolling on a frictionless horizontal surface with velocity $6 \,m / s$. 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 $=36 \,N / m )$

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

Answer 1

Answer: (B)

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