Question

As shown in the figure, if a solid sphere of mass $M$ rolling with a speed $v$ on a horizontal surface strikes a spring of force constant $k$, then the maximum compression of the spring is 

• A. $\sqrt{\frac{5Mv^2}{3k}}$
• B. $\sqrt{\frac{7Mv^2}{5k}}$
• C. $\sqrt{\frac{Mv^2}{k}}$
• D. $\sqrt{\frac{3Mv^2}{2k}}$

Hint:

When a rolling object interacts with a spring, always account for both translational and rotational kinetic energy to find the total energy converted into the spring's potential energy.

Answer 1

The Correct Option is B

The solid sphere is rolling, so it has both translational and rotational kinetic energy. 
The translational kinetic energy is $\frac{1}{2}Mv^2$, and the rotational kinetic energy is $\frac{1}{2}I\omega^2$. 
For a solid sphere, the moment of inertia $I = \frac{2}{5}MR^2$, and since it rolls without slipping, $v = R\omega$, so $\omega = \frac{v}{R}$. 
Thus, the rotational kinetic energy is $\frac{1}{2} \left(\frac{2}{5}MR^2\right) \left(\frac{v}{R}\right)^2 = \frac{1}{5}Mv^2$. 
The total kinetic energy of the sphere is: \[ \frac{1}{2}Mv^2 + \frac{1}{5}Mv^2 = \frac{5}{10}Mv^2 + \frac{2}{10}Mv^2 = \frac{7}{10}Mv^2 \] When the sphere strikes the spring, this total kinetic energy is converted into the potential energy of the spring at maximum compression $x$, given by $\frac{1}{2}kx^2$. By conservation of energy: \[ \frac{7}{10}Mv^2 = \frac{1}{2}kx^2 \] Solving for $x$: \[ x^2 = \frac{\frac{7}{10}Mv^2}{\frac{1}{2}k} = \frac{7Mv^2}{5k} \] \[ x = \sqrt{\frac{7Mv^2}{5k}} \] Thus, the maximum compression of the spring is $\sqrt{\frac{7Mv^2}{5k}}$.