Question

The left block in figure collides inelastically with the right block and sticks to it. The amplitude of the resulting simple harmonic motion is

• A. $\sqrt{\frac{m}{2k}}v$
• B. $\sqrt{\frac{m}{k}}v$
• C. $\sqrt{\frac{2m}{k}}v$
• D. $\sqrt{\frac{m}{4k}}v$

Answer 1

The correct option is A $\sqrt{\frac{m}{2k}}v$

Assuming the collision to last for a small interval only we can apply the principle of conservation of momentum The common velocity after the collision is $\frac{v}{2}$ The kinetic energy = $\frac{1}{2}\left(2m\right){\left(\frac{v}{2}\right)}^2=\frac{1}{4}m{v}^2$ This is also the total energy of vibration as the spring is unstretched at this moment If the amplitude is A the total energy can also be written as $\frac{1}{2}k{A}^2$ Thus $\frac{1}{2}k{A}^2=\frac{1}{4}m{v}^2$ giving $A=\sqrt{\frac{m}{2k}}v$