A solid cylinder of mass 3 kg is rolling on a horizontal surface with velocity 4ms-1. It collides with a horizontal spring of force constant 200 Nm-1. The maximum compression produced in the spring will be:

  • Option 1)

    0.5 m

  • Option 2)

    0.6 m

  • Option 3)

    0.7 m

  • Option 4)

    0.2 m

 

Answers (1)

As we discussed in concept

\frac{1}{2}mv^2 +\frac{1}{2}lw^2=\frac{1}{2}Kx^2

\Rightarrow \frac{1}{2}mv^2 +\frac{1}{2}\frac{mR^2}{2}\left ( \frac{V}{R} \right )^2 =\frac{1}{2}kx^2

\frac{3}{4}mv^2=\frac{1}{2}km^2\Rightarrow \frac{3}{4}\times 3\times 4^2 =\frac{1}{2}\times 200\times x^2

\frac{36}{100}=x^2 =x=\frac{6}{10} =x=0.6m


Option 1)

0.5 m

Incorrect

Option 2)

0.6 m

Correct

Option 3)

0.7 m

Incorrect

Option 4)

0.2 m

Incorrect

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