MHT CET · Physics · Center of Mass Momentum and Collision
A solid cylinder of mass \(3 \mathrm{~kg}\) is rolling on a horizontal surface with velocity \(4 \mathrm{~m} / \mathrm{s}\). It collides with a horizontal spring whose one end is fixed to rigid support. The force constant of material of spring is \(200 \mathrm{~N} / \mathrm{m}\). The maximum compression produced in the spring will be (assume collision between cylinder & spring be elastic)
- A \(0.7 \mathrm{~m}\)
- B \(0.2 \mathrm{~m}\)
- C \(0.5 \mathrm{~m}\)
- D \(0.6 \mathrm{~m}\)
Answer & Solution
Correct Answer
(D) \(0.6 \mathrm{~m}\)
Step-by-step Solution
Detailed explanation
At maximum compression, the solid cylinder will stop.
So loss in K.E. of cylinder \(=\) Gain in P.E. of spring
\( \therefore \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I \omega}^2=\frac{1}{2} \mathrm{kx}^2 \)
\( \therefore \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \frac{\mathrm{mR}^2}{2}\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^2=\frac{1}{2} \mathrm{kx}^2 \)
\( \therefore \frac{3}{4} \mathrm{mv}^2=\frac{1}{2} \mathrm{kx}^2 \)
\( \therefore \frac{3}{4} \times 3 \times(4)^2=\frac{1}{2} \times 200 \times \mathrm{x}^2 \)
\( \therefore \frac{36}{100}=\mathrm{x}^2 \Rightarrow \mathrm{x}=0.6 \mathrm{~m} \)
So loss in K.E. of cylinder \(=\) Gain in P.E. of spring
\( \therefore \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \mathrm{I \omega}^2=\frac{1}{2} \mathrm{kx}^2 \)
\( \therefore \frac{1}{2} \mathrm{mv}^2+\frac{1}{2} \frac{\mathrm{mR}^2}{2}\left(\frac{\mathrm{v}}{\mathrm{R}}\right)^2=\frac{1}{2} \mathrm{kx}^2 \)
\( \therefore \frac{3}{4} \mathrm{mv}^2=\frac{1}{2} \mathrm{kx}^2 \)
\( \therefore \frac{3}{4} \times 3 \times(4)^2=\frac{1}{2} \times 200 \times \mathrm{x}^2 \)
\( \therefore \frac{36}{100}=\mathrm{x}^2 \Rightarrow \mathrm{x}=0.6 \mathrm{~m} \)
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