Work Energy and Power - Result Question 52
55. A solid cylinder of mass $3 kg$ is rolling on a horizontal surface with velocity $4 ms^{-1}$. It collides with a horizontal spring of force constant $200 Nm^{-1}$. The maximum compression produced in the spring will be :
[2012]
(a) $0.5 m$
(b) $0.6 m$
(c) $0.7 m$
(d) $0.2 m$
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Answer:
Correct Answer: 55. (b)
Solution:
- (b) At maximum compression the solid cylinder will stop so loss in K.E. of cylinder = gain in P.E. of spring
$\Rightarrow \frac{1}{2} m v^{2}+\frac{1}{2} I \omega^{2}=\frac{1}{2} k x^{2}$ $\Rightarrow \frac{1}{2} m v^{2}+\frac{1}{2} \frac{m R^{2}}{2}(\frac{v}{R})^{2}=\frac{1}{2} k x^{2}$ $\Rightarrow \quad \frac{3}{4} m v^{2}=\frac{1}{2} k x^{2}$
$\Rightarrow \quad \frac{3}{4} \times 3 \times(4)^{2}=\frac{1}{2} \times 200 x^{2}$
$\Rightarrow \frac{36}{100}=x^{2} \Rightarrow x=0.6 m$
