InterviewSolution
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InterviewSolution
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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 `:`A. 0.7 mB. 0.2 mC. 0.5 mD. 0.6 m |
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Answer» Correct Answer - D `(1)/(2) mv^(2)[1+(K^(2))/(R^(2))]=(1)/(2)kx_("max")^(2)` Putting `m=3kg, v=4 m//s, (K^(2))/(R^(2))=1//2` (For solid cylinder), `k=200 Nm^(-1)` we get `x_("max")=0.6 m` |
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