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A body of mass 1kg falls freely from a height of 100m, on a platform of mass 3kg which is mounted on a spring having spring constant k = 1.25 X 10⁶N/m. The body sticks to the platform and the spring's maximum compression is found to be x. Given that g = 10ms^-2, the value of x will be close to:

Option 1)
8 cm
Option 2)
80 cm
Option 3)
40 cm
Option 4)
2 cm

Answers (1)

best_answer

Net work done by all the forces give the change in kinetic energy -

$W=\frac{1}{2}mv^{2}-\frac{1}{2}mv{_{0}}^{2}$

$W= k_{f}-k_{i}$

- wherein

$m=mass \: of\: the\: body$

$v_{0}= initial\: velocity$

$v= final\: velocity$

If only conservative forces act on a system, total mechnical energy remains constant -

$K+U=E\left ( constant \right )$

$\Delta K+\Delta U=0$

$\Delta K=-\Delta U$

-

Before collision After Collision

$4ks \downarrow V$

Apply momentum conservation

$-1v_{1}+o=-4\times v$

$v=\frac{v_{1}}{4}-(1)$

$v_{1}=\sqrt{2gh}=\sqrt{2\times 10\times 100}=\sqrt{2000}$

Now combine system will start compression spring and perform S.H.M

V is maximum velocity for this SH $V=A\omega ............(2)$

$\omega =\sqrt{\frac{k}{m_{net}}}=\sqrt{\frac{1.25\times 10^{6}}{4}}------(3)$

$A=\frac{V}{\omega }=\frac{v_{1}}{4\omega }=\left ( \sqrt{\frac{2000\times 4}{1.25\times 10^{6}}} \right )\times \frac{1}{4}$

$A=\left ( \sqrt{\frac{2\times 10\times 100\times 4\times 10^{2}}{125\times 10^{6}}} \right )\times \frac{1}{4}\Rightarrow$

A= 2 cm

This is not given in the question.

Option 1)

8 cm

Option 2)

80 cm

Option 3)

40 cm

Option 4)
2 cm

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