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Figure shows a block of mass m = 2 kg resting on a smooth horizontal ground attached to one end of a spring of force constant k =100 N/m in natural length. If another block of same mass and moving with a velocity u = 5 m/s toward right is placed on the block which stick to it due to friction. Find the amplitude of oscillations (in m) of the combined mass 2m.

Moderate

Solution

When second mass sticks to the lower mass, due to such an inelastic collision, the velocity of combined block is reduced by half that is u/2 to conserve momentum. Now at mean position the velocity of block can be written as

$\frac{\mathrm{u}}{2}=\mathrm{A\omega}$ [If A is the amplitude of oscillation]

or $\frac{\mathrm{u}}{2}=\mathrm{A}\sqrt{\frac{\mathrm{k}}{2\mathrm{m}}}$ [As here for combined block new angular frequency of SHM is $\mathrm{\omega}=\sqrt{\frac{\mathrm{k}}{2\mathrm{m}}}$]

or $\mathrm{A}=\frac{\mathrm{u}}{2}\sqrt{\frac{2\mathrm{m}}{\mathrm{k}}}=\mathrm{u}\sqrt{\frac{\mathrm{m}}{2\mathrm{k}}}=5\sqrt{\frac{2}{2\times 100}}=0.50\mathrm{m}$

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