JEE Main 2020 · Physics · Simple Harmonic Motion · medium

A block of mass 3 kg on a smooth horizontal table is attached to a spring of constant 300 N m^-1 and oscillates with amplitude 0.2 m. As it passes…

Numerical answer

A block of mass

3\ \mathrm{kg}

on a smooth horizontal table is attached to a spring of constant

300\ \mathrm{N\,m^{-1}}

and oscillates with amplitude

0.2\ \mathrm{m}

. As it passes through the equilibrium position, a lump of putty of mass

1\ \mathrm{kg}

drops vertically on it and sticks. The total energy of the oscillation after this (in joules) is

Show answer and solution

Answer: 4.5

Before the collision:

\omega = \sqrt{300/3} = 10\ \mathrm{rad\,s^{-1}}

and speed at equilibrium

v = A\omega = 2\ \mathrm{m\,s^{-1}}

. The putty has no horizontal velocity, so horizontal momentum is conserved:

3(2) = 4v'

giving

v' = 1.5\ \mathrm{m\,s^{-1}}

. At the equilibrium position all energy is kinetic, so the new total energy is

\tfrac12(4)(1.5)^2 = 4.5\ \mathrm{J}

(originally

6\ \mathrm{J}

). The new amplitude is

v'/\omega' = 1.5/\sqrt{75} \approx 0.17\ \mathrm{m}

, consistent with

\tfrac12 kA'^2 = 4.5\ \mathrm{J}

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