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JEE Advanced 2024 Paper 2 · Physics · Simple Harmonic Motion · medium

JEE Advanced 2024 Paper 2 Physics: Paragraph 2

A block AA of mass 2 kg2\ \mathrm{kg} is attached to a light horizontal spring of force constant 200 N m−1200\ \mathrm{N\,m^{-1}} whose other end is fixed to a wall. The block moves on a smooth horizontal table and executes simple harmonic motion with amplitude 0.10 m0.10\ \mathrm{m}. At the instant it passes through the equilibrium position moving away from the wall, a lump of putty of mass 1 kg1\ \mathrm{kg} moving horizontally toward the wall along the line of the spring with speed 1.5 m s−11.5\ \mathrm{m\,s^{-1}} strikes the block and sticks to it. The collision is instantaneous.

Question 1

Numerical answer

The new amplitude of oscillation of the combined system, in cm, is (give the answer correct to two decimal places)
Show answer and solution

Answer: 2.04

The angular frequency of block AA alone is ω0=k/mA=200/2=10 rad s−1\omega_0=\sqrt{k/m_A}=\sqrt{200/2}=10\ \mathrm{rad\,s^{-1}}, so its speed at the equilibrium position is vA=Aω0=0.10×10=1.0 m s−1v_A=A\omega_0=0.10\times10=1.0\ \mathrm{m\,s^{-1}} away from the wall. The putty moves in the opposite direction at 1.5 m s−11.5\ \mathrm{m\,s^{-1}}. Momentum conservation along the line (taking away from the wall as positive): 2(1.0)−1(1.5)=3v′2(1.0)-1(1.5)=3v', so v′=0.53=0.1667 m s−1v'=\dfrac{0.5}{3}=0.1667\ \mathrm{m\,s^{-1}}. The collision happens at the equilibrium position, so the spring is unstretched and the new motion starts at the equilibrium with this speed as its maximum speed. The new angular frequency is ω1=200/3=8.165 rad s−1\omega_1=\sqrt{200/3}=8.165\ \mathrm{rad\,s^{-1}} and the amplitude is A′=v′ω1=0.16678.165=0.02041 m=2.04 cmA'=\dfrac{v'}{\omega_1}=\dfrac{0.1667}{8.165}=0.02041\ \mathrm{m}=2.04\ \mathrm{cm}.

Question 2

Numerical answer

The kinetic energy lost in the collision, in joule, is (give the answer correct to two decimal places)
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Answer: 2.08

Kinetic energy just before the collision is the sum for block AA and the putty: 12(2)(1.0)2+12(1)(1.5)2=1.0+1.125=2.125 J\tfrac12(2)(1.0)^2+\tfrac12(1)(1.5)^2=1.0+1.125=2.125\ \mathrm{J}. Just after the collision the combined 3 kg3\ \mathrm{kg} moves at v′=16 m s−1v'=\tfrac16\ \mathrm{m\,s^{-1}} (from momentum conservation: 2(1.0)−1(1.5)=3v′2(1.0)-1(1.5)=3v'), and the spring is unstretched at that instant, so the kinetic energy is 12(3)(16)2=124=0.0417 J\tfrac12(3)\left(\tfrac16\right)^2=\dfrac1{24}=0.0417\ \mathrm{J}; this equals the total energy 12kA′2\tfrac12kA'^2 of the new oscillation (check: 12×200×(0.0204)2=0.0417 J\tfrac12\times200\times(0.0204)^2=0.0417\ \mathrm{J}). The kinetic energy lost in the collision is 2.125−0.0417=2.0833 J≈2.08 J2.125-0.0417=2.0833\ \mathrm{J}\approx2.08\ \mathrm{J}.
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