An object of mass 4 kg is attached to a spring having spring constant `100(N)/(m)`. It performs simple harmonic motion on a smooth horizontal surface with an amplitude of 2 m. A 6 kg object is dropped vertically onto the 4 kg object when it crosses the mean position, and sticks to it. the change in amplitude of oscillation due to collision isA. `1m`B. `zero`C. `2[1-sqrt((2)/(5))]`D. `2[1-(1)/(sqrt5)]`

An object of mass 4 kg is attached to a spring having spring constant

1.	An object of mass 4 kg is attached to a spring having spring constant `100(N)/(m)`. It performs simple harmonic motion on a smooth horizontal surface with an amplitude of 2 m. A 6 kg object is dropped vertically onto the 4 kg object when it crosses the mean position, and sticks to it. the change in amplitude of oscillation due to collision isA. `1m`B. `zero`C. `2[1-sqrt((2)/(5))]`D. `2[1-(1)/(sqrt5)]`
Answer» Correct Answer - C Time period of the system (object of mass 4 kg) before collision is `T_1=2pisqrt((4)/(100))` After collision, time period of the combined mass is `T_2=2pisqrt((10)/(100))` We can apply momentum conservation for just before the collision and just after the collision in the horizontal direction. `4A_1omega_1=10A_2omega_2impliesA(4xx2xxsqrt((100)/(4)))/(10xxsqrt((100)/(10)))=2sqrt((2)/(5))`m So change in amplitude, `triangleA=A_1-A_2=2[1-sqrt((2)/(5))]`m

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