Question

1. a block of massm collides perfectly inelastically with another identical block connected to a spring of force constant k. the amplitude of resulting shm will be 1) \\( \sqrt { \frac { m } { k } } \\) 2) \\( v \sqrt { \frac { m } { 2 k } } \\) 3) \\( v \sqrt { \frac { k } { m } } \\) 4) \\( v \sqrt { \frac { 2 k } { m } } \\)

Explanation:

Step1: Apply conservation of momentum

Before collision, momentum is \(p = mv\). After collision (perfectly inelastic), mass becomes \(M = m + m=2m\). Let velocity after collision be \(v'\). By conservation of momentum \(mv=(m + m)v'\), so \(v'=\frac{v}{2}\).

Step2: Apply conservation of mechanical energy

The kinetic energy of the combined mass just after collision is converted into potential energy of the spring at maximum compression (amplitude \(A\)). Kinetic energy \(K.E.=\frac{1}{2}(2m)v'^{2}\) and potential energy \(P.E.=\frac{1}{2}KA^{2}\). Substitute \(v'=\frac{v}{2}\) into \(K.E.\): \(\frac{1}{2}(2m)(\frac{v}{2})^{2}=\frac{1}{2}KA^{2}\).
Simplify the left - hand side: \(\frac{1}{2}(2m)\frac{v^{2}}{4}=\frac{mv^{2}}{4}\). Then from \(\frac{mv^{2}}{4}=\frac{1}{2}KA^{2}\), solve for \(A\).
Cross - multiply: \(mv^{2}=2KA^{2}\), so \(A = v\sqrt{\frac{m}{2K}}\).

Answer:

1. \(V\sqrt{\frac{m}{2K}}\)