Modelling the physical behaviour of solids, which often have complicated internal structures, can be quite challenging using conventional means alone. If we wish to simplify the problem while still taking the interactions between constituent particles into account, we may use the concept of quasiparticles, for which the energy-momentum relation may be different from the one which usually applies to real particles. When external electric or magnetic fields are applied on a solid, the motion of quasiparticles can typically be treated with the methods of classical mechanics. One type of quasiparticle of effective mass \(m\) and carrying charge \(q\) exists within some two-dimensional interface-like structure. Its motion is constrained to the \(x y\)-plane. Its kinetic energy \(K\) can be expressed in terms of the magnitude of its momentum \(p\) by the equation \[ K=\frac{p^{2}}{2 m}+\alpha p \] where \(\alpha\) is a positive constant.

1 For a real particle of mass \(m\) in free motion, its kinetic energy \(K\) can be expressed in terms of the magnitude of its momentum \(p\) by \(K=p^{2} / 2 m\). Express its velocity \(\vec{v}\) in terms of its momentum \(\vec{p}\) using the work-energy theorem.

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2 Using a similar method, express the velocity \(\vec{v}\) of a quasiparticle in terms of its momentum \(\vec{p}\).

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3 Express \(v=|\vec{v}|\) in terms of \(K\).

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4 We now place the two-dimensional interface within a uniform magnetic field of magnitude \(B\) and pointing in the \(+z\) direction. For a quasiparticle of kinetic energy \(K\), which will undergo uniform circular motion, find the radius of its trajectory, the period of its motion, and the magnitude of its angular momentum.

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5 We replace the magnetic field with a uniform electric field of magnitude \(E\) and pointing in the \(+x\) direction. Note that the component of the quasiparticle's acceleration perpendicular to the electric field may be nonzero. Find the components of the quasiparticle's acceleration \(a_{x}\) and \(a_{y}\) when it moves with speed \(v\) and its velocity makes an angle \(\theta\) with the electric field.

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