202104150814 Homework 1 (Q2)

A particle is thrown with speed v_0 and an elevated angle \theta on the floor. The air resistance is negligible.

(a) During the flight, the following quantities are investigated. Determine whether the following items are constants.

i. \displaystyle{\frac{\mathrm{d}v}{\mathrm{d}t}}, where v is the speed of the particle.

ii. \displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}, where \mathbf{v} is the velocity of the particle.

(b) What is the radius of curvature of the path when the particle reaches the highest point?


Answer.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

(a)

i. \displaystyle{\frac{\mathrm{d}v}{\mathrm{d}t}} varies with time.

ii. \displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=\mathbf{a}=\mathbf{g}=\textrm{Const.}


Explanation.

\begin{aligned} \mathbf{v} & =v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}} \\ & = v_0\cos\theta\,\hat{\mathbf{i}}+\big(v_0\sin\theta -gt\big)\,\hat{\mathbf{j}}\\ v & = |\mathbf{v}| \\ & = \sqrt{(v_0\cos\theta )^2+(v_0\sin\theta -gt)^2} \\ & = \sqrt{v_0^2-2gtv_0\sin\theta +g^2t^2}\\ \frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} & = \bigg( \frac{\mathrm{d}}{\mathrm{d}t}(v_0\cos\theta ) \bigg)\,\hat{\mathbf{i}} + \bigg(\frac{\mathrm{d}}{\mathrm{d}t}(v_0\sin\theta -gt)\bigg)\,\hat{\mathbf{j}}\\ & = -g\,\hat{\mathbf{j}}\\ & = \mathbf{g} \\ \frac{\mathrm{d}v}{\mathrm{d}t} & = \bigg( \sqrt{v_0^2-2gtv_0\sin\theta +g^2t^2}\bigg)'\\ & = \frac{1}{2}\Big( \sqrt{v_0^2-2gtv_0\sin\theta +g^2t^2}\Big)^{-1} \cdot (-2gv_0\sin\theta + 2g^2t)\\ & \propto t \end{aligned}


(b)

\begin{aligned} a=\frac{v^2}{r} & \Rightarrow g=\frac{v_0^2\cos^2\theta}{r} \\ & \Rightarrow r=\frac{v_0^2\cos^2\theta}{g} \end{aligned}

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