Answers to Mechanics Practical 2

Question 1 §

Question 1a §

$$\text{LHS=}\sqrt{x^2+y^2},\text{RHS}=2\text{m}$$

$$⟹\text{LHS=}[L{]}^2+[L^2],\text{RHS}=[L{]}^2$$

$$⟹\text{LHS}=\text{RHS}\therefore \text{complete equation}$$

Question 1b §

$$\text{LHS}=\frac{v_x^2}{y},\text{RHS}=(65\text{km}\cdot {\text{h}}^{-3})t$$

$$⟹\text{LHS=}[L{]}^2[T{]}^{-2}[L{]}^{-1},\text{RHS}=[L][T{]}^{-3}[T]$$

Question 1c §

$$\text{LHS}=y^{-\frac{1}{2}}\sqrt{v_x^3},\text{RHS}=(12\text{cm})t^{\frac{3}{2}}$$

$$⟹LHS=[L][L{]}^3[T{]}^{-3},RHS=[L][T{]}^3$$

$$⟹\text{LHS}\ne \text{RHS}\therefore \text{incomplete equation}$$

Question 1d §

$$\text{LHS}=\frac{\sqrt{v_x^2}}{v_y},\text{RHS}=3$$

$$⟹\text{LHS}=[L][T{]}^{-1}[L{]}^{-1}[T],\text{RHS}=0$$

$$⟹\text{LHS}=\text{RHS}\therefore \text{complete equation}$$

Question 1e §

$$LHS=1\text{ light year},\text{RHS}=(1.17\text{ rad})x$$

$$⟹\text{LHS}=[L],\text{RHS}=[L]$$

$$⟹\text{LHS}=\text{RHS}\therefore \text{complete equation}$$

Question 1f §

$$\text{LHS}={\theta }^2{v}_x^2,\text{RHS}=\frac{y^2}{t^2}$$

$$⟹\text{LHS}=[L{]}^2[T{]}^{-2},\text{RHS}=[L{]}^2[T{]}^{-2}$$

$$⟹\text{LHS}=\text{RHS}\therefore \text{complete equation}$$

Question 2 §

Given that $v_x(t=0)=0$ and $x(t=0)=0$,

Question 2a §

$$a_x(t)=a_0$$

$$⟹v_x(t)=\int a_x(t)dt=a_{0}t+c_1:c_1=0⟹v_x(t)=a_{0}t$$

$$⟹x(t)=\int v_x(t)dt=\frac{1}{2}{a}_0{t}^2+c_2:c_2=0⟹x(t)=\frac{1}{2}{a}_0{t}^2$$

$$\therefore \text{the acceleration is constant, so it probably describes freefall or something similar.}$$

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y=2

Question 2b §

$$a_x(t)=\gamma t$$

$$⟹v_x(t)=\int a_x(t)dt=\frac{1}{2}\gamma t^2+c_1:c_1=0⟹v_x(t)=\frac{1}{2}\gamma t^2$$

$$⟹x(t)=\int v_x(t)dt=\frac{1}{6}\gamma t^3+c_2:c_2=0⟹x(t)=\frac{1}{6}\gamma t^3$$

$$\therefore \text{the acceleration is linearly increasing, but I don’t know what this means.}$$

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y=2x

Question 2c (integrated incorrectly, tau should be multiplied because it’s dividing a quotient) §

$$a_x(t)=\frac{v_{\infty }}{\tau }{e}^{-\frac{t}{\tau }}$$

$$⟹v_x(t)=\int a_x(t)dt=-\frac{v_{\infty }}{{\tau }^2}{e}^{-\frac{t}{\tau }}+c_1:c_1=\frac{v_{\infty }}{\tau }⟹v_x(t)=\frac{v_{\infty }}{\tau }-\frac{v_{\infty }}{\tau }{e}^{-\frac{t}{\tau }}$$

$$⟹x(t)=\int v_x(t)dt=\frac{v_{\infty }}{\tau }t+\frac{v_{\infty }}{{\tau }^2}{e}^{-\frac{t}{\tau }}+c_2:c_2=-\frac{v_{\infty }}{\tau }$$

$$⟹x(t)=\frac{v_{\infty }}{\tau }t+\frac{v_{\infty }}{{\tau }^2}{e}^{-\frac{t}{\tau }}-\frac{v_{\infty }}{\tau }$$

$$\therefore \text{I have no idea what physical conditions this could be under.}$$

Question 2d §

Note that this follows a different, safer method

$$a_x(t)=a_0\cos (\omega t)$$

$$⟹v_x(t)=v_x(0)+{\int }_0^t{a}_x(t^{\prime })d{t}^{\prime }={\left[\frac{a_0}{\omega }\sin (\omega t^{\prime })\right]}_0^t=\frac{a_0}{\omega }\sin (\omega t)$$

$$⟹x(t)=x(0)+{\int }_0^t{v}_x(t)dt={\left[-\frac{a_0}{{\omega }^2}\cos (\omega t^{\prime })\right]}_0^t=\frac{a_0}{{\omega }^2}-\frac{a_0}{{\omega }^2}\cos (\omega t)$$

$$\therefore \text{I have no idea what physical conditions this could be under.}$$

Question 3 §

Question 3a §

$$\vec{r}(t)=\left[(10\text{m}\cdot {\text{s}}^{-1})t\right]\hat{\text{i}}+\left[(-2.5\text{m}\cdot {\text{s}}^{-2})t^2\right]\hat{\text{j}}$$

$$⟹\vec{v}(t)=\left[10\text{m}\cdot {\text{s}}^{-1}\right]\hat{\text{i}}+\left[(-5\text{m}\cdot {\text{s}}^{-2})t\right]\hat{\text{j}}$$

$$⟹\vec{a}(t)=\left[0\text{m}\cdot {\text{s}}^{-1}\right]\hat{\text{i}}+\left[-5\text{m}\cdot {\text{s}}^{-2}\right]\hat{\text{j}}$$

Question 3b §

$$\vec{r}(t)=\left[R\cos (\omega t)\right]\hat{\text{i}}+\left[R\sin (\omega t)\right]\hat{\text{j}}$$

$$⟹\vec{v}(t)=\left[-R\omega \sin (\omega t)\right]\hat{\text{i}}+\left[R\omega \cos (\omega t)\right]\hat{\text{j}}$$

$$⟹\vec{a}(t)=\left[-R{\omega }^2\cos (\omega t)\right]\hat{\text{i}}+\left[-R{\omega }^2\sin (\omega t)\right]\hat{\text{j}}$$

Question 3c §

$$\vec{r}(t)=\left[R(\omega t-\sin (\omega t))\right]\hat{\text{i}}+\left[R(\omega t-\cos (\omega t))\right]\hat{\text{j}}$$

$$⟹\vec{v}(t)=\left[R(\omega -\omega \cos (\omega t))\right]\hat{\text{i}}+\left[R(\omega +\omega \sin (\omega t))\right]\hat{\text{j}}$$

$$⟹\vec{a}(t)=\left[{\omega }^{2}R\sin (\omega t)\right]\hat{\text{i}}+\left[{\omega }^{2}R\cos (\omega t)\right]\hat{\text{j}}$$

Question 4 §

Question 4a §

$$v_{avg}=\frac{\Delta s}{\Delta t}=\frac{s_f-s_i}{t_f-t_i}$$

Question 4b §

Given the aforementioned definition that $v_{avg}=\frac{\Delta s}{\Delta t}$, we can say that:

$$v_{x,avg}=\frac{x_2-x_1}{t_2-t_1}$$

$$⟹v_{x,avg}=\frac{1}{t_2-t_1}\cdot x_{avg}$$

$$⟹v_{x,avg}=\frac{1}{t_2-t_1}\cdot {\int }_{t_1}^{t_2}{v}_x(t)dt$$

Question 4c §

Given that acceleration is constant, we can say that

$$v_x(t)=v_{x0}+a(t-t_0)$$

Thus we can calculate the velocity at $t_1$ and $t_2$:

$$(1)v_x(t_1)=v_{x0}+a(t_1-t_0)$$

$$(2)v_x(t_2)=v_{x0}+a(t_2-t_0)$$

Now, integrating $v_x(t)$ over $[t_1,t_2]$:

$${\int }_{t_1}^{t^2}{v}_x(t)dt={\int }_{t_1}^{t^2}\left(v_{x0}+a(t-t_0)\right)dt$$

$$⟹{\int }_{t_1}^{t^2}{v}_x(t)dt=v_{x0}(t_2-t_1)+\frac{1}{2}a((t_2-t_0{)}^2-(t_1-t_0{)}^2)$$

Then we can simplify the equation by using the equations $(1),(2)$ from earlier:

$$⟹{\int }_{t_1}^{t^2}{v}_x(t)dt=\frac{1}{2}(v_x(t_1)+v_x(t_2))(t_2-t_1)$$

$$⟹\frac{1}{t_2-t_1}{\int }_{t_1}^{t^2}{v}_x(t)dt=\frac{1}{2}(v_x(t_1)+v_x(t_2))$$

Then by using the equation derived in part (b):

$$⟹v_{x,avg}=\frac{1}{2}(v_x(t_1)+v_x(t_2))\square$$