Answers to Mechanics Practical 3

Question 1 §

$$\vec{a}(t)=a_x\hat{i}+a_y\hat{j}$$

$$\vec{v}(0)=v_{x,0}\hat{i}+v_{y,0}\hat{j}$$

$$\vec{r}(0)=x_0\hat{i}+y_0\hat{j}$$

Question 1a §

$$\vec{v}(t)=\vec{v}(0)+{\int }_0^t{a}_x(t^{\prime })d{t}^{\prime }$$

$$⟹\vec{v}(t)=v_{x,0}\hat{i}+v_{y,0}\hat{j}+{\left[a_x{t}^{\prime }\hat{i}+a_y{t}^{\prime }\hat{j}\right]}_0^t$$

$$⟹\vec{v}(t)=v_{x,0}\hat{i}+v_{y,0}\hat{j}+a_{x}t\hat{i}+a_{y}t\hat{j}$$

$$⟹\vec{v}(t)=(v_{x,0}+a_{x}t)\hat{i}+(v_{y,0}+a_{y}t)\hat{j}$$

Question 1b §

$$\vec{r}(t)=\vec{r}(0)+{\int }_0^t{v}_x(t^{\prime })d{t}^{\prime }$$

$$⟹\vec{r}(t)=x_0\hat{i}+y_0\hat{j}+{\left[(v_{x,0}+a_x{t}^{\prime })\hat{i}+(v_{y,0}+a_y{t}^{\prime })\hat{j}\right]}_0^t$$

$$⟹\vec{r}(t)=x_0\hat{i}+y_0\hat{j}+(a_{x}t)\hat{i}+(a_{y}t)\hat{j}$$

$$⟹\vec{r}(t)=(x_0+a_{x}t)\hat{i}+(y_0+a_{y}t)\hat{j}$$

Question 2 §

$$\vec{v}=(\sqrt{3}\text{m}\cdot {\text{s}}^{-1})\hat{i}+(-1\text{m}\cdot {\text{s}}^{-1})\hat{j}$$

Map this to a new frame tilted by an angle $\alpha =30^{\circ }$ such that $\hat{i}\mapsto \cos \alpha \hat{i}-\sin \alpha \hat{j}$ and $\hat{j}\mapsto \sin \alpha \hat{i}+\cos \alpha \hat{j}$.

$$\cos 30^{\circ }=\frac{\sqrt{3}}{2},\sin 30^{\circ }=\frac{1}{2}$$

Using the relationships between orthonormal vectors,

$$\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}={\hat{i}}^{\prime }\cdot {\hat{j}}^{\prime }={\hat{j}}^{\prime }\cdot {\hat{i}}^{\prime }=0$$

$$\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}={\hat{i}}^{\prime }\cdot {\hat{i}}^{\prime }={\hat{j}}^{\prime }\cdot {\hat{j}}^{\prime }=1$$

Question 2a §

Change of basis from $R\to R^{\prime }$

$$\vec{v}\cdot {\hat{i}}^{\prime }={\vec{v}}_i^{\prime }$$

$$⟹{\vec{v}}_i^{\prime }=\sqrt{3}\cdot \frac{\sqrt{3}}{2}-(-1)\frac{1}{2}$$

$$⟹{\vec{v}}_i^{\prime }=\frac{3}{2}+\frac{1}{2}=2\text{m}\cdot {\text{s}}^{-1}$$

Question 2b §

Change of basis from $R\to R^{\prime }$

$$\vec{v}\cdot {\hat{j}}^{\prime }={\vec{v}}_j^{\prime }$$

$$⟹{\vec{v}}_j^{\prime }=\frac{1}{2}\cdot \sqrt{3}+\frac{\sqrt{3}}{2}(-1)$$

$$⟹{\vec{v}}_j^{\prime }=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=0\text{m}\cdot {\text{s}}^{-1}$$

Question 2c §

The tilted frame $R^{\prime }$ is more suitable as it removes one component, the $j$ component.

Question 3 §

$$\text{Law of composition of velocities}:\vec{v}(A|B)=\vec{v}(A|M)+\vec{v}(M|B)$$

Given a basis $(\hat{i},\hat{j})$, Bob has velocity $\vec{v}(B|M)=(3\text{m}\cdot {\text{s}}^{-1})\hat{i}$ relative to Matt with velocity $0$. Tim has velocity $\vec{v}(T|M)=-(12\text{m}\cdot s^{-1})\sin [6{\text{s}}^{-1}t]\hat{i}+(12\text{m}\cdot {\text{s}}^{-1})\cos [(6{\text{s}}^{-1})t]\hat{j}$ relative to Matt with velocity $0$.

Question 3a §

By using the above law of composition of velocities, determine the relative velocities of Matt and Tim relative to Bob.

$$\vec{v}(M|B)=-\vec{v}(B|M)=(-3\text{m}\cdot {\text{s}}^{-1})\hat{i},$$

$$\vec{v}(T|B)=\vec{v}(T|M)+\vec{v}(M|B)$$

$$⟹\vec{v}(T|B)=-(12\text{m}\cdot s^{-1})\sin [6{\text{s}}^{-1}t]\hat{i}+(12\text{m}\cdot {\text{s}}^{-1})\cos [(6{\text{s}}^{-1})t]\hat{j}+(-3\text{m}\cdot {\text{s}}^{-1})\hat{i}$$

$$⟹\vec{v}(T|B)=-(3\text{m}\cdot {\text{s}}^{-1}+(12\text{m}\cdot s^{-1})\sin [(6{\text{s}}^{-1})t])\hat{i}+(12\text{m}\cdot {\text{s}}^{-1})\cos [(6{\text{s}}^{-1})t]\hat{j}$$

Question 3b §

By using the above law of composition of velocities, determine the relative velocities of Matt and Bob relative to Tim.

$$\vec{v}(M|T)=-\vec{v}(T|M)=(12\text{m}\cdot s^{-1})\sin [6{\text{s}}^{-1}t]\hat{i}-(12\text{m}\cdot {\text{s}}^{-1})\cos [(6{\text{s}}^{-1})t]\hat{j}$$

$$\vec{v}(B|T)=\vec{v}(B|M)+\vec{v}(M|T)$$

$$⟹\vec{v}(B|T)=(3\text{m}\cdot {\text{s}}^{-1})\hat{i}+(12\text{m}\cdot s^{-1})\sin [6{\text{s}}^{-1}t]\hat{i}-(12\text{m}\cdot {\text{s}}^{-1})\cos [(6{\text{s}}^{-1})t]\hat{j}$$

$$⟹\vec{v}(B|T)=(3\text{m}\cdot {\text{s}}^{-1}+12\text{m}\cdot s^{-1})\sin [6{\text{s}}^{-1}t]\hat{i}-(12\text{m}\cdot {\text{s}}^{-1})\cos [(6{\text{s}}^{-1})t]\hat{j}$$

Question 4 §

Two blocks $B_1$ and $B_2$ with respective masses $m_1=5\text{ kg}$ and $m_2=20\text{ kg}$ are being pushed by a constant force \vec{F}=F_\hat{i} applied on $B_1$. The two blocks moves with the same acceleration $\vec{a}=1\text{ m}\cdot {\text{s}}^{-2}\hat{i}$ as see from a Galilean frame $(O,\hat{i},\hat{j})$. As shown below, Answers to Mechanics Practical 3 2023-11-03 16.48.56.excalidraw

Question 4a §

By considering $B=B_1+B_2$, we can determine the force \vec{F}=F_\hat{i}.

$$\Sigma F=ma$$

$$⟹\vec{F}=(m_1+m_2)\cdot \vec{a}$$

$$⟹\vec{F}=(25\text{ N})\hat{i}$$

Question 4b §

By considering $B_1$, we can determine the force ${\vec{F}}_{B_2\to B_1}$ exerted by $B_2$ on $B_1$.

\vec{F}_\hat{i}-\vec{F}_{B_2\to B_1}=m_1\cdot\vec{a}

$$⟹(25\text{ N})\hat{i}-{\vec{F}}_{B_2\to B_1}=5\hat{i}$$

$$\therefore {\vec{F}}_{B_2\to B_1}=(20\text{ N})\hat{i}$$

Question 4c §

By Newton’ third law, we can determine the force ${\vec{F}}_{B_1\to B_2}$ exerted by $B_1$ on $B_2$.

$${\vec{F}}_{B_1\to B_2}=-{\vec{F}}_{B_2\to B_1}$$

$$\therefore {\vec{F}}_{B_1\to B_2}=-(20\text{ N})\hat{i}$$

Etc. §