Answers to Mechanics Practical 1

Question 1 §

$$x=3\text{ inches}=\frac{3}{39.37}\text{ metres}=0.0762\text{ metres}$$

$$t=5\text{ days}=(5\cdot 86400)\text{ seconds}=432000\text{ seconds}$$

$$v=30\text{km}\cdot {\text{h}}^{-1}=\frac{30}{3.6}{\text{ms}}^{-1}=8.\dot{3}{\text{ ms}}^{-1}$$

$$j=72{\text{mm}}^3\cdot {\text{ms}}^{-1}=\frac{72}{10^9}{\text{m}}^3\cdot {\text{ms}}^{-1}=(7.2\cdot 10^{-8}){\text{m}}^4\cdot {\text{s}}^{-1}$$

$$f=21{\text{ week}}^{-1}=\frac{21}{604800}{\text{s}}^{-1}=(3.47\dot{2}\cdot 10^{-5}){\text{s}}^{-1}$$

$$d=1.4\text{ light year}\approx 1.4\cdot (9.461\cdot 10^{15})\approx (1.325\cdot 10^{16})\text{m}$$

Question 2 §

Question 2a §

$$LHS=x^3,RHS=(2{\text{s}}^{-1})t^2$$

$$⟹LHS={\text{m}}^3,RHS={\text{s}}^{-1}\cdot {\text{s}}^2=s$$

$$\therefore LHS\ne RHS\text{, Equation (a) is not complete}$$

Question 2b §

$$LHS=(1{\text{m}}^{-2}\cdot \text{s})x^2,RHS=t$$

$$⟹LHS=\text{s},RHS=\text{s}$$

$$\therefore LHS=RHS\text{, Equation (b) is complete}$$

Question 2c §

$$LHS=\sqrt{t},RHS=(6{\text{mm}}^{-\frac{3}{2}}\cdot {\text{s}}^{-\frac{1}{2}})x^{\frac{3}{2}}$$

$$⟹LHS={\text{s}}^{\frac{1}{2}},RHS={\text{s}}^{-\frac{1}{2}}$$

$$\therefore LHS\ne RHS\text{, Equation (c) is not complete}$$

Question 2d §

$$LHS=(3{\text{s}}^{-2})x^2+(20{\text{m}}^{-3}\cdot {\text{s}}^{-2})x^5,RHS=(x/t{)}^2$$

$$⟹LHS={\text{s}}^{-2}{\text{m}}^2,RHS={\text{s}}^{-2}{\text{m}}^2$$

$$\therefore LHS=RHS\text{, Equation (d) is complete}$$

Question 2e §

$$LHS=x,RHS=3$$

$$⟹LHS=\text{m},RHS=\text{dimensionless}$$

$$\therefore LHS\ne RHS\text{, Equation (e) is not complete}$$

Question 2f §

$$LHS=1,RHS=\cos (\frac{t}{x}(12\text{m}\cdot {\text{s}}^{-1}))$$

$$⟹LHS=\text{dimensionless},RHS=\text{dimensionless}$$

$$\therefore LHS=RHS\text{, Equation (e) is complete}$$

Question 3 §

$$x=5.6\cdot 10^{10}\text{m},t=1.296\cdot 10^7\text{s},\text{at a constant velocity}$$

$$⟹v=\frac{x}{t}=\frac{5.6\cdot 10^{10}\text{m}}{1.296\cdot 10^7\text{s}}=\frac{350}{81}\cdot 10^3{\text{ms}}^{-1}$$

$$⟹\frac{350}{81}\cdot 10^3{\text{ms}}^{-1}\approx 4.32{\text{ms}}^{-1}$$

Question 4 §

$$v=\frac{125}{9}\text{m}\cdot {\text{s}}^{-1},x_0=100\text{m}$$

Question 4a §

$$t=2\text{s}:x_1=v\cdot t=\frac{125}{9}{\text{ms}}^{-1}\cdot 2\text{s}$$

$$⟹x_1=\frac{250}{9}\text{m}=27.\dot{7}\text{m}$$

Question 4b §

$$v=0{\text{ms}}^{-1},u=\frac{125}{9}{\text{ms}}^{-1},a=a_x{\text{ms}}^{-2},s<100\text{m}$$

$$v^2=u^2+2as⟹0{\text{m}}^2{\text{s}}^2=\frac{15625}{81}{\text{m}}^2{\text{s}}^{-2}+2(a_x\text{m}{\text{s}}^{-2})(100\text{m})$$

$$⟹a_x{\text{ms}}^{-2}=\frac{-\frac{15625}{81}{\text{m}}^2{\text{s}}^{-2}}{200\text{m}}$$

$$⟹a_x{\text{ms}}^{-2}=-\frac{625}{648}{\text{ms}}^{-2}$$

$$\therefore a_x\approx <-0.96{\text{ms}}^{-2}$$

Question 5 §

$$a=a_x$$

$$x(t)=x_0+\frac{1}{2}(v_x(t)+v_0)t$$

$$v_x(t)=v_0+a_{x}t$$

Question 5a §

Derive the following by using the above equations:

$$x(t)=x_0+v_0+\frac{1}{2a_x{t}^2}$$

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Question 5b §

Derive the following by using the above equations:

$$v_x(t{)}^2-v_0^2=2a_x(x(t)-x_0)$$

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Question 5c §

Derive the following by using the above equations:

$$x(t)=x_0+v(t)t-\frac{1}{2a_x{t}^2}$$

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