Author: mrchuakh

24P1Q10

10 A
$\displaystyle \omega =\frac{2\pi }{24\times 3600}=7.272\times {{10}^{-5}}\text{ rad }{{\text{s}}^{-1}}$

$\displaystyle \begin{aligned}a&=r{{\omega }^{2}} \\ &=(1.3\times {{10}^{7}}\div 2){{(7.272\times {{10}^{-5}})}^{2}} \\ &=0.0339\text{ m }{{\text{s}}^{-1}} \end{aligned}$

The acceleration is centripetal, i.e. directed towards the centre of the circular motion.

24P1Q09

9 A

$\displaystyle \omega =\frac{\Delta \theta }{\Delta t}=\frac{\mathrm{l}\div r}{t}$

24P1Q08

8 D

Work is done on the gas whenever the gas is compressed.

24P1Q07

7 C

$\displaystyle \begin{aligned} {{F}_{engine}}&=600+900(9.81)\sin 10{}^\circ \\ &=2133\text{ N} \end{aligned}$

$\displaystyle \begin{aligned} P&=Fv \\ &=(2137)(16) \\ &=34.1\text{ kW} \end{aligned}$

24P1Q06

6 B

Positive work done while loading = area OPQT
Negative work done while unloading = area SRQT
Net work done = OPQT – SRQT = OPRS

24P1Q05

5 D

When there were 3 springs: $\displaystyle W=3kx$

When there were 2 springs: $\displaystyle W=2kx'$

Dividing the two equations:

$\displaystyle \begin{aligned} 2&=\frac{2}{3}\frac{x'}{x} \\ x'&=3x \end{aligned}$

24P1Q04

4 B

F_X acts perpendicularly to plane X.

F_Y acts perpendicularly to plane Y.

W acts vertically downward.

These three force vectors must also form a closed triangle since net force is zero ( $\displaystyle {{F}_{x}}+{{F}_{Y}}+W=0$ ).

24P1Q03

3 C

Method 1

Solve for the time to reach the peak t_p using the vertical motion.

$\displaystyle \begin{aligned} (v&=u+at) \\ 0&=15\sin 40{}^\circ -1.62{{t}_{p}} \\ {{t}_{p}}&=5.952\text{ s} \end{aligned}$

Time of flight, $\displaystyle {{t}_{f}}={{t}_{p}}=2\times 5.952=11.9\text{ s}$

Method 2

$\displaystyle \begin{aligned} (s&=ut+\frac{1}{2}a{{t}^{2}}) \\ 0&=15\sin 40{}^\circ {{t}_{f}}-\frac{1}{2}(1.62){{t}_{f}}^{2} \\ 0&={{t}_{f}}[15\sin 40{}^\circ {{t}_{f}}-\frac{1}{2}(1.62){{t}_{f}}] \\ {{t}_{f}}&=0\text{ (rejected) or 11}\text{.9 s} \end{aligned}$

24P1Q02

2 B

Gradient of the v-t graph is acceleration.

Option B is the only one with constant positive gradient followed by constant negative gradient, matching the constant positive acceleration followed by constant negative acceleration.

24P1Q01

1 A

$\displaystyle \begin{aligned} A&=\pi {{r}^{2}} \\ &=\pi {{(4\times {{10}^{2}}\times {{10}^{6}})}^{2}} \\ &=5.02\times {{10}^{17}}\text{ }{{\text{m}}^{2}} \end{aligned}$

Option A: $\displaystyle 5\times {{10}^{-1}}\text{ G}{{\text{m}}^{2}}=5\times {{10}^{-1}}\times {{({{10}^{9}})}^{2}}\text{ }{{\text{m}}^{2}}=5\times {{10}^{17}}\text{ }{{\text{m}}^{2}}$

Option B: $\displaystyle 5\times {{10}^{2}}\text{ G}{{\text{m}}^{2}}=5\times {{10}^{2}}\times {{({{10}^{9}})}^{2}}\text{ }{{\text{m}}^{2}}=5\times {{10}^{20}}\text{ }{{\text{m}}^{2}}$

Option C: $\displaystyle 5\times {{10}^{-2}}\text{ T}{{\text{m}}^{2}}=5\times {{10}^{-2}}\times {{({{10}^{12}})}^{2}}\text{ }{{\text{m}}^{2}}=5\times {{10}^{22}}\text{ }{{\text{m}}^{2}}$

Option D: $\displaystyle 5\times {{10}^{0}}\text{ T}{{\text{m}}^{2}}=5\times {{10}^{0}}\times {{({{10}^{12}})}^{2}}\text{ }{{\text{m}}^{2}}=5\times {{10}^{24}}\text{ }{{\text{m}}^{2}}$