(a)(i) Correct Answer: \( \boxed{0.16\,\mathrm{MPa}} \)
The airbag is approximately cylindrical, so its volume is
\(V=\pi r^{2}h\)
The radius is \(r=0.90\,\mathrm{m}\) and \(h=24.0\,\mathrm{m}\).
Therefore,
\(V=\pi(0.90)^{2}(24.0)\)
\(V\approx61.1\,\mathrm{m^{3}}\)
Using the ideal gas equation,
\(PV=nRT\)
The temperature is
\(T=273+15=288\,\mathrm{K}\)
Hence,
\(P=\dfrac{nRT}{V}\)
\(P=\dfrac{(4200)(8.31)(288)}{61.1}\)
\(P\approx1.64\times10^{5}\,\mathrm{Pa}\)
Therefore, \( \boxed{P\approx0.16\,\mathrm{MPa}} \), which is about \(0.2\,\mathrm{MPa}\).
(a)(ii) Correct Answer: \( \boxed{4.3\times10^{7}\,\mathrm{kg}} \)
The maximum safe pressure is four times the pressure found in (a)(i):
\(P_{\max}=4(1.64\times10^{5})\)
\(P_{\max}\approx6.56\times10^{5}\,\mathrm{Pa}\)
The contact area of one airbag is
\(A=24(1.8)=43.2\,\mathrm{m^{2}}\)
For fifteen airbags, the total supporting force is
\(F=15PA\)
\(F=15(6.56\times10^{5})(43.2)\)
\(F\approx4.25\times10^{8}\,\mathrm{N}\)
The component of the weight perpendicular to the slope is
\(F=mg\cos4.0^{\circ}\)
Therefore,
\(m=\dfrac{4.25\times10^{8}}{(9.8)\cos4.0^{\circ}}\)
\(m\approx4.3\times10^{7}\,\mathrm{kg}\)
Hence, the maximum safe mass is \( \boxed{4.3\times10^{7}\,\mathrm{kg}} \).
(a)(iii) Correct Answer:
An appropriate assumption is that the weight or mass of the boat is distributed evenly among the airbags.
Other acceptable assumptions include that the airbags have the stated contact area and that they do not touch each other.
(a)(iv) Correct Answer:
When the airbag expands adiabatically, there is negligible heat transfer between the gas and its surroundings.
Thus, \(Q=0\).
From the first law of thermodynamics,
\(\Delta U=Q-W\)
Therefore, since \(Q=0\) and the gas does work while expanding, the internal energy decreases.
Hence, \( \boxed{\Delta U<0} \), and the temperature of the gas decreases.
(b)(i) Correct Answer: \( \boxed{0.44\,\mathrm{m\,s^{-2}}} \)
The boat starts from rest, so \(u=0\).
Using
\(s=ut+\dfrac{1}{2}at^{2}\)
\(14=\dfrac{1}{2}a(8.0)^{2}\)
Therefore,
\(\boxed{a=0.44\,\mathrm{m\,s^{-2}}}\)
(b)(ii) Correct Answer: \( \boxed{1.1\,\mathrm{MN}} \)
The component of the weight down the slope is
\(F_g=mg\sin4.0^{\circ}\)
\(F_g=(4.5\times10^{6})(9.8)\sin4.0^{\circ}\)
\(F_g\approx3.08\times10^{6}\,\mathrm{N}\)
The resultant force is
\(F_{\mathrm{net}}=ma\)
\(F_{\mathrm{net}}=(4.5\times10^{6})(0.44)\)
\(F_{\mathrm{net}}\approx1.98\times10^{6}\,\mathrm{N}\)
Hence, the resistive force is
\(F_{\mathrm{R}}=F_g-F_{\mathrm{net}}\)
\(F_{\mathrm{R}}=3.08\times10^{6}-1.98\times10^{6}\)
\(\boxed{F_{\mathrm{R}}\approx1.1\times10^{6}\,\mathrm{N}=1.1\,\mathrm{MN}}\)
(b)(iii) Correct Answer:
The free-body diagram should show:
• The weight \(mg\) vertically downwards.
• The normal reaction force \(N\) perpendicular to the slope.
• The resistive force \(F_{\mathrm{R}}\) acting up the slope, opposite to the motion.
(c)(i) Correct Answer: \( \boxed{7.8\,\mathrm{rad}} \)
The diameter of the rod is \(1.80\,\mathrm{m}\), so its radius is
\(r=0.90\,\mathrm{m}\)
The distance travelled by the boat is \(14\,\mathrm{m}\). For rolling without slipping,
\(s=r\theta\)
Therefore,
\(\theta=\dfrac{s}{r}=\dfrac{14}{0.90}\)
\(\boxed{\theta\approx7.8\,\mathrm{rad}}\)
(c)(ii) Correct Answer:
The frictional forces at the top and bottom of the rod act in opposite directions.
Because these forces act at different positions relative to the central axis, they produce torques in the same rotational sense.
The resultant torque is therefore non-zero, producing the angular acceleration of the rod.
(d)(i) Correct Answer: \( \boxed{2.22\,\mathrm{rad\,s^{-1}}} \)
The linear speed and angular velocity are related by
\(v=\omega r\)
Therefore,
\(\omega=\dfrac{v}{r}\)
\(\omega=\dfrac{2.0}{0.90}\)
\(\boxed{\omega=2.22\,\mathrm{rad\,s^{-1}}}\)
(d)(ii) Correct Answer: \( \boxed{0.35\,\mathrm{s}} \)
At \(t=0\), point P is \(45^{\circ}\) above the horizontal. To reach the top of the circle, P rotates through
\(\Delta\theta=90^{\circ}-45^{\circ}=45^{\circ}=\dfrac{\pi}{4}\,\mathrm{rad}\)
Using
\(\Delta\theta=\omega t\)
\(t=\dfrac{\pi/4}{2.22}\)
\(\boxed{t\approx0.35\,\mathrm{s}}\)
(d)(iii) Correct Answer: \( \boxed{0.79\,\mathrm{m\,s^{-1}}} \)
The vertical displacement is modelled by
\(x=x_{0}\sin(\omega t+\phi)\)
Since P is initially \(45^{\circ}\) above the horizontal,
\(\phi=\dfrac{\pi}{4}\)
Differentiating gives
\(v_x=\omega x_{0}\cos(\omega t+\phi)\)
Since \(x_{0}=r=0.90\,\mathrm{m}\) and \(\omega=2.22\,\mathrm{rad\,s^{-1}}\), at \(t=3.0\,\mathrm{s}\),
\(v_x=(2.22)(0.90)\cos\left[(2.22)(3.0)+\dfrac{\pi}{4}\right]\)
\(v_x\approx0.79\,\mathrm{m\,s^{-1}}\)
Hence, \( \boxed{v_x\approx0.79\,\mathrm{m\,s^{-1}}} \).
