2020-Nov-Physics_paper_2__TZ0_HL – All Questions with detailed solution.
Q.1[(a) (i)].2020-Nov-Physics_paper_1__TZ0_HL
Topic: Forces
Given: A company delivers packages to customers using a small unmanned aircraft. Rotating horizontal blades exert a force on the surrounding air. The air above the aircraft is initially stationary.
The air is propelled vertically downwards with speed $v$. The aircraft hovers motionless above the ground. A package is suspended from the aircraft on a string. The mass of the aircraft is $0.95 \mathrm{~kg}$ and the combined mass of the package and string is $0.45 \mathrm{~kg}$. The mass of air pushed downwards by the blades in one second is $1.7 \mathrm{~kg}$.
Calculation: State the value of the resultant force on the aircraft when hovering.
Answer/Explanation
Solution:
Q.1[(a) (ii)].2020-Nov-Physics_paper_1__TZ0_HL
Topic: Forces
Discuss: Outline, by reference to Newton’s third law, how the upward lift force on the aircraft is achieved.
Answer/Explanation
Solution:
- Blades exert a downward force on the air
- air exerts an equal and opposite force on the blades by Newton’s third law
$\text{OR}$
Q.1[(a) (iii)].2020-Nov-Physics_paper_1__TZ0_HL
Topic: Momentum and impulse
Calculate: Determine $v$. State your answer to an appropriate number of significant figures.
Answer/Explanation
Solution:
Force $=$ rate of change of momentum
$
\begin{aligned}
& \Rightarrow M g=\frac{m v}{t} \\
& \Rightarrow \text { we have } m=1.4 \mathrm{~kg}(0.95+0.45), \quad g=9.8 \mathrm{~ms}^{-2}, \frac{\mathrm{m}}{\mathrm{t}}=1.7 \mathrm{~kg} \\
&\Rightarrow 1.4 \mathrm{~kg}\times 9.8\mathrm{~ms}^{-2}=1.7 \mathrm{~kg}\times v\\
& \Rightarrow \quad \therefore v=\frac{1.4 \times 9.8}{1.7}=8.1 \mathrm{~ms}^{-1}
\end{aligned}
$
Q.1[(a) (iv)].2020-Nov-Physics_paper_1__TZ0_HL
Topic: Work, energy and power
Calculate: Calculate the power transferred to the air by the aircraft.
Answer/Explanation
Solution:
$
\begin{aligned}
\text { Power transferred to air } & =\frac{K . E . \text { transferred }}{\text { time }}=\frac{1}{2} \frac{m v^2}{t} \\
& =\frac{1}{2} \times 1.7 \times(8.1)^2=55.77 \mathrm{~W}
\end{aligned}
$
$\therefore \text{power transferred} \mathrm{P} \cong \mathbf{5 6 W}$
Q.1(b).2020-Nov-Physics_paper_1__TZ0_HL
Topic: Forces
Given: The package and string are now released and fall to the ground. The lift force on the aircraft remains unchanged.
Calculate: the initial acceleration of the aircraft.
Answer/Explanation
Solution:
$\text{vertical force= lift force – weight}$
After release of the packet, mass of the craft $=0.95 \mathrm{~kg}$.
$\text{Net upward force} =1.4 \times 9.81-0.95 \times 9.81$
$
=4.45 \times 9.81
$
So acceleration (upward) $=\frac{0.45 \times 9.81}{0.95}=4.6 \mathrm{~ms}^{-2}$
Q.2(a).2020-Nov-Physics_paper_1__TZ0_HL
Topic: Circular motion
Given: The Rotor is an amusement park ride that can be modelled as a vertical cylinder of inner radius $R$ rotating about its axis. When the cylinder rotates sufficiently fast, the floor drops out and the passengers stay motionless against the inner surface of the cylinder. The diagram shows a person taking the Rotor ride. The floor of the Rotor has been lowered away from the person.
Discuss: Draw and label the free-body diagram for the person.
Answer/Explanation
Solution:
