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:

Scroll to Top