**2020-Nov-Physics_paper_2__TZ0_SL – All Questions with detailed solution**### Q.1[(a) (i)].2020-Nov-Physics_paper_1__TZ0_SL

**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:**

When the aircraft is hovering motionless, the upward force exerted by the air on the blades must balance the weight of the aircraft and the suspended package. Therefore, the resultant force on the aircraft is zero.

### Q.1[(a) (ii)].2020-Nov-Physics_paper_1__TZ0_SL

**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

$\text{OR}$

- air exerts a reaction force on the blades by Newton’s third law}

### Q.1[(a) (iii)].2020-Nov-Physics_paper_1__TZ0_SL

**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}

$

$\colorbox{yellow}{8.1 m/s}$

### Q.1(b).2020-Nov-Physics_paper_1__TZ0_SL

**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)** $=\frac{0.45 \times 9.81}{0.95}=4.6 \mathrm{~ms}^{-2}$

$\colorbox{yellow}{ $4.6~\mathrm{ms}^{-2}$ }$

### Q.2(a).2020-Nov-Physics_paper_1__TZ0_SL

**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:**