Question.1[(a) (i)].2019-May-Physics_paper_2__TZ1_SL
Topic:
Given: A girl rides a bicycle that is powered by an electric motor. A battery transfers energy to the electric motor. The emf of the battery is $16 \mathrm{~V}$ and it can deliver a charge of $43 \mathrm{kC}$ when discharging completely from a full charge.
The maximum speed of the girl on a horizontal road is $7.0 \mathrm{~m} \mathrm{~s}^{-1}$ with energy from the battery alone. The maximum distance that the girl can travel under these conditions is $20 \mathrm{~km}$
Prove: that the time taken for the battery to discharge is about $3 \times 10^3 \mathrm{~s}$.
▶️Answer/Explanation
Solution:
$V=\frac{d}{t}=>t=\frac{d}{v} ; d=20 \mathrm{~km}=20000 \mathrm{~m}, v=7.0 \mathrm{~m} / \mathrm{s}: \quad V=\frac{20000}{7.0}=2.85 \times 10^3 \approx 3 \times 10^3 \mathrm{~s}$
Question.1[(a) (ii)].2019-May-Physics_paper_2__TZ1_SL
Topic:
Show: that the average power output of the battery is about $240 \mathrm{~W}$.
▶️Answer/Explanation
Solution:
Using the formula $P = \frac{W}{t}$ and the equation for the voltage $V = \frac{W}{Q}$ to obtain $W = VQ$. Plugging in the given values, we get:
$$
P = \frac{W}{t} = \frac{VQ}{t} = \frac{(16 \mathrm{~V})(43 \times 10^3 \mathrm{~C})}{2.85 \times 10^3 \mathrm{~s}} \approx 241 \mathrm{~W} \approx 240 \mathrm{~W}
$$
So the average power output of the battery is about $240 \mathrm{~W}$, as previously calculated. Well done!
Question.1[(a) (iii)].2019-May-Physics_paper_2__TZ1_SL
Topic:
Given: Friction and air resistance act on the bicycle and the girl when they move. Assume that all the energy is transferred from the battery to the electric motor.
Calculate: the total average resistive force that acts on the bicycle and the girl.
▶️Answer/Explanation
Solution:
$F_{\text {res }}=F_{\text {engine: }}$
Power $=\mathrm{Fv} \Rightarrow \mathrm{F}=\frac{\text { power }}{\mathrm{v}}=\frac{241}{7.0}=34 \mathrm{~N}$
Question.1[(b) (i)].2019-May-Physics_paper_2__TZ1_SL
Topic:
Given: The bicycle and the girl have a total mass of $66 \mathrm{~kg}$. The girl rides up a slope that is at an angle of $3.0^{\circ}$ to the horizontal.
Calculate: the component of weight for the bicycle and girl acting down the slope.
▶️Answer/Explanation
Solution:
The weight of the bicycle and the girl is given by:
$$
mg = (66\text{ kg})(9.81\text{ m/s}^2) \approx 648\text{ N}
$$
The component of weight acting down the slope is given by:
$$
mg\sin\theta = (648\text{ N})\sin(3.0^\circ) \approx 33.7\text{ N}
$$
where $\theta = 3.0^\circ$ is the angle of the slope with respect to the horizontal.
Question.1[(b) (ii)].2019-May-Physics_paper_2__TZ1_SL
Topic:
Given:The battery continues to give an output power of $240 \mathrm{~W}$. Assume that the resistive forces are the same as in (a)(iii).
Calculate: the maximum speed of the bicycle and the girl up the slope.
▶️Answer/Explanation
Solution:
Sum of backward forces $=F_{\text {res }}+F_{\mathrm{mg}}=34+34=68 \mathrm{~N}$
$$
\text { power }=\mathrm{Fv} \Rightarrow \mathrm{v}=\frac{\text { power }}{\mathrm{F}}=\frac{240 \mathrm{~W}}{68 \mathrm{~N}}=3.5 \mathrm{~m} / \mathrm{s}
$$