IB DP Physics Mock Exam SL Paper 2 Set 1 - 2025 Syllabus
IB DP Physics Mock Exam SL Paper 1B Set 1
Prepare for the IB DP Physics Mock Exam SL Paper 2 Set 1 with our comprehensive mock exam set 1. Test your knowledge and understanding of key concepts with challenging questions covering all essential topics. Identify areas for improvement and boost your confidence for the real exam
Question
Airboats are used for transport across a river. To move the boat forward, air is propelled from
the back of the boat by a fan blade.
(a) Outline why a force acts on the airboat due to the fan blade. [3]
(b) In a test the airboat is tied to the river bank with a rope normal to the bank. The fan
propels the air at its maximum speed. There is no wind.
(i) Show that a mass of about 240kg of air moves through the fan every second. [2]
(ii) Show that the tension in the rope is about 5kN. [1]
(c) The rope is untied and the airboat moves away from the bank. The variation with time t
of the speed v of the airboat is shown for the motion.
(i) Estimate the distance the airboat travels to reach its maximum speed. [2]
(ii) Deduce the mass of the airboat. [3]
(d) The fan is rotating at 120 revolutions every minute. Calculate the centripetal
acceleration of the tip of a fan blade. [2]
Answer/Explanation
Ans.
(a) ALTERNATIVE 1
there is a force «by the fan» on the air / air is accelerated «to the rear» $\checkmark$ by Newton $3 \checkmark$
there is an «equal and» opposite force on the boat $\checkmark$
ALTERNATIVE 2
air gains momentum «backward» $\checkmark$
by conservation of momentum / force is rate of change in momentum $\checkmark$ boat gains momentum in the opposite direction $\checkmark$
(b)(i) $\pi R^2 O R$ «mass of air through system per unit time $= A v \rho$ seen $\checkmark$
$244 \mathrm{kg} \mathrm{s}^{-1} \checkmark$
(b)(ii) «force $=$ Momentum change per sec $=A v^2 \rho=» 244 \times 20$ OR $4.9 « \mathrm{kN} » \checkmark$
(c)(i)recognition that area under the graph is distance covered $\checkmark$
«Distance $=» 480-560$ «m» $\checkmark$
(c)(ii)calculation of acceleration as gradient at $t=0 \ll=1 \mathrm{~m} \mathrm{~s}^{-2} \mathrm{\|} \checkmark$
use of $F=$ ma OR $\frac{4900}{1}$ seen $\checkmark$
$4900 \pi \mathrm{kg} » \checkmark$
(d) ALTERNATE 1
« $\omega=» 4 \pi \mathrm{rad} \mathrm{s}^{-1} \checkmark$
« $a=r \omega^2=» 280 \ll \mathrm{m} \mathrm{s}^{-2} \| \quad \checkmark$
ALTERNATE 2
$
\begin{aligned}
& 《 v=\frac{2 \pi r}{T} »=22.6 \mathrm{~m} \mathrm{~s}^{-1} \checkmark \\
& 《 a=\frac{v^2}{r} »=280 \ll \mathrm{m} \mathrm{s}^{-2} » \\
&
\end{aligned}
$