A car of mass $$\(1600 \mathrm{~kg}\)$$ is pulling a caravan of mass $$\(800 \mathrm{~kg}\)$$. The car and the caravan are connected by a light rigid tow-bar. The resistances to the motion of the car and caravan are $$\(400 \mathrm{~N}\)$$ and $$\(250 \mathrm{~N}\)$$ respectively. The car and caravan are travelling along a straight horizontal road. The engine's power is now suddenly increased to $$\(39 \mathrm{~kW}\)$$. Find the instantaneous acceleration of the car and caravan and find the tension in the tow-bar. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................
Exam No：9709_w20_qp_43 Year：2020 Question No：6(a)(ii)
Answer:
$\mathrm{DF}=\frac{39000}{25}(=1560)$
For applying Newton's $2^{\text {nd }}$ law to the system to form an equation in $a$, or to the caravan or the car to form an equation in $T$ and $a$
$\begin{array}{l}1560-650=2400 a \\ T-250=800 a \\ 1560-400-T=1600 a\end{array}$
$\left[a=\frac{(1560-650)}{2400}\right]$
$\begin{array}{l}a=0.379 \mathrm{~ms}^{-2}(0.37916 \ldots) \\ T=553 \mathrm{~N} \quad(553.33 \ldots)\end{array}$
For applying Newton's $2^{\text {nd }}$ law to the system to form an equation in $a$, or to the caravan or the car to form an equation in $T$ and $a$
$\begin{array}{l}1560-650=2400 a \\ T-250=800 a \\ 1560-400-T=1600 a\end{array}$
$\left[a=\frac{(1560-650)}{2400}\right]$
$\begin{array}{l}a=0.379 \mathrm{~ms}^{-2}(0.37916 \ldots) \\ T=553 \mathrm{~N} \quad(553.33 \ldots)\end{array}$
Knowledge points:
4.4.1 apply Newton’s laws of motion to the linear motion of a particle of constant mass moving under the action of constant forces, which may include friction, tension in an inextensible string and thrust in a connecting rod If any other forces resisting motion are to be considered (e.g. air resistance) this will be indicated in the question.
4.4.4 solve simple problems which may be modelled as the motion of connected particles. e.g. particles connected by a light inextensible string passing over a smooth pulley, or a car towing a trailer by means of either a light rope or a light rigid tow- bar.
4.5.4 use the definition of power as the rate at which a force does work, and use the relationship between power, force and velocity for a force acting in the direction of motion Including calculation of (average) power;P = Fv
4.5.5 solve problems involving, for example, the instantaneous acceleration of a car moving on a hill
Solution:
Download APP for more features
1. Tons of answers.
2. Smarter Al tools enhance your learning journey.
IOS
Download
Download
Android
Download
Download
Google Play
Download
Download