Laws of Motion MCQs for NEET — Physics Questions with Answers

For a vehicle on a banked road, the maximum safe velocity is given by √(gR((μs + tanθ)/(1 - μs tanθ))), where g is the acceleration due to gravity, R is the radius of curvature, μs is the coefficient of static friction, and θ is the angle of banking. This expression takes into account the centripetal force and the frictional force acting on the vehicle.

(d) Given, mass of particle m=0.01 kg.
Radius of circle along which particle is moving, r=6.4cm.

 ∴ Kinetic energy of particle, K.E=8x10^-4 J

=>  $\frac{1}{2}$mv²=8x10^-4 J

=> v²=$\frac{16\times {10}^{-4}}{0.01}$=16x10^-2 …(i)

As it is given that K.E of particle is equal to 8x10^-4 J by the end of second revolution after the beginning of motion of particle. It means, it’s initial velocity (u) is 0 m/s at this moment.

∴ By Newton’s 3^rd equation of motion,

v²= u²+2$a_{t}s$

v²= 2$as$ or v²= 2$a$ (4πr)
(∴ particle covers 2 revolutions)

a= v²/8πr = 16x10^-2/8x3.14x6.4x10^-2
(∴ from equation (i), v²=16x10^-2)

∴ $a_t$=0.1m/s²

Velocity v= $\sqrt{2\mathrm{gh}}$

and momentum p=mv

From Eqs. (i) and (ii), we have 

                 $p\propto \sqrt{h}$

Hence          $$\begin{gathered} \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\sqrt{\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}} \\ \mathrm{so}, \frac{{\mathrm{P}}_2}{{\mathrm{P}}_1}=\sqrt{\frac{2\mathrm{h}}{\mathrm{h}}}=\sqrt{2} \\ {\mathrm{P}}_2=1.414 {\mathrm{p}}_1 \end{gathered}$$

% change =$\frac{{\mathrm{P}}_2-{\mathrm{P}}_1}{{\mathrm{P}}_1}\times 100=41\%$

In this condition, centripetal force is equal to 

static frictional force road and tyres,

so           

                 $$\begin{gathered} {\mathrm{\mu }}_{\mathrm{s}}\mathrm{mg}=\frac{{\mathrm{mv}}^2}{\mathrm{R}} \\ {\mathrm{v}}_{\max }=\sqrt{{\mathrm{\mu }}_{\mathrm{s}}\mathrm{Rg}} \end{gathered}$$

Given, r= 5 cm=$5\times {10}^{-2}$m

and $\mathrm{T}=0.2 \mathrm{\pi s}$

We know that acceleration

              $$\begin{gathered} \mathrm{a}={\mathrm{r\omega }}^2 \\ =\frac{4{\mathrm{\pi }}^2}{{\mathrm{T}}^2}\mathrm{r} \\ =\frac{4\times {\mathrm{\pi }}^2\times 5\times {10}^{-2}}{(0.2\mathrm{\pi }{)}^2}=5 {\mathrm{ms}}^{-2} \end{gathered}$$

The tension in the supporting cable is the force required to provide the necessary acceleration to the combined mass of the lift and the person. Using Newton's second law, F = ma = (60 + 940) kg × (10 + 1) m/s^2 = 11000 N.

The tension in the supporting cable is the force required to accelerate the lift upwards. Using Newton's second law, F = ma, where F is the tension (28000 N), m is the mass of the lift (2000 kg), and a is the acceleration. Therefore, a = 28000 N / 2000 kg = 14 m/s^2 upwards.

According to Newton's second law of motion, force = mass $\times$ acceleration.

Here,  $\overrightarrow{F}=6\hat{i}-8\hat{j}+10\hat{k}$

         $\left|F\right|=\sqrt{36+64+100}$

             = $10\sqrt{2}N$

        $a=1 m{s}^{-2}$

$\therefore m=\frac{10\sqrt{2}}{1}=10\sqrt{2}kg$

Balancing the forces, we get

        $Mg-N=M\frac{v^2}{R}$

For weightlessness, N = 0

$\therefore \frac{M{v}^2}{R}=Mg$

where R is the radius of curvature and v is the speed of car.

Therefore,                $v=\sqrt{Rg}$

Putting the values, R=20m, g=10.0m/$s^2$

So,  $v=\sqrt{20\times 10.0}=14.14m/s^2$

Thus, the speed of the car at the top of the hill is between 14m/s and 15m/s.

Note: The roller coaster is a popular amusement ride developed for amusement parks and modern theme parks.