Physics 4 Marks Questions

m → mass of child

$\text{R}-\text{mg}\cos45^{\circ}=0$

$\Rightarrow\text{R = mg}\cos45^{\circ}=\text{mg/v}^2 \ ..(1)$

Net force acting on the boy due to which it slides down is $\text{mg}\sin45^{\circ}-\mu\text{R}$

$=\text{mg}\sin45^{\circ}-\mu\text{mg}\cos45^{\circ}$

$=\text{m}\times10\Big(\frac{1}{\sqrt{2}}\Big)-0.6\times\text{m}\times10\times\Big(\frac{1}{\sqrt{2}}\Big)$

$=\text{m}\Big[\Big(\frac{5}{\sqrt{2}}\Big)-0.6\times\Big(\frac{5}{\sqrt{2}}\Big)\Big]$

$=\text{m}(2\sqrt{2})$

acceleration $=\frac{\text{Force}}{\text{mass}}=\frac{\text{m}(2\sqrt{2})}{\text{m}}=2\sqrt{2}\text{m/s}^2$

Let, a maximum acceleration produced in car.

$\therefore\text{ma}=\mu\text{R}$ [For more acceleration, the tyres will slip]

$\Rightarrow\text{ma}=\mu\text{mg}\Rightarrow\text{a}=\mu\text{g}=1\times10=10\text{m/s}^2$

For crossing the bridge in minimum time, it has to travel with maximum acceleration

$\text{u = 0, s = 500m, a = 10m/s}^2$

$\text{s = ut}+\frac{1}{2}\text{at}^2$

$\Rightarrow500=0+\Big(\frac{1}{2}\Big)10\text{t}^2\Rightarrow\text{t}=10\text{sec}.$

If acceleration is less than 10m/s², time will be more than 10sec. So one can’t drive through the bridge in less than 10sec.

To reach in minimum time, he has to move with maximum possible acceleration.

Let, the maximum acceleration is ‘a’

$\therefore\text{ma}-\mu\text{R}=0\Rightarrow\text{ma}=\mu\text{mg}$

$\Rightarrow\text{a}=\mu\text{g}=0.9\times10=9\text{m/s}^2$

1. Initial velocity $\text{u = 0, t = ?}$

$\text{a}=9\text{m/s}^2,\text{s}=50\text{m}$

$\text{s = ut}+\frac{1}{2}\text{at}^2\Rightarrow50=0+\Big(\frac{1}{2}\Big)9\text{t}^2$

$\Rightarrow\text{t}=\sqrt{\frac{100}{9}}=\frac{10}{3}\text{sec}.$

2. After overing 50m, velocity of the athelete is

$\text{V = u + at}=0+9\times\Big(\frac{10}{3}\Big)=30\text{m/s}$

He has to stop in minimum time. So deceleration ia $-\text{a}=-9\text{m/s}^2(\text{max})$

$\text{R = ma}$

$\text{ma}=\mu\text{R}$ (max frictional force)

$\Rightarrow\text{a}=\mu\text{g}=9\text{m/s}^2$ (Deceleration)

$\text{u}^1=30\text{m/s},\text{ v}^1=0$

$\text{t}=\frac{\text{v}^1-\text{u}^1}{\text{a}}=\frac{0-30}{-\text{a}}=\frac{-30}{-\text{a}}=\frac{10}{3}\text{sec}.$

Hardest brake means maximum force of friction is developed between car’s type & road.

Max frictional force $=\mu\text{R}$

From the free body diagram

$\text{R}-\text{mg}\cos\theta=0$

$\Rightarrow\text{R}=\text{mg}\cos\theta \ ...(\text{i})$

and $(\mu\text{R + ma}-\text{mg}\sin)=0 \ ...(\text{ii})$

$\Rightarrow\mu\text{mg}\cos\theta+\text{ma}-\text{mg}\sin\theta=0$

$\Rightarrow\mu\text{g}\cos\theta+\text{a}-10\times\Big(\frac{1}{2}\Big)=0$

$\Rightarrow\text{a}=5-\{1-(2\sqrt{3})\}\times10\Big(\frac{\sqrt{3}}{2}\Big)=2.5\text{m/s}^2$

When, hardest brake is applied the car move with acceleration 2.5m/s²

$\text{S = 12.8m, u = 6m/s}$

So, velocity at the end of incline

$\text{V}=\sqrt{\text{u}^2+2\text{as}}=\sqrt{6^2+2(2.5)(12.8)}$

$=\sqrt{36+64}=10\text{m/s}=36\text{km/h}$

Hence how hard the driver applies the brakes, that car reaches the bottom with least velocity 36km/h.