Physics Guru

What minimum kinetic energy should be imparted to a projectile of mass 12 g, launched from point on the Moon's surface, so that it hits a target P, where OP=200m? The height of the target above the lunar surface is 50m. Take the acceleration due to gravity on the Moon to be one-sixth of that on Earth, with $g_{Earth} = 9.8 m /{s^2}$. Solution Using the typical symbols associated with projectile motion, we have, $x = u\cos \theta .t,\cos \theta  = \frac{x}{{ut}}$ and $y = u\sin \theta .t - \frac{1}{2}g{t^2},\sin \theta  = \frac{{y + \frac{1}{2}g{t^2}}}{{ut}}$ ${\sin ^2}\theta  + {\cos ^2}\theta  = 1 = \frac{{{x^2}}}{{{u^2}{t^2}}} + \frac{{{{\left( {y + \frac{1}{2}g{t^2}} \right)}^2}}}{{{u^2}{t^2}}}$ $\Rightarrow {u^2} = \frac{{{x^2}}}{{{t^2}}} + \frac{{{y^2} + yg{t^2} + \frac{1}{4}{g^2}{t^4}}}{{{t^2}}}$ $ \Rightarrow {u^2} = \frac{{{x^2}}}{{{t^2}}} + \frac{{{y^2}}}{{{t^2}}} + yg + \frac{1}{4}{g^2}{t^2}$ $ \Rightarrow {u^2} = \frac{{{r^2}}}{{{t^2}}} + \frac{{{g^2}{t^2}...

Question A fat man trying to touch his toes while keeping his legs straight is more inclined to fall in comparison to a lean man. True/False Answer True. A man with a prominent abdominal bulge may have his centre of mass displaced farther forward while bending to touch his toes with his legs straight. This can increase the perpendicular distance between the line of action of his weight and the ankle joints, thereby increasing the gravitational torque. Consequently, maintaining rotational equilibrium becomes more difficult, increasing the likelihood of falling forward.

Student: Normally there is a perception that, "speed kills" and that high speed is dangerous. But, planes fly at high speed and are generally considered safe. So, how do you address this paradox? Teacher: It all depends on the context. So, sometimes high speed can be dangerous but sometimes high speed is desired. In fact, sometimes high speed is essential. A very common case is that of circular motion. Consider an electron moving in a circular orbit in the Bohr model of the hydrogen atom. Electrostatic force, $\frac {k.e.e}{r^2}$ = centripetal force, $\frac {mv^2}{r}$ So, $v = e \sqrt \frac {k}{rm} = 1.6\times 10^{-19} \sqrt \frac {9\times 10^9}{0.53\times 10^{-10}\times 9.1\times 10^{-31}}$                                                        $= 1.6\times 3\times 10^{-19} \times 10^{24} \sqrt \frac {1000}{48.23}$         ...

Consider an experiment involving a supercar having virtually no upper limit on its capacity to run at high speeds fitted with wheels whose rotational inertia can be neglected. Let the coefficient of friction between the tyres of the car and the ground to be $0.25$. Find the minimum time $t$ it will take to race $500 m$ starting from rest on a horizontal stretch.  At the end of $500 m$ stretch there is incline plane at angle $\theta$ very close to $15^\circ$ ($sin \theta = 0.25$). The engine of car is turned of at the end of $500 m$ stretch and car is allowed to move freely. After how much time will the car come to a halt? [$g=10 m/s^2$] Solution For horizontal motion, we have maximum frictional force $= \mu N = \mu mg = ma$ So, maximum $a = \mu g = 0.25  \times 10 = 2.5 m/s^2$ $s=ut+\frac {1}{2}at^2$ So, $500=\frac {1}{2}\times 2.5 \times t^2$ $\Rightarrow t = 20s$ For motion in incline plane, Deceleration $= gsin\theta $ Deceleration $=10\times 0.25 = 2.5 m/s^2$ Since deceler...

One edge of a square conducting loop made of uniform wire is mounted on a vertical non-conducting light pole. The light pole is free to rotate about the vertical axis passing through it. A uniform magnetic field $B$ points towards the right side such that loop's plane is parallel to the field as shown in the figure. The current $I_0$ is switched on at $t=0$. Find the instantaneous angular acceleration of the loop at $t=0$. The mass per unit length of the uniform wire is $\mu$ and length of one edge is $l$. Solution Force $F$ on the right edge $F=I_0lB$. This force will create torque which will rotate the loop. Torque $\tau =l.F=I_0 l^2B$ We have $\tau = I\alpha$ $I=\mu l.l^2+\mu l.\frac {l^2}{3}+\mu l.\frac {l^2}{3}$ $\therefore I=\frac {5}{3} \mu l^3 $ Now, $\tau =l.F=I_0 l^2B = I \alpha = \frac {5}{3} \mu l^3 \alpha$ $\therefore \alpha = \frac {3I_0B}{5\mu l}$

During an excavation in the southern part of India, a bundle of old leaves bearing inscriptions was discovered. During carbon dating, a sample of the leaves was found to have a C-14 activity of 14.40 disintegrations per gram per minute. A fresh, identical sample of the leaves has a C-14 activity of 15.12 disintegrations per gram per minute. Determine the age of the old leaves. The half-life of C-14 is 5730 years. Solution We have, $t = 3.32{t_{1/2}}{\log _{10}}\frac{{{{[A]}_0}}}{{[A]}}$ $\Rightarrow t = 3.32 \times 5730 \times {\log _{10}}\frac{{15.12}}{{14.40}}$ $ \Rightarrow t = 19023.6 \times {\log _{10}}1.05$ $ \Rightarrow t \approx 19023.6 \times 0.02119 \approx 403$ years

A small projectile is launched with velocity $u$ in $xz$ vertical plane in a gravity-free large hall from origin $(0, 0, 0)$ at an angle of $60^\circ$ from $x-axis$. At what position on $x-axis$ should another projectile be launched at the same instant with identical speed $u$ in the $xz$ plane at an angle of $120^\circ$ from $x-axis$ so that the two projectiles collide after time $t$? (ignore air resistance) Answer:  As can be seen in the figure, we have an equilateral triangle.  The distance between the two, $OA = OP = AP = ut$.  So, the position of A $\equiv (ut, 0, 0)$.