Physics Guru

Why are small steps advisable while walking on wet path? Answer Let F be the force that the person exerts on the ground. The ground exerts equal and opposite force F on the person. Let $\theta$ be the angle that the force F exerted on the person by the ground makes with the vertical. The force F that ground exerts on the person can be resolved into two components: $F cos \theta = N$ (normal reaction) $F sin \theta = Friction \leq \mu. N$ So, $F sin \theta \leq \mu. F cos \theta $ So, $tan \theta \leq \mu $ or $(tan \theta)_{max} = \mu $ The coefficient of friction $\mu $ decreases when the surface is wet. So, the maximum permissible $\theta $ also decreases. Hence, the legs should be kept more nearly vertical, which is achieved by taking smaller steps.

A projectile is projected from the origin with a speed of $20m{s^{ - 1}}$ at an angle of ${63^\circ }$ above the horizontal. After a delay of 1s, a drone is launched from the same point with constant velocity. The motion of both the projectile and the drone takes place in the vertical xz-plane. If the drone strikes the projectile 2s after the projectile is launched, find the velocity of the drone. Ignore air resistance. [$g = 10m{s^{ - 2}}$, $\sin {63^\circ } = \frac{4}{5}$ and $\cos {63^\circ } = \frac{3}{5}$] Solution For projectile motion, $x = 20\cos {63^\circ }.2 = 24m$ $z = 20\sin {63^\circ }.2 - \frac{1}{2}{.10.2^2} = 32 - 20 = 12m$ Position vector of the projectile when the drone strikes it,  $\vec r = 24\hat i + 12\hat k$ m Velocity of drone = $\frac{{\vec r}}{1} = 24\hat i + 12\hat k$ $m{s^{ - 1}}$

Find the minimum value of the function $f(x, y)$ for $x, y > 0:$ $f(x,y)=\sqrt{x^{2}+1}+\sqrt{\frac{y^{2}}{x^{2}}+4}+\sqrt{\frac{9}{y^{2}}+9}$ Solution Let, $a = x,b = \frac{y}{x},c = \frac{3}{y}$ Then $a,b,c > 0$ and $abc = x.\frac{y}{x}.\frac{3}{y} = 3$ So $f = \sqrt {{a^2} + 1}  + \sqrt {{b^2} + 4}  + \sqrt {{c^2} + 9} $ Now, Minkowski inequality (in its 2-dimensional form) for ${a_i},{b_i} \ge 0$, $\sum\limits_{i = 1}^n {\sqrt {a_i^2 + b_i^2} }  \ge \sqrt {{{\left( {\sum\limits_{i = 1}^n {{a_i}} } \right)}^2} + {{\left( {\sum\limits_{i = 1}^n {{b_i}} } \right)}^2}} $ $\therefore f = \sqrt {{a^2} + {1^2}}  + \sqrt {{b^2} + {2^2}}  + \sqrt {{c^2} + {3^2}}  \ge \sqrt {{{(a + b + c)}^2} + {{(1 + 2 + 3)}^2}} $ $\therefore f \ge \sqrt {{{(a + b + c)}^2} + 36} $ Now, $A.M. \ge G.M$ or $\frac{{a + b + c}}{3} \ge {(abc)^{1/3}}$ $ \Rightarrow a + b + c \ge {3^{4/3}}$ $\therefore f \ge \sqrt {{3^{8/3}} + 36} $ $ \Rightarrow f \ge \sqrt {{{9.9}^{1/3}} + 9.4} ...

What should be the minimum value of amplitude so that the particle speed can match or exceed the wave speed in sinusoidal progressive wave $y = A sin (ωt − \frac {2\pi}{\lambda}x)$? (A) Particle speed can never exceed wave speed (B) $A_{min} = \lambda$ (C) $A_{min} = \frac {\lambda}{2\pi}$ (D) Particles are not present as the wave travels in vacuum Solution Particle speed = $|\frac {\partial y}{\partial t}| = Aω |cos (ωt − \frac {2\pi}{\lambda}x)|$ So, particle speed is less than or equal to Aω. Wave speed = $\nu.\lambda = \frac {ω}{2\pi}.\lambda$ If particle speed were to match or exceed wave speed, $Aω |cos (ωt − \frac {2\pi}{\lambda} x)| \geq \frac {ω}{2\pi}.\lambda$ $\therefore A \geq \frac{{\frac{\lambda }{{2\pi }}}}{{\left| {cos\left( {\omega t - \frac{{2\pi }}{\lambda }x} \right)} \right|}}$ For A to be min. cosine should be maximum = 1 $A_{min} \geq \frac {\lambda}{2\pi}$ When $A_{min} = \frac {\lambda}{2\pi}$, particle speed can match wave speed periodically and when $A_{min} ...

A stationary man standing on a circular platform that is free to rotate is holding an umbrella starts rotating it anticlockwise. The platform starts rotating clockwise due to conservation of angular momentum. Man also rotates clockwise along with the platform.

Q: As an exercise for wrist, palm and fingers, a person squeezes and releases a rubber ball. Is the work done on the ball positive, negative or zero during (a) compression and (b) during expansion? A: Work done on ball = Increase in elastic potential energy Potential energy increases during compression (think of analogy with spring) and thus the work done is positive. Potential energy decreases during expansion and thus the work done is negative.

Q: In beta decay of a nucleus initially at rest, the recoiling daughter nucleus and the emitted beta particle are sometimes observed not to move along the same straight line. How can this be explained in view of the law of conservation of linear momentum?  A: In beta decay, a nucleus emits a beta particle (electron or positron) and a neutrino (or antineutrino). The neutrino carries away some of the momentum. Since, initial linear momentum is $\vec 0$, conservation of linear momentum requires $\vec 0 = \overrightarrow {{p_R}}  + \overrightarrow {{p_\beta }}  + \overrightarrow {{p_\nu }} $. ​If beta particle and neutrino move in straight line, $\overrightarrow {{p_\nu }}  = k\overrightarrow {{p_\beta }} $ where k is a scalar and therefore $\overrightarrow {{p_R}}  =  - (1 + k)\overrightarrow {{p_\beta }} $ which means the daughter nucleus and the emitted beta particle move along the same straight line.  However, if beta particle and neutrino do not move ...