Physics Guru

A long lorry is moving with constant velocity along a horizontal road. Another identical lorry is moving ahead of it with the same velocity. A small object is projected from a point with position vector $\vec r_1$ on the first lorry at an angle of $45^\circ$ above the horizontal. What should be the horizontal-plane component of the object's velocity relative to the first lorry so that the object reaches a point with position vector $\vec r_2$ on the second lorry? Neglect air resistance. Solution Since both lorries have the same constant velocity, the second lorry is at rest relative to the first. Thus, in the first-lorry frame, the projectile has to cover the horizontal displacement $$\vec d=\vec r_2 - \vec r_1$$ Its horizontal range is $$R=|\vec r_2 - \vec r_1|$$ For a projectile launched at $45^\circ$, $$R=\frac {u^2sin90^\circ}{g}=\frac {u^2}{g}$$ $$u=\sqrt {g|\vec r_2 - \vec r_1|}$$ The horizontal-plane velocity vector is therefore in the direction from $\vec r_1$ to $\vec r_2$...

A parallel-plate capacitor containing a polar dielectric with a dielectric constant $k$, connected to a battery with an e.m.f. $E$, is moved from room temperature ($25\text{ }^\circ\text{C}$) into a freezer. What happens to the electrostatic potential energy of the capacitor? (Assume thermal expansion of the plates is neglected and the battery's e.m.f. remains constant.) Answer At room temperature (25 °C), thermal energy causes molecules to vibrate and collide violently and randomly. This thermal chaos fights against the electric field from the battery, constantly knocking the molecular dipoles out of alignment. When capacitor is placed in a cool freezer, the thermal kinetic energy of the molecules drops significantly. With less thermal agitation disrupting them, the electric field becomes much more effective at lining up those molecular dipoles in an orderly fashion. Because more dipoles successfully align with the field, the material's internal polarization increases. A highe...

A dishonest carnival shop owner bends the barrel of an air rifle downward by $15^\circ$ near the middle, causing unsuspecting visitors to miss the balloon target. To compensate and successfully burst the balloon, at what approximate angle above the target should a clever visitor aim? (Neglect air resistance and effect of gravity considering short distance firing) Answer To understand how to compensate for the tampered rifle, we can break down the mechanics of the modification and the required adjustment: The Defect: Bending the rifle barrel downward by $15^\circ$ near its middle introduces a permanent angular offset to the muzzle. When the user aligns the sights of the rifle directly with the target ($0^\circ$), the altered trajectory of the barrel causes the projectile to exit at a $15^\circ$ downward angle relative to the line of sight, resulting in a miss below the balloon. The Compensation: To successfully hit the target, the visitor must counteract the $15^\circ$ downward deflecti...

Four persons are placed at the vertices of ABCD square frame of side a mounted on xy-plane having its center as the origin. The square frame is revolving with angular velocity $\vec \omega = \omega \hat k$. With what velocity vector should A throw a light ball with respect to himself when the square frame has all its sides parallel to x and y axis with point A lying in the second quadrant, so that the ball just falls downward as seen from the ground as if it were free falling? Solution The square frame is rotating anti-clockwise. The velocity of person A w.r.t. ground = $\frac {a\sqrt 2}{2}\omega (-cos 45^\circ \hat i - sin 45^\circ \hat j)$ = $-\frac {a\omega}{2} (\hat i + \hat j)$ The person should throw the ball at velocity = $+\frac {a\omega}{2} (\hat i + \hat j)$ This way, the initial velocity of ball would be 0 w.r.t. the ground and it will just fall downward like a free fall. Note that whether ABCD is labelled clockwise or anti-clockwise on square, it does not matter.

Do rockets on lunar missions, like Chandrayaan, take off at escape velocity? Answer Chandrayaan wants to reach the Moon, not simply escape Earth and fly off into deep space. So, the answer is no—not directly. Reaching Earth's escape velocity means having enough speed and energy to leave Earth on an unbound trajectory. But a lunar spacecraft isn't trying to escape Earth completely. It is trying to leave Earth orbit and enter a trajectory that intersects the Moon's orbit. The Moon itself orbits Earth, so even when Chandrayaan reaches the Moon, it is still within Earth's gravitational influence. Earth's gravity doesn't suddenly stop at the Moon. The spacecraft is still being pulled by Earth, while the Moon's gravity also acts on it. That's why a lunar mission is more accurately described as leaving Earth orbit, rather than escaping Earth's gravity altogether. Chandrayaan is first placed into Earth orbit. It then performs additional engine burns to raise...

$I = \int {\frac{{1 + x}}{{{e^{ - x}}{{(1 + x{e^x})}^n}}}dx = ?} ,n \ne 1$ Solution Let, $1 + x{e^x} = t$ $(x{e^x} + {e^x})dx = dt$ $ \Rightarrow {e^x}(1 + x)dx = dt$ Now, $I = \int {\frac{{{e^x}(1 + x)dx}}{{{{(1 + x{e^x})}^n}}}}  = \int {\frac{{dt}}{{{t^n}}}}  = \frac{{{t^{ - n + 1}}}}{{ - n + 1}} + C$ $\therefore I = \frac{{{{(1 + x{e^x})}^{1 - n}}}}{{1 - n}} + C$

Case (I) A video image of a man, carried by light, enters a spherical planet made of a transparent material with refractive index 𝜇 > 1. The light emerges from the diametrically opposite side of the planet, where it is viewed. The diameter of the planet is 𝑑. Case (II) The same video image is viewed after the light has travelled a distance d through vacuum. In which case will the man appear younger to the observer? Assume that the man continues to age normally while the light is travelling. Answer The key idea is that we see an object as it was when the light left it. The light slows down in case (I) so that the observer sees the image from an earlier time. Image from an earlier time means younger man. Mathematically, speed of light in case (I), $v=\frac {c}{𝜇}$ Time taken to cross diameter, $t= \frac {𝜇d}{c}$ In case (II), time taken to cover the same distance d in vacuum, $t'=\frac {d}{c}$ Since, 𝜇 > 1, t > t' So, in Case (I), the light takes longer to reach the...