Physics Guru

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...

If $a\sin x = b\sin \left( {x + \frac{{2\pi }}{3}} \right) = c\sin \left( {x + \frac{{4\pi }}{3}} \right) \neq 0$, prove that $(a+b+c)^2=a^2+b^2+c^2$. Solution Dividing by $abc$ ($\because a,b,c \neq 0$), $\frac{{\sin x}}{{bc}} = \frac{{\sin \left( {x + \frac{{2\pi }}{3}} \right)}}{{ca}} = \frac{{\sin \left( {x + \frac{{4\pi }}{3}} \right)}}{{ab}} = \frac{{\sin x + \sin \left( {x + \frac{{2\pi }}{3}} \right) + \sin \left( {x + \frac{{4\pi }}{3}} \right)}}{{bc + ca + ab}}$ $\therefore (ab + bc + ca)\sin x = bc\left\{ {\sin x + \underbrace {\sin \left( {x + \frac{{2\pi }}{3}} \right) + \sin \left( {x + \frac{{4\pi }}{3}} \right)}_{}} \right\}$ $ \Rightarrow (ab + bc + ca)\sin x = bc\left\{ {\sin x + 2\sin \left( {x + \pi } \right)\cos \left( {\frac{\pi }{3}} \right)} \right\}$ $ \Rightarrow (ab + bc + ca)\sin x = bc\left\{ {\sin x - 2\sin x.\frac{1}{2}} \right\} = 0$ $\therefore ab + bc + ca = 0 \because \sin x \ne 0$ Now, ${(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2(ab + bc + ca) = {a^2...

A small drone is launched from a helicopter flying at a height of 720 m and moving horizontally at a constant velocity of 180 km/h. The velocity of the drone is 90 km/h with respect to the helicopter having same direction as that of helicopter and remains constant until it loses power after flying for 4 minutes. How far away from its launching point will the drone land? [Ignore air resistance. Take $g=10 m/s^2$.] Solution Velocity of drone w.r.t. the ground when it is launched=180+90=270 kmph Horizontal distance covered in 4 minutes = 270 kmph . 4 min. = 18 km Time to fall t after it loses power can be obtained from the equation, $720 m = \frac {1}{2} gt^2$ $144 = t^2$ t = 12 sec Horizontal distance covered after it loses power = 270 kmph. 12 sec = 0.9 km Total horizontal distance = 18 km + 0.9 km = 18.9 km