Physics Guru

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 $\equiv (ut, 0, 0)$.

A small decorative 50 mW LED lamp is connected to a 3.0 V DC button cell. The circuit is left open, with a 2.0 cm gap between two exposed copper terminals inside a terracotta diya. The LED is mounted at the top of the diya, while the button cell is hidden at the bottom. Person X proposes pouring ordinary tap water into the diya so that the water bridges the gap between the two terminals. He says that distilled water is generally considered safe but is of no use here. Person Y says tap water causes a short circuit. So, pouring it would burn the circuit. Who is correct? Answer Usually, it is considered that water and electricity do not play well together, and water is a no-no as far as electricity is concerned. This perception has a lot to do with high voltages, including domestic supply voltages, and the dangers posed by water during the rainy season or in bathrooms. In this problem, however, we are dealing with a low-voltage, small-current situation. Tap water is not a short circuit in...

In the circuit shown, the battery is connected across AB. Will the drift velocity of electrons be maximum in the smallest resistance $R$ as there would be least resistance for the electrons to move in $R$? (all resistors are made of wires with identical cross sectional area and identical material) Answer The given circuit is actually balanced Wheatstone bridge as $\frac {4R}{2R} = \frac {8R}{4R}$ which means no current flows through the resistance $R$. So, even though the resistance $R$ is smallest, the drift velocity is not going to be maximum in $R$. In fact the drift velocity would be zero in $R$.  In the top we would have $2R+4R=6R$ and in the bottom we would have $4R+8R=12R$. $6R$ and $12R$ would be in parallel, so the higher current would flow through $6R$. So, the current would be maximum in $2R$ and $4R$ both.  Now, $I=nAev_d$ So, drift velocity is same for resistive wires having identical current, cross-sectional area and identical material. Hence, the drift velocity ...

Q: Why does the axis of rotation of a spinning top tilt further away from the vertical after some time? (Neglect air friction. Assume real-life situation otherwise.) A: Friction at the pivot point drains the top's rotational kinetic energy, causing its spin speed to drop over time. The spin angular momentum $L=Iω$ therefore decreases as the top slows down. Reduced angular momentum means less resistance to a change in the direction of the axis. In a real situation, there are always small moments when the axis is not absolutely vertical. In such instances, gravity exerts a torque that tends to tip the top over. The top responds to this torque by precessing—the axis rotates around the vertical. As the spin speed decreases, the rate of precession increases and the motion becomes increasingly pronounced. As the top tilts farther, the gravitational torque also increases because the perpendicular distance from the pivot to the line of action of gravity increases. With less angular momentu...

Consider a charged parallel plate capacitor without dielectric. An electric field exists directed from the positive plate toward the negative plate. When a dielectric slab is brought in, its molecular response depends on its nature: If the molecules are non-polar, the external field displaces their electron clouds toward the positive plate, inducing dipole moments. If the molecules are already polar, the existing dipoles rotate to align with the field, with the negative end of each dipole orienting toward the positive plate and the positive end toward the negative plate. After this reorientation, the dielectric can be visualized as a chain of aligned dipoles. A layer of bound negative charge accumulates adjacent to the positive plate, and a layer of bound positive charge accumulates adjacent to the negative plate. These bound charges produce an internal electric field within the dielectric that opposes the original field. The superposition of the two fields results in a reduced net ele...

A large horizontal platform, open at the top, moves vertically upward with a constant velocity $v$. A small object is launched from the platform at an angle of $\theta = 30^\circ$ with a launch velocity $u$ (measured relative to the platform). Determine the height above the launch point at which the object strikes the platform. (Ignore air resistance.) Solution We can solve this problem by analyzing the motion in the stationary ground reference frame as a function of time $t$. Step 1: Position of the Platform Since the platform moves upward with a constant velocity $v$, its vertical position $z_p$ at any time $t$ is given by: $$z_p = vt$$ Step 2: Position of the Object The initial vertical velocity of the object in the ground frame is the sum of the platform's velocity and the vertical component of the object's launch velocity: $u \sin\theta + v$. Accounting for the acceleration due to gravity $g$, the vertical position $z_o$ of the object at time $t$ is: $$z_o = (u...

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 $\vec r_2$...