Day: January 30, 2024

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202401301506 Solution to 2000-CE-PHY-II-28

Two insulated uncharged metal spheres X and Y are placed in contact. A positively-charged rod is brought near X as shown below. X is then earthed momentarily. The charged rod is removed and the two spheres are then separated. Describe the charges on X and Y. (modified)

Roughwork.

Have some mental pictures (*make drawings if you don’t have ample RAM).

First, when a positively-charged rod is brought near X, charges are induced:

Notice that on surface areas X and Y making contact, as boxed below:

the net charge is zero, Q=0. Hence, the plus and the minus signs (representing positive and negative charges) are erased from the drawing:

Then, X is earthed momentarily:

it is convenient to treat the two spheres X and Y as one body X+Y:

The process of earthing enables a transfer of negative free charges (electrons being the carrier) unidirectionally between two bodies:

here from the earth to the body, but no reversed (why?); the plus and minus signs, as boxed below:

should cancel off each other, and are thus erased from the drawing:

once the charged rod is removed:

the negative charges will be redistributed over X+Y such that the electrostatic repulsive forces between them are kept to a minimum:

after the two spheres are separated,

both spheres X and Y will be negative charged. The descriptions go complete.

This problem is not to be attempted.

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202401300926 Exercise 81.b.1

A particle is projected at an angle of 45^\circ with the horizon from a point on a horizontal plane, with a velocity \textrm{1,000} feet per second. Find its range, and find its distance from the point of projection at the end of \textrm{5} seconds.

Gage, Alfred Payson. (1887). The High School Physics.

Roughwork.

Draw a picture for the scenario.

We set up the notation system below:

\begin{aligned} \textrm{Displacement} &:\mathbf{s}(x(t),y(t))=s_x\,\hat{\mathbf{i}}+s_y\,\hat{\mathbf{j}} \\ \textrm{Velocity} &: \mathbf{v}(t)=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}} \\ \textrm{Acceleration} &: \mathbf{a}=-10\,\mathrm{m\,s^{-1}}\,\hat{\mathbf{j}} \\ \end{aligned}

and resolve them into their components:

\begin{aligned} v^2(t) & = v_x^2(t)+v_y^2(t) \\ v_x(t_0) & = v_y(t_0)=\frac{\sqrt{2}}{2}v(t_0) \\ v_y(t) & = (v_y(t_0))+(- a)(t)\\ s(t) & = s_x^2(t) + s_y^2(t) \\ s_x(t) & = (v_x(t_0))(t) \\ s_y(t) & = h+(v_y(t_0))(t)+\frac{1}{2}(-a)(t)^2 \end{aligned}

its range being s_x(t=5) and its distance from the point of projection s(t=5)=|\mathbf{s}(t=5)| at the end of \textrm{5} seconds.

This problem is not to be attempted.