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## Will the ball hit the man ?

dk_ch
Featured 2 months ago

The ball will not hit the man.

#### Explanation:

Given that

• the velocity of projection $u = 8 m \text{/} s$ at the roof edge
• the angle of projection from roof $\alpha = {40}^{\circ}$ downward.
• horizontal component of the velocity $u \cos \alpha = 8 \cos {40}^{\circ} m \text{/} s$

• vertical component of the velocity $u \sin \alpha = 8 \sin {40}^{\circ} m \text{/} s$

Let the ball reaches at the 2m vertical height after $t$ sec of its leaving the roof edge. So in t sec it covers a vertical displacement of $12 m$ from roof edge at 14m height from the ground.

So we can write
$12 = 8 \sin 40 t + \frac{1}{2} \times g \times {t}^{2}$
Taking acceleration due to gravity $g = 9.8 m \text{/} {s}^{2}$

$4.9 {t}^{2} + 8 \sin 40 t - 12 = 0$

$\implies t = \frac{- 8 \sin 40 + \sqrt{{\left(8 \sin 40\right)}^{2} + 4 \cdot 12 \cdot 4.9}}{2 \times 4.9} = 1.1 \sec$

During this time the horizontal displacement of the ball becomes

$40 \cos 40 \times t = 40 \cos 40 \times 1.1 m \approx 6.7 m$

This horizontal shift being greater than the distance (4m) of the man from the wall , the ball will not hit the man,

## Gravitational force acts on all objects in proportional to their masses. Why then, a heavy object does not fall faster than a light object?

Bill K.
Featured 1 month ago

Mathematically-speaking, the mass cancels out in the equation of motion if we assume that gravity is the only force acting (so no air resistance).

#### Explanation:

It can be experimentally observed (in a vacuum) that the acceleration due to gravity near the surface of the Earth is about $g = 9.8$ meters per second per second (in saying this I am implicitly assuming that the acceleration is independent of the object...so I'm kind of "cheating"...but I thought it was worth it so that you actually see the mass canceling out in the equation below).

If $m$ is the mass of the object, in kg, then the product $m \cdot g$ is the gravitational force on the object (in Newtons). This can also be thought of as the weight of the object.

If this is the only force acting (no air resistance), then Newton's 2nd Law says that this equals the product of the mass $m$ and the (downward) acceleration $a$. In other words, $m \cdot g = m \cdot a$. But now the value of $m$ cancels out, leaving a constant acceleration $a = g$, independent of the mass.

## Why is the symbol G there? What role does it perform?

Steve J.
Featured 1 month ago

G is the constant of proportionality.

#### Explanation:

The universal law of gravitation says that the force of attraction between 2 bodies is proportional to the product of their masses, ${m}_{1} \mathmr{and} {m}_{2}$, divided by the square of the distance, r, between them. It is not equal to that, it is proportional to that.

It is similar to the weight of a mass here on Earth. The weight is proportional to the mass. But, to get the weight in Newtons, you have to multiply by a constant of proportionality, which in that case is g, $9.8 \frac{m}{s} ^ 2$.

So to get the the force of attraction between 2 bodies in Newtons, you have to multiply $\frac{{m}_{1} \cdot {m}_{2}}{r} ^ 2$ by G.

G has both a number as part of it and a combination of units. The constant G that makes the formula come out in Newtons is
$G = 6.67 \cdot {10}^{-} 11 \frac{N \cdot {m}^{2}}{\text{kg}} ^ 2$

The units that are part of that constant are necessary to allow the units from the 2 masses and the square of the radius to be cancelled leaving Newtons as the only units on the result.

I hope this helps,
Steve

## A car enters the freeway with a speed of 5.3 m/s and accelerates uniformly for 3.6 km in 4.4 min. How fast (in m/s) is the car moving after this time?

John D.
Featured 1 month ago

${v}_{f} = 22 \text{ m/s}$

#### Explanation:

This problem gives us the initial velocity ${v}_{i}$ of the car, the time $t$ in which it accelerates, and the distance $\Delta x$ it travels. Our goal is to find the final velocity ${v}_{f}$ of the car.

Which UAM formula uses ${v}_{i}$, ${v}_{f}$, $\Delta x$, and $t$?

$\Delta x = \left(\frac{{v}_{i} + {v}_{f}}{2}\right) t$

This equation is perfect -- it gives us a way to plug in everything we know and only have our one variable ${v}_{f}$ left to solve for! So, let's do just that:

$\Delta x = \left(\frac{{v}_{i} + {v}_{f}}{2}\right) t$

#3.6 " km" = ((5.3" m/s" + v_f)/2)(4.4 " min")#

Before we continue to solve this, you may notice a few units that are out of place. We should convert $\text{km}$ to $\text{m}$, and $\text{min}$ to $\text{s}$.

$3.6 \text{ km" * (1000 " m")/"km" = 3600 " m}$

$4.4 \text{ min" * (60 " s")/"min" = 264 " s}$

Now that we have our converted values, we can continue to solve the equation.

#3600 " m" = ((5.3" m/s" + v_f)/2)(264 " s")#

#3600 " m" = (5.3" m/s" + v_f)(132 " s")#

#3600 " m" = 699.6 " m" + (132" s")(v_f)#

#2900.4 " m" = (132 " s")(v_f)#

$21.97 \text{ m/s} = {v}_{f}$

This is the final velocity of the car after the 4.4 minutes. However, since the problem only gave us an accuracy of 2 significant figures, our answer should also only have 2 significant figures.

${v}_{f} = 22 \text{ m/s}$

## What are the mathematical formulations of quantum mechanics?

Truong-Son N.
Featured 2 weeks ago

They are:

• Schrodinger formulation (wave mechanics)
• Heisenberg formulation (matrix mechanics)
• Feynman Path Integral formulation

The major differences are:

• Schrodinger formulated time-dependent wave functions and time-independent operators.
• Heisenberg formulated time-independent ket vectors and time-dependent operators.
• Feynman formulated an integration over all quantum paths, which could be described as the convolution of the Green's function #G(x,x'; t)# response to an impulse with a weight given by the stationary state $\psi \left(x , {t}_{0}\right)$.

For simplicity we do this in one dimension.

SCHRODINGER FORMULATION

The time-dependent Schrodinger equation is:

#iℏ(delPsi)/(delt) = hatHPsi#

where #Psi(x,t) = e^(-iEt//ℏ)psi(x)# is the time-dependent wave function and $\hat{H}$ is the time-independent Hamiltonian of the system.

As you can see, $\Psi = \Psi \left(x , t\right)$, but $\hat{H}$ is not time-dependent. Schrodinger postulated that:

1. The wave function $\Psi$ was time-dependent, but the operators associated with it are time-independent.
2. The wave function $\psi$ was an electric charge density spread out over allspace.
3. With the input of Max Born, $\psi$ is the probability amplitude, while ${\int}_{\text{allspace" psi^"*}} \psi d \tau$ is the probability of finding the electron somewhere.

In essence, Schroedinger was a guy who believed in wave mechanics and particle-wave duality, i.e. that quantum particles could be described by the de Broglie relation:

$\lambda = \frac{h}{m v}$

This is probably the easiest to understand out of all the formulations.

HEISENBERG FORMULATION

Heisenberg, for the life of him, saw it this way:

1. A system is described by some arbitrary ket vector #| n >>#, known as a state vector, independent of time.
2. The inner product of #| n >># with #<< x |# describing all the possible position bra vectors gives a complete set of eigenstates $\left\langlex | n\right\rangle = {\psi}_{n} \left(x\right)$.
3. The operators could be functions of time, and any time translation operator is unitary.

Under this formulation, the time-dependent Schrodinger equation becomes:

#iℏ (d << x | n >>)/(dt) = hatH << x | n >>#

From this, one could formulate the following relationships:

Overlap of definite position and momentum

#<< x | p_x >> = 1/sqrt(2piℏ) e^(ip_x x//ℏ)#

Matrix product of position and momentum with themselves

${\int}_{- \infty}^{\infty}$# | x >> << x |#$\mathrm{dx} = {\int}_{- \infty}^{\infty}$# | p_x >> << p_x |#${\mathrm{dp}}_{x}$ $\equiv 1$

Dirac Delta function

${\int}_{- \infty}^{\infty} \left\langlex | {p}_{x}\right\rangle \left\langle{p}_{x} | x '\right\rangle {\mathrm{dp}}_{x} = \left\langlex | x '\right\rangle$

#= 1/(2piℏ) int_(-oo)^(oo) e^(ip_x(x-x')//ℏ)dp_x = delta(x-x')#

${\int}_{- \infty}^{\infty} \left\langle{p}_{x} | x\right\rangle \left\langlex | {p}_{x} '\right\rangle \mathrm{dx} = \left\langle{p}_{x} | {p}_{x} '\right\rangle$

#= 1/(2piℏ) int_(-oo)^(oo) e^(ix(p_x-p_x')//ℏ)dx = delta(p_x-p_x')#

Relation between wave functions in position/momentum representations

${\phi}_{n} \left({p}_{x}\right) = \left\langle{p}_{x} | n\right\rangle$

$= {\int}_{- \infty}^{\infty} \left\langle{p}_{x} | x\right\rangle \left\langlex | n\right\rangle \mathrm{dx}$

#= 1/sqrt(2piℏ) int_(-oo)^(oo) e^(-ip_x x//ℏ) psi_n(x) dx#

${\psi}_{n} \left(x\right) = \left\langlex | n\right\rangle$

$= {\int}_{- \infty}^{\infty} \left\langlex | {p}_{x}\right\rangle \left\langle{p}_{x} | n\right\rangle {\mathrm{dp}}_{x}$

#= 1/sqrt(2piℏ) int_(-oo)^(oo) e^(ip_x x//ℏ) phi_n(p_x) dp_x#

FEYNMAN PATH INTEGRAL FORMULATION

Probably the hardest one to understand, I think... Feynman wanted to describe the transition probability amplitude of the quantum particle by summing over all possible quantum paths:

#| psi (x,t') >># $= {\int}_{- \infty}^{\infty} \left\langle\psi \left(x ' , t '\right) | \psi \left({x}_{0} , {t}_{0}\right)\right\rangle \mathrm{dx} ' \cdot$ #| psi(x',t') >>#

Under this formulation, define the unitary time translator

#hatU_t = e^(-ihatH(t-t_0)//ℏ)Theta(t-t_0)#,

where $\Theta \left(t - {t}_{0}\right) = {\int}_{- \infty}^{t} \delta \left(t ' - {t}_{0}\right) \mathrm{dt} '$ is the Heaviside step function and $\hat{H}$ is the Hamiltonian of the system.

Then a Green function can be defined in terms of ${\hat{U}}_{t}$ in position space:

#G(x,x'; t) = << x | hatU_t | x' >>#

where $x '$ is the position at which an initial impulse to the system was imparted at time ${t}_{0}$, and $x$ is the propagated position at time $t$.

The Feynman Path Integral is then the Green's function formulation of a system. For instance, if

#iℏ(delPsi)/(del t) = hatH Psi#,

then since $\frac{\partial}{\partial t} \left(\Theta \left(t - {t}_{0}\right)\right) = \delta \left(t - {t}_{0}\right)$,

#(hatH - iℏ(del)/(delt)) G(x,x'; t)#

#= (hatH - iℏ(del)/(delt)) << x | e^(-ihatH(t-t_0)//ℏ)Theta(t-t_0) | x' >>#

#= [ . . . ] = -iℏ delta(x-x')delta(t-t_0)#

As it turns out, if the Hamiltonian of a system is given by, for example,

$\hat{H} = {p}_{x}^{2} / \left(2 m\right) + V \left(\hat{x}\right)$,

then for a small enough time step $\frac{t}{N} \to \mathrm{dt}$, for a single time step ($N = 1$):

#G(x,x'; t) = [ . . . ]#

#= 1/(2piℏ) int_(-oo)^(oo) e^(-ip_x^2(t-t_0)//2mNℏ)e^(-iV(hatx)(t-t_0)//Nℏ) e^(ip_x(x-x')//ℏ) dp_x#

#= sqrt((Nm)/(2pi iℏ(t-t_0))) "exp"[(i(t-t_0))/(Nℏ) (N^2/2 m((x - x')/(t-t_0))^2 - V(x'))],#
$N = 1$

This does indeed relate back to the wave function:

#Psi_n(x,t) = int_(-oo)^(oo) underbrace(G(x,x'; t))_(<< x | hatU_t | x' >>) underbrace(psi_n(x',t_0))_(<< x' | n >>) dx'#

In a way (and this made the most sense to me):

The wave function can be given by the integral over the path described by the convolution of the impulse on the system with a weight given by the initial state.

## Kinetic energy equation in cylindrical coordinates?

Cesareo R.
Featured 2 weeks ago

$K = \frac{1}{2} m \left({\dot{r}}^{2} + {\left(\theta \dot{r}\right)}^{2} + {\dot{z}}^{2}\right)$

#### Explanation:

Considering $K = \frac{1}{2} m \left({\dot{x}}^{2} + {\dot{y}}^{2} + {\dot{z}}^{2}\right)$

and the pass relationship

$\left\{\begin{matrix}x = r \cos \theta \\ y = r \sin \theta \\ z = z\end{matrix}\right.$

and also

$\left\{\begin{matrix}\dot{x} = \cos \theta \dot{r} - r \sin \theta \dot{\theta} \\ \dot{y} = \sin \theta \dot{r} + r \cos \theta \dot{\theta} \\ \dot{z} = \dot{z}\end{matrix}\right.$

and substituting, we get

$K = \frac{1}{2} m \left({\dot{r}}^{2} + {\left(\theta \dot{r}\right)}^{2} + {\dot{z}}^{2}\right)$

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