32

Conservation of momentum implies $$mv+MV=0$$ where $m$ and $v$ are the mass and speed of the bullet, $M$ and $V$ of the rifle. Of course, it is $0$ in total as the system is at rest at the beginning. This implies $$|v|=MV/m$$ so the smaller the bullet, the bigger its speed. In terms of KE $$ke={1\over 2} m v^2={1\over 2} M^2V^2/m$$ for the bullet whereas $$...


7

In effect you are asking what the nature of kinetic energy is. After the rifle has been fired the bullet and the rifle have the same momentum - in opposite direction. In that sense we can say that momentum is equally shared. To get a closer look at what kinetic energy is I propose the following demonstration: you set up a series of ribbons that can be ...


5

I think it might help to consider a case of uniform circular motion. The OP is already aware about its general derivation through differentiation, but uniform circular motion may help develop some intuition. Consider that an object undergoes circular motion with constant angular velocity $\omega$ and thus constant speed $v$. A complete angle is $2 \pi$, so ...


5

The comments made by @Ben51 and @David White are both correct that the second answer is incorrect unless the mass $M$ is fixed, which it is not since it's on a frictionless surface, or that $M$ is infinitely large (or at least $M>>>>>m$) so that its movement after the collision is immeasurably small. But the problem does not say that either. ...


4

The bullet and rifle have the same, but oppositely directed momentum, say p. The kinetic energy of each object is $p^2/2m$. The smallest mass has the highest kinetic energy. On a pedantic note, I assumed that the rifle was at rest before it fired. For a sensible rifle this should hold. If someone holds the rifle the mass of that person should be added and ...


4

That is probably referring to the standard definition of the four-force: https://en.m.wikipedia.org/wiki/Four-force The four-force is the derivative of the four-momentum with respect to proper time. Since the four-momentum is covariant and since the proper time is a Lorentz scalar, the four-force is manifestly covariant. It is commonly regarded as the ...


4

Momentum is the quantity of motion. Word soup. You can stir them around all you like, but those words won't expain anything. The language of physics is math. $\vec{p}=m\vec{v}$ The momentum of a particle in classical mechanics, $\vec{p}$, is equal to its mass, $m$, times its velocity, $\vec{v}$. why isnt momentum and velocity the same? Because $m\vec{v}$ ...


4

You're increasing the pressure of the air on the back of your car, hence forcing it to accelerate!


3

You can write down the velocity in cm/s, or m/s, or km/h... for example, the physical velocity $v$ may be expressed as $$ v = v_{km/h} \, \frac{km}{h}= v_{m/s} \, \frac{m}{s}= v_{L/T} \, \frac{L}{T} $$ where $L$ and $T$ are generic lenght and time quantities, while $v_{L/T} $ is the "value" of $v$ when you take $L$ and $T$ as units. If you are ...


3

The question is not stupid at all. Momentum is involved in Newton's 2nd law, $$ \boldsymbol{F}=\frac{d}{dt}\left(m\boldsymbol{v}\right) $$ But more intuitively, it is a measure of the push that an object will transfer you if it hits you. Even if it hits you very slowly, a heavy object can push you very significantly due to its mass. So it makes sense that ...


3

The four-force is the only kind of force that makes physical sense. You can always obtain the three-force from the four-force by discarding the $t$ component (projecting onto the $xyz$ hyperplane) and dividing the rest by $dt/dτ$. Neither of those operations is physically sensible, because $t$ is a coordinate on the same footing as $x,y,z$, and they shouldn'...


3

You can calculate the velocity the instant the ball hits the ground using $$v_f^2 = v_i^2 + 2gh$$ You have the impulse which is given by $$F\Delta t = m\Delta v$$ This means you can actually calculate the change in momentum. The change in momentum is obviously $m(v_f - v_i)$. You should be able to do the rest. Good luck!


2

Same time interval does not mean same distance. It means the same impulse which results in the same change in momentum (in magnitude ) As you say, the momentum is conserved. On the other hand, the distances travelled by gun and bullet depend on the accelerations of the two objects and these are not the same, as you already mentioned. So the work done on ...


2

knzhou and Dale are both correct. Here is how to express it in standard notation. Four-force $$ F^a \equiv \frac{dP^a}{d\tau} $$ where $P^a$ is 4-momentum and $\tau$ is proper time. Three-force $$ {\bf f} \equiv \frac{d{\rm p}}{dt} $$ where ${\rm p}$ is three-momentum and $t$ is time in the chosen inertial frame in which the three-force is being calculated....


2

This is a meaningless argument. There are different ways of setting up relativistic dynamics. You can write your equations either in terms of the three-force $d\mathbf{p}/dt$ or the four-force $dp^\mu / d\tau$. One approach doesn’t invalidate the other, because both will give the exact same results if applied consistently. It is also not true that one ...


2

It's a good question, especially since p=mv and T = m$v^2$/2, where T is the kinetic energy, we can write T = $p^2$/2m. So it appears there is a relationship between p and T, and when one is conserved so should the other be conserved so why use both? But this relationship only deals with the magnitude of p. As has been pointed out, p is a vector, p, so has ...


2

Usually q quantity in physics that one bothered to name, it's because that quantity is very important. Force is defined in terms of momentum, $\vec F =\dfrac{ \mathrm{d} \vec p}{\mathrm d t}$. This leads to the conservation of momentum, which states that if the net external force on a system is zero, the total momentum of that system will be conserved. The ...


2

I'll do it step by step. First lower all the indices: \begin{equation} -\frac{1}{4}F_{\mu \nu}F^{\mu\nu}=-\frac{1}{4}F_{\mu \nu}g^{\mu\rho}g^{\nu\sigma}F_{\rho\sigma} \end{equation} then expand the products, \begin{equation} -\frac{1}{4}(\partial_{\mu}A_{\nu}\partial_{\rho}A_{\sigma}-\partial_{\mu}A_{\nu}\partial_{\sigma}A_{\rho}-\partial_{\nu}A_{\mu}\...


2

Here is what I tried. Let the initial velocity of the croquet ball be $v$ and the final velocity be $x$. Now, let the initial velocity of the other ball be $0$, and the final velocity be $v/2$ and the mass be $m$. Ok. This part is fine. Using the conservation of momentum, I got $0.28v + 0m = .28x +mv/2$. Now, I rewrite this as $0.28v = \frac m2v +0.28x$. ...


2

The simple explanation i can think of is that momentum describe how hard is it to stop that object. The more momentum an object have, the harder it is to stop. Momentum is not only about velocity, but also mass, it's stated in the definition of memontum $$\vec{p}=m\vec{v}$$ for example, compare a small ball with a mass of $5$ g and a big boulder with a mass ...


2

Newton laws are actually valid for a point particle. However, the bad news is that we never have point particles in real life. All of them are extended bodies. For example, the typical rectangle that represents a car moving. We've been always placing the weight on it center (it is correct), the friction next to the floor, and so on. Well, all this is correct ...


2

This is why helicopters hovering close to the ground in ground effect require less power to do so than when they are hovering high off the ground. A "bubble" of compressed air piles up in the space between the ground and the rotor blades and pushes back on the rotor blades, increasing the amount of lift. The same thing happens when you put your ...


2

Kinetic energy is not conserved when you fire a bullet. Both the rifle and bullet are at rest until the explosive in the shell provides energy to accelerate the bullet down the barrel. Momentum is, however, always conserved. If we align our coordinates with the barrel and compare the system before the shot (when the system is at rest) and just after, $$ \...


2

Simple answer From geometry, you know the length of an arc is $s = r\, \theta$ where $r$ is the radius of the arc, and $\theta$ the angle it spans. Take the time derivative of the above as the arc describes the motion of a particle riding on a rotating body and you have $$v = r \, \omega $$ where $\omega = \frac{\rm d}{{\rm d}t} \theta$. Also $r$ is the ...


2

Note that $\sin(px)=\frac{1}{2i}\left(e^{ipx}-e^{-ipx}\right)$ is a superposition of a state with positive momentum and a state with negative momentum. The average momentum is $p-p=0$. Your square well problem is a special case of this. Mathematically, the boundary conditions mean that you only get $\sin$ solutions, and not complex exponentials. Physically, ...


2

Note also that the expectation value of momentum for any real eigenfunction is zero. Also, while the accepted answer by @Andrew does give you the right intuition as to why the average momentum is zero, there is a slight subtlety due to the fact that the wavefunction isn't $\sin(px)$ over all space, but rather $\sin(px)$ within the box and zero outside it. As ...


2

How much would the astronaut swing the bat, and how much would the bat swing the astronaut? In terms of angular momentum, the bat and the astronaut will be exactly equal and opposite; and in terms of angular velocity, this will be the ratio of their moments of inertia, which I'd guess would be something like 1:1, so these would be approximately equal and in ...


1

What you call linear velocity is just the classical velocity of the test mass in m/s. The angular velocity tells you how much "angle" you cover per second when moving in a circle (so rotating). Assume you do one full rotation per second. The bigger the radius of the revolution is the faster you have to move to cover the full circumference. That is ...


1

Electrons and positrons can each have any momentum or spin. It is when you demand that the total momentum and spin of a system of electrons and positrons is, for example, zero that there is a relation. Forget the Dirac sea. As an explanation of positrons it is a flawed concept. As an explanation of positively charged holes in matter, it is a brilliant idea.


1

Breaking the vectors into components are not necessary, and may cloud the picture. Essentially, you are trying to prove that, for an elastic collision with two identical-mass particles with one initially at rest, the triangle of momentum-vectors (formed from the single nonzero initial momentum vector and final particle momenta) is a [possibly degenerate] ...


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