New answers tagged momentum
0
I think I have figured out the answer, hopefully.
Firstly, let us begin by stating the First Law:
$$\sum \vec{F} = \frac{\delta \vec{p}}{\delta t}$$
When the net force is constant, we it means that there is no change of momentum, in other words,
$$ \frac{\delta \vec{p}}{\delta t} = 0$$
In this case, we know that the function of momentum is a constant, ...
2
This problem has a recursive flavor that we'll not try to avoid.
Conservation of momentum tells us that
$$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$
Imposing the boundary condition $v(0)=0$ we find
$$v(n) = \frac{n}{n+p}v_0$$
as claimed.
Let $a_n$ be the time at which the $n$th bullet strike occurs.
We have $a_1=x_0/v_0$ and
$$v_0 (a_n - T) = v_0 ...
-2
The time taken between the N-1 collision and the N collision is $T-T\frac{N-1}{p}=T\frac{p-1+N}{p}$
Edit:
Reasoning:
The difference in T is due to the N-1 collision and is given by: $T\frac{N-1}{p}$
0
This equation works but for those components of velocities in direction of contact of two bodies i.e in the direction of forces they exert on each other,in the direction perpendicular to the force the velocities won't change.
0
In regards to the current question, for a single charge going in a loop, i = qf. f = 1/(2pi x sq root(LC)), w = 1/(sq root LC) = V/R. After the corresponding substitutions, i = qV/2piR
1
You're not doing anything wrong, the objects will have different momenta in different reference frames. What should be the same in every reference frame is the forces acting on the objects during the collision. The laws of physics are the same in every reference frame, but not necessarily the numbers that go into the equations.
By way of example, lets ...
1
The problem is that your frame of reference, if you put it in an object that is accelerating is not an inertial one; the discrepancies are due to inertial forces that you're not taking into account. Why don't you just observe things from a fixed, absolute, inertial reference frame, instead?
3
Not much sense. Your "center of charge" is nothing but the dipole moment divided by the net total charge. "Normalised dipole moment, if you will".
If you take $q|\vec v|$ instead of $q\vec v$, you get something related to current (generally current times a factor). Current is conserved at a junction.
Regarding your equal-and-opposite situation, the closest ...
0
When a particle is deflected by gravity the gravitational field will also be modified by the particle. To form a conservation law for momentum you need to take into account the momentum in the gravitational field as well as the particle. This can be done e.g. using pseudo-tensor methods.
This works but remember that momentum is a relative concept. Even in ...
1
energy is always positive or 0. And these are just numbers we associate with a body due to its motion according to different set of mathematical rules so that we can study these particles . As such they have no physical meaning . For example momentum is just $m$ x $\vec v$ . It is just a number we associated with a body by the quantities we defined ourselves ...
6
Imagine that you have just two particles with the same mass and same speed, but going in opposite directions. They have opposite momenta, so the total momentum is zero. But they each have energy, and the total energy is not zero. The reason is because kinetic energy is just $\frac{1}{2} m v^2$. That square means that the kinetic energy can never be ...
0
The cart and the ramp have to be able to move at different speeds from one another, otherwise the cart could never move up the ramp or back down it again. So they can't both have the same velocity at all times. However, the centre of mass of the system always keeps moving with the same velocity in the $x$ direction (both before and after the collision), and ...
0
With your current assumptions you do not have enough equations to solve this problem, since it is two dimensional, which gives you 4 unknown variables: $u_{1}',v_{1}',u_{1}',v_{2}'$ where $u$ and $v$ are the speeds in the respectively $x$ and $y$ direction, the indexes $1$ and $2$ indicate which object it is and the apostrophe ($'$) indicates that these ...
0
Here's a formula to help you find final velocities from initial velocities:
$$v_1=\frac{u_1(m_1-m_2)+2m_2u_2}{m_1+m_2}$$
$$v_2=\frac{u_2(m_2-m_1)+2m_1u_1}{m_1+m_2}$$
These formulas you get from combining momentum and energy equations. You have to apply both of the above formulas separately in 2 dimensions: $x$ and $y$.
So you should get ...
4
Notice that
\begin{align}
i\frac{d(\psi^*\psi)}{dx}
&=\frac{d\big[(-i\psi)^*\psi\big]}{dx} \\
&= \frac{d(-i\psi)^*}{dx}\psi + (-i\psi)^*\frac{d\psi}{dx} \\
&= \left(-i\frac{d\psi}{dx}\right)^*\psi + \psi^*\left(i\frac{d\psi}{dx}\right) \\
\end{align}
Now subtract the second term on the right from both sides to get
\begin{align}
...
1
Ashish already essentially said this, but using only the equations of motion we can show conservation of total momentum with the following calculation:
\begin{align}
\frac{d}{dt}(m_1\dot{\vec x_1} +m_2\dot{\vec x_2})
&= m_1\ddot{\vec x_1} +m_2\ddot{\vec x_2} \\
&= m_1\left(- G m_2 \frac{\vec{x}_1-\vec{x}_2}{|\vec{x}_1-\vec{x}_2|^3}\right) + ...
1
OK, the first part: net force $\vec{F}=m_1 \ddot{\vec{x}}_1+m_2 \ddot{\vec{x}}_2=\vec{0}$, so by newton's second law total momentum is conserved(or force is rate of change of momentum, if total force is zero, total momentum doesn't change), and the second part: there is no sort of non conservative force acting, hence mechanical energy is conserved.
Non ...
0
There are two things in this context :
First taking k =1 is OK as long as you use F = ma to define unit of F as 1 Newton. (All proportionality constants such as the one in Coulomb law are not taken to be unity.)
When done as above, k can be taken as unit-less But if you opt to use another force equation such as Hookes law (F = C.x to define unit of F, ...
1
Brief explanation:
When going from classical Lagrangian (say, non-relativistic point-)mechanics to quantum mechanics, there is an intermediate step known as classical Hamiltonian mechanics.
To reach the intermediate step, one has to perform a Legendre transformation $(q,\dot{q}) \longrightarrow (q, p)$, where $(q, p)$ are (generalized) canonical phase ...
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