# Tag Info

6

You're confusing the acceleration of your car with the acceleration in a collision. You actually have to look at it "backwards" from what you've described above. That is, in the collision you don't do a $F = ma$ calculation where $a$ is the acceleration of your gas pedal. Instead in the collision you have a force $F$ resulting from the collision and you ...

4

Another way to think about Newton's second law (and the way he originally defined it) is $F=\dfrac{d\rho}{dt}$, where $\rho=mv$ is momentum and $\dfrac{d\rho}{dt}$ is the rate of change of momentum. I think you meant to say that the obstacle will exert a force on you - and that is correct. If you could calculate your change in velocity, and the amount of ...

3

Most gamma rays from $pp$ collisions come from neutral pions ($p+p\to p+p+\pi^0$), you'd first have to do some relativistic momentum & energy conservation to determine the energy of the neutral pion. It's easiest if you consider the two subsequent reactions: $$p+p\to p+\Delta^+ \\ \Delta^+\to p+\pi^{0}$$ (it's up to you to figure out the kinematics). ...

2

The window and the force have to have the same force but on each other. This is dictated by Newton's third Law of Motion. The rock exerted a force on the window which was not enough to break it. The window, in turn, exerted an equal and opposite force on the rock which was quite enough to make it reverse its direction of motion. This is what is happening in ...

2

The most general relationship is $$c(b) = \frac{\int_0^b \frac{\mathrm{d}\sigma}{\mathrm{d}b}\mathrm{d}b}{\int_0^\infty \frac{\mathrm{d}\sigma}{\mathrm{d}b}\mathrm{d}b} = \frac{1}{\sigma_\text{inel}}\int_0^b \frac{\mathrm{d}\sigma}{\mathrm{d}b}\mathrm{d}b\tag{1}$$ (source, one of many). In practice, we usually use the Glauber model to describe heavy ion ...

2

2 dimensional collision can be reduced to a 1-dimensional problem in the case of spheres--see here. The $\pm$ you encounter when solving the kinetic energy is likely because there are two solutions and the equations are satisfied by either one. One solution is simply where the particles pass right through eachother, which you can discard.

1

Pairs of charged particles and/or objects attract via the $Q_1Q_2/R^2$ Coulomb's law. This is a classical approximation that quantifies how their velocities are changing when the objects are large or distances are much longer than the Compton wavelength etc. When the particles get really close, there are new effects that are neglected by the laws of ...

1

Positrons can be easily produced in pair production reactions, when gamma rays with energy more than 1 MeV interact in the field of a nucleus, so there is no problem in producing them at accelerator sites or even from decay products of reactor cores. For LEP in particular the positrons were generated by using an electron beam hitting a target that would ...

1

Since you are dealing with an inelastic collision, energy is not conserved when the bullet hits the block. You should try to find a relation between the initial velocity of the bullet and the velocity of the combined system (bullet+block) after the collision from conservation of momentum.

1

When you hit the obstacle (if you don't destroy it) your speed goes down to zero in a quite small time. That gives you the acceleration. It starts when you start hitting and it ends when you come to a complete stop. This acceleration is due to the force that the obstacle generates on the car. Think about: \$\displaystyle a=\frac{\Delta v}{\Delta ...

1

When I read about this question, I was thinking, well, "a" can't be zero because it should represent your deceleration from 900 miles/hour. The resulting F would be the force you experience when you decelerate. (The deceleration depends on how you hit the obstacle, how long did it take you to stop/slow, across what distance, and the degree of deformity of ...

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