Tag Info

5

Regardless of the physical undefinability of "painfulness", I'd like to plug some numbers in a particular scenario: Let's have a momentum of $p = 1000$m$\cdot$kg/s, A 0.25kg bullet would be fatal, moving at $v = 1000/0.25 = 4000$m/s, while a 2000kg car moves at $v=0.5$m/s, So at least in this scenario and particularly for relatively low momentum ...

3

This is an answer to the question version 1. Later versions invalidate the details of this answer, but some of the ideas are still valid. Will edit to current version if I get a chance. I'll define "pain" as the change in momentum, or the energy delivered (the two are related by your velocity after the impact, provided your mass is unchanged, so unless you ...

3

Momentum is conserved in magnitude and direction. So in order to analyze any situation of momentum conservation, you should always start with $$\sum \mathbf p_{i}=\sum\mathbf p_f$$ where the subscripts denote the initial and final momenta. As to the ball & wall, you are correct that momentum is not conserved if you are only looking at the ball. If you ...

2

In inelastic collisions, kinetic energy is not conserved, so I'm going to assume you mean a totally elastic collision since you say energy is conserved. O.K, so when the ball hits the wall, the speed of the wall before and after is 0, so that means the kinetic energy of the ball is conserved and thus the magnitude of the velocity is the same before and after ...

2

For calculations of a collision, you look at (kinetic) energy and momentum equations. Momentum is conserved; energy may be dissipated. There is a direct exchange of momentum in the form of $F\Delta t$ - the same impulse that slows one object down accelerates the other, by the same amount. But the force may result in a deformation of the ball such that ...

2

Let us first calculate how much energy carried by an object of mass $m$ and momentum $p$. So the velocity is $v=p/m$, and therefore, the kinetic energy is $T=mv^2/2=p^2/2m$ Therefore, if $p$ is a constant, the heavier the object is, the less the kinetic energy $T$ it carries. This means that a heavier object hits you and you receive less energy, which ...

2

It is the momentum of the entire system that is conserved. The fundamental reason for this is that the laws of physics are the same everywhere in space. This argument for momentum conservation is called Noether's Theorem. So where did you go wrong in your original example? Well you assumed that the wall was completely rigid. In reality that isn't actually ...

1

Coefficient of restitution and hardness are not the same thing. The COR basically tells us how much energy gets lost in the collision process. In the case of a soft/hard ball with identical v1/v2 the COR=1, i.e. there is no energy loss. However, the collision of a perfectly elastic ball with a perfect wall will take a different amount of time, depending on ...

Only top voted, non community-wiki answers of a minimum length are eligible