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/sci/ - Science & Math

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4230956 No.4230956 [Reply] [Original]

I was taught that for an object striking a wall in a perfectly elastic collision with momentum mv, (1) the object will return with momentum 2mv. (2) Now I'm wondering why and I had an idea: that it's to do with (3). I thought maybe the object exerts a force on the wall, giving the wall, and anything it's attached to, gains momentum mv and moves slightly, while at the same time exerting a force back on the object, giving the object its momentum mv, and having the overall effect of having each object move in opposite directions which, for an observer in a position where the wall appears not to move, the object is moving with 2mv away?

If so, how does this conserve momentum? Surely we've gone from 1mv-> to the right to having 1mv both left and right, making 0mv overall, and not conserving momentum?

>> No.4230966

The change of momentum is 2mv but it returns with mv

>> No.4230976

>>4230956
That should read:

exerts a force on the wall, giving the wall, and anything it's attached to, momentum mv and moving it slightly

>> No.4230992

kinetic energy is conserved

ergo: Ek init = Ek final

so the ball will lose velocity and the wall will gain velocity

>> No.4230996

>>4230992
Yes, but this should occur in an inelastic collision too. Momentum will still be conserved.

>> No.4231004

>>4230996

no, it won't. In a non-elastic collision, energy is lost as heat, sound, anything.

assuming we're tlaking about truly elastic collisions here

>> No.4231016

>>4231004
I guess if energy is lost some momentum will be lost, but anyway - assume it's elastic - it seems like momentum is not being conserved here and I can't work out how it works.

>> No.4231028

>>4231016

I just told you.
so, for objects A and B, with momentum P

Initially, p = pA = mv
after collision, p = pA + pB
so clearly pA must decrease in order for pB to exist, thus conserving momentum

>> No.4231030

>>4230992
>EK

>> No.4231040

>>4231030

Ek = 1/2 mv^2 = 1/2 mp

the two are interconnected, clearly, you fuck

>> No.4231042

>>4231040
>doesn't get reference to EK

>> No.4231046

>>4230956
Good job, you are actually thinking about physics.

In reality the wall would have to move, the thing is that since the wall is attached to the earth and the mass is so large compared to that of the ball, any velocity the wall gains would be so infinitely small it is not worth doing the mathematics over.

So yes the wall does move, but it is almost unnoticeable, and likewise the ball doesn't exactly go back up to the same velocity, but it is so close that it is not worth making a distinction over.

>> No.4231047

>>4230956
YOU WERE TAUGHT WRONG.

\thread

>> No.4231050

>>4231028
But I was told that it doesn't decrease, I was told that the momentum of the object striking the wall doubles.

>> No.4231058

>>4231050
Right, so if we assume the whole 2mv coming back thing is wrong, and it's as simple as >>4231046, that's it?

>> No.4231077

>>4231050
your teacher is bad at teaching. The wall is a metaphor for an object with an extremely large mass.

So take for instance if you had a 1 kg ball going 5 m/s, and a wall that was 10,000kg. in an elastic collision, the wall would be going .001 m/s, and the ball would be going 4.999 m/s the other way. it is just that when assuming the wall is immovable(infinite mass), the velocity of the wall goes to 0, so the velocity of the ball goes to 5

>> No.4231080
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4231080

>>4231050
YOU WERE TAUGHT WRONG! Your teacher is either shit or YOU DIDN'T ACTUALLY UNDERSTAND WHAT HE/SHE WAS TRYING TO TELL YOU.

\thread

>> No.4231083

>>4231077
Yeah, to be honest I understood that already, which is why I came to /sci/ as >>4231080 is probably right, I probably misunderstood a point my lecturer was making. He's a PhD so it's probably my fault.

>> No.4231151

The wall is fixed to the ground fixed to the earth.

The earth gains this momentum.