Explaining Newton's First Law & the Dropped Weight Trolley

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pkc111 said:
OK, so wouldn't a loosely held wall (exaggerated example - on ice) react differently to a tightly held wall ?

My explanation isn't phrased correctly, arildino's is phrased much better, I'm going to have to work on my explanations. :frown:

~H
 
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arildno said:
Momentum is definitely NOT conserved, neither for the tennis ball system, or the tennis ball+brick wall system.

In both cases, non-negligible external forces acts upon the system:
1. In the tennis ball system, there is a huge external force acting upon it from the brick wall, effectively REVERSING, not conserving, the ball's momentum.

2. In the wall+ball system, the force from the ground keeps the wall at rest.
This force must balance the ball's huge collision force upon the wall, and is therefore not negligible to the magnitude of the internal forces in the collision.
Hence, momentum conservation does not apply here either.

ENERGY, however, might well be conserved; in that case, we have an elastic collision.

Thanks arildno, that makes sense.

I guess if you think of a fixed wall as an extension of the Earth then momentum may be conserved in the tennis ball + wall/earth system but the resulting velocity of the wall/earth would only be undetectably small for conservation to occur ?
 
In the ball+EARTH system, momentum is, indeed, conserved (neglecting any external forces from stars and such).
 
pkc111 said:
Thanks arildno, that makes sense.

I guess if you think of a fixed wall as an extension of the Earth then momentum may be conserved in the tennis ball + wall/earth system but the resulting velocity of the wall/earth would only be undetectably small for conservation to occur ?

Yes, if you consider the Earth as part of the system, then you are correct, but as you stated only the tenis ball and the wall is part of the 'system', hence momentum for that system is not conserved.

~H

Edit: arildno has got there before me :smile:
 
Thanks again guys, nite nite.:smile:
 
As a note, in the elastic collision between an object A (approaching B with velocity V) and B (at rest initially) where external forces are negligible, the final velocities are:
For Object A: (m-M)/(M+m)V, For B: 2mV/(M+m)
where m is the mass of A and M the mass of B.

Thus, if M>>m, Object A will effectively reverse its velocity, whereas B remains practically at rest.