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# Ball at rest. Need physics help. - Page 2

Quote

Quote:
Originally Posted by Rulesman

The object (ball) was not at rest, it was moving and therefore had kinetic energy. It also had potential energy in its elasticity (the coefficient of restitution indicates to what extent this is utilised by the rebound).

Non-technical.

1) A body on a horizontal plane is travelling along a straight line from A to B.

2)It encounters something solid (B)

3) and then travels in the opposite direction towards A (ie rebounds).

In step 1 it was moving towards B. In step 3 it is moving towards A. In what direction is it moving in step 2?

Technical

## Ball bouncing from massive wall.

Most physics textbooks consider the case of a ball bouncing from a massive object, say the floor, or a wall. They consider the case where the collision is nearly or totally elastic. In the totally elastic collision, the ball loses no kinetic energy in the collision, so its speed after collision is the same as before the collision. The student thinks, "Of course that must be the case, because of conservation of energy." Seldom does the textbook, or the student, consider how conservation of momentum is satisfied in this problem. They should, for the analysis is instructive. Analyzing this may also give some insight into why energy, momentum and the conservation laws took so long to be formulated, since the concepts are subtle.

Consider a ball of mass m colliding elastically with a stationary object of larger mass M. Draw the picture before and after the collision. The conservation equations are:

 [1]

mv1 = mv2 + MV2

 [2]

(1/2)mv12 = (1/2)mv22 + (1/2)MV2

where v1 is the initial velocity of the smaller ball, v2 is its final velocity after collision, and V2 is the velocity of the larger mass after the collision.

Multiply the energy equation by 2 to eliminate the (1/2) factors.

 [3]

mv12 = mv22 + MV2

Divide this by m on both sides.

 [4]

v12 = v22 + (M/m)V22

Rearrange.

 [5]

v12 - v22 = (M/m)V22

Divide the momentum equation by m on both sides.

 [6]

v1 = v2 + (M/m)V2

Rearrange and square both sides.

 [7]

(v1 - v2)2 = (M/m)2V22

Multiply Eq. [5] by (M/m) and combine with [7] to eliminate V22.

 [8]

(M/m)(v12 - v22) = (v1 - v2)2

Multiply both sides by (m/M)

 [9]

(v12 - v22) = (m/M)(v1 - v2)2

Take the limit as (m/M) goes to zero.

 [10]

(v12 - v22) = 0

So one solution of this is v1 = -v2. Another solution is v1 = v2, corresponding to the case where the two objects do not collide at all.

One can graph the values of v2 and V2 against (m/M) and show that as (m/M) goes to zero, the values of the final velocities do indeed smoothly go to the limiting case values. In words, the reason this can happen is that kinetic energy is a scalar, and momentum is a vector, and kinetic energy varies as the square of the speed, while momentum varies as only the first power of speed. Therefore the quantity momentum/energy varies with speed as (1/v). So when v goes to zero, the momentum/energy can be infinite.

Those who have had calculus will recognize that this is a case where an indeterminate form arises when you take the limit of the value of momentum of a body whose mass increases to infinity.

Do elastic collisions actually exist in reality? Anywhere on earth? atomic or sub-atomic level?

So, if I throw tennis ball at a wall, obviously the ball pushes against the wall and wall pushes against the ball?

so obviously, wall has more momentum, after having collided with the ball, compared to the wall momentum before having collided with the ball? (ball's force against wall, and wall's force against ball)

But of course this is beside the point indeed.

so it seems sensible, from Newtonian point of view, that ball's momentum before impact, is the same as ball's momentum after impact + wall's momentum after impact.

What causes elastic collisions? The elastic quantity of a collision, the efficiency if you will?

Obviously it would make sense, the ball actually pushes the wall back, but the effect is almost unmeasurable. Wall is built upon foundations to the ground etc...

If you played golf and the surface of the course was completely elastic, always providing elastic collision with a bouncing ball, I suppose your teeshot with a driver would never stop bouncing?

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Quote:
Originally Posted by Rulesman

Why?

The ball is not at rest. Any talk of a ball changing direction is essentially attempting to pervert the definition of "at rest" to consider only a moment that is infinitely short.

Quote:
Originally Posted by iacas

Why?

The ball is not at rest. Any talk of a ball changing direction is essentially attempting to pervert the definition of "at rest" to consider only a moment that is infinitely short.

The link related to the point in the previous post about a drive bouncing down the course.

A ball which drops vertically and rebounds vertically is at rest for a very small amount of finite time.

I'm afraid the physics or mathematical mechanics is too complicated and lengthy to post here.

Quote:
Originally Posted by Rulesman

#### Problem:

A 0.1 kg ball is thrown straight up into the air with an initial speed of 15 m/s.  Find the momentum of the ball
(a) at its maximum height and
(b) half way up to its maximum height.

Solution:
(a) At the ball's maximum height its velocity is zero, and therefore its momentum is zero.

(b) The ball's total energy E = K + U is constant.
Initially U = 0 and K = (1/2)mv=(0.05 kg)(15 m/s)= 11.25 J.
At its maximum height K = 0 and U = mgh = 11.25 J.
Halfway up to its maximum height K = U = (1/2)11.25 J.
Therefore v= 2K/m = (11.25 J)/(0.1 kg) = 112.5 (m/s)2
v = 10.6 m/s, p =( 0.1 kg)(10.6 m/s) = 1.06 kgm/s

(a) At the ball's maximum height its velocity is zero,

This is of course the same for any object which reverses its direction.

Only as long as the rebound vector is exactly 180° from the original trajectory.  If there is any deviation from that perfect case, then the ball is never completely at rest, because it will always still carry some angular momentum.

Quote:
Originally Posted by Fourputt

Only as long as the rebound vector is exactly 180° from the original trajectory.  If there is any deviation from that perfect case, then the ball is never completely at rest, because it will always still carry some angular momentum.

And that is what I was saying when I wrote "The laws of physics says it could possibly be. But only in the cases described above".

In fact the case you mentioned

Quote:
Originally Posted by Fourputt

Only as long as the rebound vector is exactly 180° from the original trajectory.  If there is any deviation from that perfect case, then the ball is never completely at rest, because it will always still carry some angular momentum.

That also ignores any spin the ball may have.

Either way, c'mon, this isn't physics class. We all know what "at rest" means, so let's stop pretending it means for an infinitely small period of time.

A golf ball is on a train to Chicago traveling at 60 mph.  If Chicago is 1003 miles away......

Quote:
Originally Posted by boogielicious

A golf ball is on a train to Chicago traveling at 60 mph.  If Chicago is 1003 miles away......

Quote:
Originally Posted by Missouri Swede

Oh Christ! Don't get that shit started again!!!

Everyone must now take 125 yards of their drive because
at the apex of the ball' s flight a photograph of it being stationary can be taken. OK.
This is officially the stupidist thread ever, and it's not even interesting from a semantic standpoint.
It is not a scientific or hypothetical debate/discussion. And don't pretend that it is. The OP emphasised that he NEEDs physics help.
I think he got that right. ;-/
He even stated that the rules state something that they don't state.

Is that Schrodinger's cat in your avatar?

Quote:
Originally Posted by boogielicious

Is that Schrodinger's cat in your avatar?

Very clever. You get points for that.

Quote:
Originally Posted by Fourputt

Only as long as the rebound vector is exactly 180° from the original trajectory.  If there is any deviation from that perfect case, then the ball is never completely at rest, because it will always still carry some angular momentum.

I agree with this and this is the reason I deleted my initial response, which cited the Intermediate Value Theorem.  Just because something is at rest with respect to one of the coordinates of 3-space doesn't mean it is necessarily at rest with respect to the other coordinates.

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