Elastic collision between ball and block

[TSny]

I'm not following your statement that v grows by 2v+1. If m << M, then the values of v for the first few collisions are approximately 2, 4, 6, 8, 10, etc. if the block was initially moving at 1 unit of speed. And I don't understand how you are getting some of your other numbers.

The following table shows what I get for the exact closest distance of approach and the approximate distance of closest approach according to L_closest ≈ √(m/M), assuming the first collision occurs at 1 unit of distance from the wall:

$$ \frac{M}{m} = 3 \;\;\;\;\;\;\; L^{closest}_{exact} = 0.50000 \;\;\;\;\;\; L^{closest}_{approx} = 0.577$$ $$ \frac{M}{m} = 25 \;\;\;\;\;\; L^{closest}_{exact} = 0.19967 \;\;\;\;\;\; L^{closest}_{approx} = 0.200$$ $$ \frac{M}{m} = 100 \;\;\;\; L^{closest}_{exact}= 0.09979 \;\;\;\;\;\; L^{closest}_{approx} = 0.100$$ $$ \frac{M}{m} = 10000 \;\;\; L^{closest}_{exact}= 0.0099995 \;\;\; L^{closest}_{approx} = 0.010$$

 

[bobie]

I'm not following your statement,$$ \frac{M}{m} = 3 \;\;\;\;\;\;\; L^{closest}_{exact} = 0.50000 \;\;\;\;\;\; L^{closest}_{approx} = 0.577$$

The KE parameter is not reliable as it's unlikely that M ever stops, when M reaches its closest distance, if still has a lot of positive Ke that at next clash is turned into negative
If M/m =3 the closest L is 1/2, L_closest = .5 m after the initial collision, even if M has still 1/2 V_{0 and therefore} 1/4 of its KE. At the first real clash M bounces back.
Isn' that so?
as to 2v+1 you are right, it's v+2!

So , we must conclude that if a 10-ton-steel-slab is running at 100 Km/h and meets a 1-gr/3cm-pingpongball on its way at 10 m from a wall , the ball will not be crushed but it will stop the slab

 

[TSny]

The KE parameter is not reliable as it's unlikely that M ever stops, when M reaches its closest distance, if still has a lot of positive Ke that at next clash is turned into negative
If M/m =3 the closest L is 1/2, L_closest = .5 m after the initial collision, even if M has still 1/2 V_{0 and therefore} 1/4 of its KE. At the first real clash M bounces back.
Isn' that so?

Yes, that's right. But when m << M, the block will be approximately at rest at the point of closest approach and essentially all of the KE will be in the ball.

So , we must conclude that if a 10-ton-steel-slab is running at 100 Km/h and meets a 1-gr/3cm-pingpongball on its way at 10 m from a wall , the ball will not be crushed but it will stop the slab

Let's assume it's 10 metric tons, so M/m = 10⁷. If the first collision takes place 100 m from the wall, then the closest the slab will make it to the wall is about 3.16 cm from the wall. The initial speed of the slab does not affect this result. (If the initial distance is only 10 m, then the ball will get crushed.)

If the initial speed of the slab is 100 km/h, then at the distance of closest approach, the ball will have a speed of about 316,000 km/h.

 

[bobie]

Thanks for your attention, TSny.