The discussion focuses on the dynamics of a cannon firing bullets into a log, specifically analyzing the log's velocity after multiple bullets impact. The log, with a mass of 100m0, experiences a change in velocity after each bullet, calculated using conservation of momentum. Key equations derived include the log's velocity after N bullets, expressed as v_N = (N * v0) / (100 + N), and the time elapsed between bullet impacts, which is T_2 = (t0 * (N + 99)) / 100. The analysis also addresses the heat energy generated in the log and the relationship between the number of bullets and the log's velocity.
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If you take the sum from 1 to N-1Karol said:The sum should indeed be from 2 to N since, i think, for example if i have 3 bullets then i have to add 2 intervals:
$$\sum_2^N \frac{N+99}{100}=\sum_2^N N+(N-1)99=\frac{(N-1)(N+2)}{2}+99(N-1)=\frac{(n-1)(N+200)}{200}$$
But it's strange that if i take the sum from 1 to (N-1) it gives a different thing:
$$\sum_1^{N-1} \frac{N+99}{100}=\frac{N(N+197)}{200}$$
I thought only the number of members in the sum counts, but no.