Trying to calculate the forces in a rugby scrum

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SUMMARY

The discussion focuses on calculating the forces involved in a rugby scrum, specifically between two teams weighing 850kg and 910kg. The players accelerate at an estimated rate of 2 m/s² over a distance of 0.5 meters before impact, resulting in a final speed of approximately 1.4 m/s (5 km/h) at the moment of collision. The calculations utilize fundamental physics principles, including F=ma and the equations of motion, to derive these values. The analysis highlights the complexity of accurately modeling the dynamics of a rugby scrum compared to simpler scenarios like a truck collision.

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Zacski
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Hi there, my wife just asked me how hard the impact is when two sides of a rugby scrum come together. I think she's looking for a "like when a truck hits a slightly smaller truck" answer, but now I'm really interested. We're watching Japan v New Zealand at the world cup and the two scrum weights are 850kg and 910kg respectively. They stand a metre apart (at least according to the rules) and so each pack accelerates over half a metre then come to a relatively complete stop on impact.

I understand F=ma and the conservation of momentum, but I trip up when I try to work out the rate of acceleration and the final speed reached at point of impact.
 
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According to https://sportsscientists.com/2008/08/beijing-2008-men-100m-race-analysis/a top athlete running the 100-m dash covers the first 10 meters in 1.85 s. Assuming constant acceleration from zero initial speed, you have
$$d=\frac{1}{2}at^2~\rightarrow~a=\frac{2d}{t^2}=\frac{2\times 10~\rm{m}}{1.85^2 \rm{s^2}}=5.8 \rm{m/s^2}$$
Of course the rugby players are not top Olympic athletes starting from a sprinter's stance and furthermore they all start moving as one. So let's guess that their acceleration is about, ##2~\rm{m/s^2}##.
Using the equation relating distance, acceleration and speed you can find the final speed after moving by 0.5 m:
$$2a\Delta x=v^2-0^2~\rightarrow~v=\sqrt{2a\Delta x}=\sqrt{2\times 2~\rm{m/s^2}\times 0.5~\rm{m}}=1.4~\rm{m/s}=5~\rm{km/h}.$$
 

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