Question #213243

In 1896 in Waco,Texas,William Crush parked two locomotives at opposite ends of a 6.4-km-long track, fired them up, tied their throttles open,and then allowed them to crash head-on at full speed (Fig. 7-1) in front of 30,000 spectators. Hundreds of people were hurt by flying debris; several were killed. Assuming each locomotive weighed 1.2  106 N and its acceleration was a constant 0.26 m/s2, what was the total kinetic energy of the two locomotives just before the collision?


1
Expert's answer
2021-07-06T09:42:47-0400

We know that

Newton's second law

v2=u2+2asv^2=u^2+2as

u=0


v2=2as,v2=2×0.26×64002=1164m/secv^2=2as,v^2=2\times0.26\times\frac{6400}{2}=1164m/sec

Kinetic energy

KE=12mv2KE=\frac{1}{2}mv^2


KE=12wgv2=12(1.2×1069.8)×1664=1.019×108JKE=\frac{1}{2}\frac{w}{g}v^2=\frac{1}{2}(\frac{1.2\times10^6}{9.8})\times1664=1.019\times10^8J


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