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Published byHayden MacGregor Modified over 3 years ago

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Q 3 Revisited Why did the two different methods produce different results?

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The question was inconsistent! Initial momentum was kgms -1 One of the momentum vectors was at an angle of 50 degrees. The other vector was at 90 degrees to that one. Now the triangle is fixed! All quantities can be calculated. There is no freedom to make up results.

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Calculate what the velocities should have been. Initial momentum was kgms degrees. 90 degrees The masses have a mass of kg. 40 degrees

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Trigonometry. Initial momentum was kgms degrees. 90 degrees Sin 40 = p 2 / p 2 = x sin 40 = kgms degrees p2p2 p1p1 Cos 40 = p 1 / p 1 = x cos 40 = kgms -1 Pythagoras : sqrt of ( ) = – Agrees!!

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Velocities p 2 = kgms -1 = 0.065kg x V 2 V 2 = p 2 / = ms -1 = 1.2 ms -1 (2 s.f) p 1 = kgms -1 = 0.065kg x V 1 V 1 = p 1 / = ms -1 = 1.4 ms -1 (2 s.f)

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