178x Filetype PPTX File size 1.24 MB Source: www.animatedscience.co.uk
Momentum m = 5 kg A rocket of mass 5 kg is travelling u = 200 ms-1 -1 horizontally with a speed of 200 ms when it m = 3 kg explodes into two parts. One part of mass 3 1 m = m – m = 5 – 3 = 2 kg kg continues in the original direction with a 2 1 speed of 100 ms-1. The other part also v = 100 ms-1 1 continues in this same direction. Calculate v = v its speed. 2 2 momentum before the explosion = momentum after the explosion mu = m v + m v 1 1 2 2 (5 x 200) = (3 x 100) + (2 x v ) 2 1000 = 300 + 2v2 1000 – 300 = 2v2 700 = 2v2 v = 700/2 2 v = 350 ms-1 2 Animated Science 2016 D (mv-mu )/t so kgms-2 W = mg or F=ma A Cons of Momentum 0 = m v – m v 1 1 2 2 Animated Science 2015 3 A If 100% elastic momentum is passed through to T so T 2 1 must be at rest. As same mass the velocity is unaltered Animated Science 2015 Ft = mv-mu (mv-mu) /F = t mv/F = t 584.6 x 10-6 A p = momentum = mv = m/V V = m p = (V) * v -4 3 -4 -1 -1 v = 2 x 10 m / 7.2 x 10 ms = 0.277ms -3 -4 3 -1 -1 -1 1000kgm x 2.0 x 10 m x 0.277ms = 0.05555 kgms = 0.056 kgms B
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