()()()()()112 215 kg 3.0 m/s10 kg2.0 m/s(5 kg)10 kg 2.5 m/siifffmvv+=++=+GGGGGwhich yields 12.0 m/sfv=G. Thus, the speed of the 5.0 kg block immediately after the collision is 20.ms.(b) We find the reduction in total kinetic energy: ()()()22225 kg 3 m/s10 kg 2 m/s5 kg 2 m/s10 kg 2.5 m/s1.25 J 1.3 J.ifKK−=+−−=−≈−(c) In this new scenario where Gvf240=. m s , momentum conservation leads to Gvf110. m s and we obtain 40 JKΔ=+.(d) The creation of additional kinetic energy is possible if, say, some gunpowder were on
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