The diagram shows the vertical cross-section of a surface. $$\(A, B\)$$ and $$\(C\)$$ are three points on the cross- section. The level of $$\(B\)$$ is $$\(h \mathrm{~m}\)$$ above the level of $$\(A\)$$. The level of $$\(C\)$$ is $$\(0.5 \mathrm{~m}\)$$ below the level of $$\(A\)$$. A particle of mass $$\(0.2 \mathrm{~kg}\)$$ is projected up the slope from $$\(A\)$$ with initial speed $$\(5 \mathrm{~m} \mathrm{~s}^{-1}\)$$. The particle remains in contact with the surface as it travels from $$\(A\)$$ to $$\(C\)$$. Given that the particle reaches $$\(B\)$$ with a speed of $$\(3 \mathrm{~m} \mathrm{~s}^{-1}\)$$ and that there is no resistance force, find $$\(h\)$$. .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... .................................................................................................................................................... ....................................................................................................................................................
Exam No：9709_m20_qp_42 Year：2020 Question No：3(a)
Answer:
Initial \(\mathrm{KE}=1 / 2 \times 0.2 \times 5^{2}\)
or Final KE \(=1 / 2 \times 0.2 \times 3^{2}\)
\(1 / 2 \times 0.2 \times 5^{2}=0.2 g h+1 / 2 \times 0.2 \times 3^{2}\)
\(h=0.8\)
or Final KE \(=1 / 2 \times 0.2 \times 3^{2}\)
\(1 / 2 \times 0.2 \times 5^{2}=0.2 g h+1 / 2 \times 0.2 \times 3^{2}\)
\(h=0.8\)
Knowledge points:
4.5.2 understand the concepts of gravitational potential energy and kinetic energy, and use appropriate formulae
4.5.3 understand and use the relationship between the change in energy of a system and the work done by the external forces, and use in appropriate cases the principle of conservation of energy Including cases where the motion may not be linear (e.g. a child on a smooth curved ‘slide’), where only overall energy changes need to be considered.
Solution:
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