Conservation of Mechanical Energy

실제 시험에서는 이렇게 출제됩니다

이 개념을 다룬 실제 CSCA 기출문제입니다. 답을 펼치기 전에 먼저 직접 풀어 보세요.

1. CSCA Physics December 2025 Past Paper
   Question: A basketball with mass $\mathrm { m } = 0.5 \mathrm {~kg}$ bounces from the ground to a height $\mathrm { h } = 1 \mathrm {~m}$ before falling back to the ground. The change in gravitational potential energy during this entire process is ( ). $\left( \mathrm { g } = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)$
   • A．10 J
   • B．0.5 J
   • C．5 J
   • D．0 J
   정답 및 풀이 보기

   정답: D

   1 Initial and final positions are both on the ground, so gravitational potential energy change over the whole bounce cycle is $\Delta \mathrm { U } = 0$. 2 Therefore the net change in gravitational potential energy is 0 J .
   CSCA Physics December 2025 Past Paper 전체에서 보기
2. CSCA Physics March 2026 Past Paper
   Mass $m$ is thrown upward at v from ground level, no air resistance. Mechanical energy at the highest point is (ground $=$ zero PE)
   • A．$\frac { 1 } { 2 } m v ^ { 2 }$
   • B．mgh
   • C．$1 / 2 m v ^ { 2 } + m g h$
   • D．0
   정답 및 풀이 보기

   정답: A

   No air resistance → mechanical energy conserved. At launch: $\mathrm { E } = 1 / 2 \mathrm { mv } ^ { 2 } + 0 = 1 / 2 \mathrm { mv } ^ { 2 }$. At top: $\mathrm { KE } = 0$, $\mathrm { PE } = \mathrm { mgh }$, but total E still $= 1 / 2 \mathrm { mv } ^ { 2 }$. B and C are not correct expressions; D is clearly wrong. Answer: A．
   CSCA Physics March 2026 Past Paper 전체 풀기 — 시간 측정 및 채점
3. CSCA Physics April 2026 Past Paper
   A 0.2 kg ball falls freely from a height of 2 m to the ground. The change in its gravitational potential energy is $\left( g = 10 \mathrm {~m} / \mathrm { s } ^ { 2 } \right)$
   • A．0 J
   • B．- 4 J
   • C．4 J
   • D．2 J
   정답 및 풀이 보기

   정답: B

   As the ball descends, its gravitational PE decreases. $\Delta E _ { p } = - m g h = - 0.2 \times 10 \times 2 = - 4 \mathrm {~J}$.
   CSCA Physics April 2026 Past Paper 전체 풀기 — 시간 측정 및 채점