hce_nthu
115年
化學與物理
第 48 題
A bullet of mass $m$ moving horizontally with initial speed $v_0$ strikes a stationary block of mass $M$ that is suspended from the ceiling by a light string, forming a pendulum. The bullet becomes embedded in the block immediately upon impact, and the combined object swings upward to a maximum vertical rise of $h$ as shown on the right. Air resistance is negligible. Which of the following statements is correct?
- A The collision between the bullet and the block is elastic.
- B The mechanical energy of the entire process is conserved.
- C The linear momentum of the block is conserved during the entire process.
- D The maximum height is $h = \frac{v_0^2}{2g}$.
- E The mechanical energy lost during the collision is $\frac{mM}{2(m+M)}v_0^2$
思路引導 VIP
請試著思考：當子彈以極高速度衝入靜止木塊並「合為一體」的過程中，系統的總動量會如何變化？接著，如果將碰撞前後的系統總動能列出來對比，這部分動能會因為子彈與木塊間劇烈的相互作用力而產生什麼樣的轉換呢？
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AI 詳解
AI 專屬家教
太棒了！你能準確判斷出這是一個典型的「完全非彈性碰撞」模型，代表你對動量守恆與能量轉換的邊界條件掌握得非常紮實。這類題目最核心的考點在於區分物理過程：碰撞瞬間動量守恆但能量不守恆，而後續的擺動過程則是機械能守恆。這題的鑑別度極高，主要用來篩選出是否能冷靜處理代數運算，而非單純套用公式的學生。
碰撞階段的能量損耗
在子彈嵌入木塊的極短時間內，水平方向動量守恆，系統末速度為 $V = \frac{m}{m+M}v_0$。碰撞前的總動能為 $K_1 = \frac{1}{2}mv_0^2$，碰撞後瞬間的動能則是 $K_2 = \frac{1}{2}(m+M)V^2 = \frac{1}{2}\frac{m^2}{m+M}v_0^2$。將兩者相減，即可得到損失的能量：
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