本题目来源于试卷: 2012美国US F=MA物理竞赛，类别为 美国F=MA物理竞赛
[单选题]
A $12\,\mathrm{kg}$ block moving east at $4\,\mathrm{m/s}$ collides head on with a $6\,\mathrm{kg}$ block that is moving west at $2\,\mathrm{m/s}$. The two blocks move together after the collision. What is the loss in kinetic energy in this collision?
A. $36\,\mathrm{J}$
B. $48\,\mathrm{J}$
C. $60\,\mathrm{J}$
D. $72\,\mathrm{J}$
E. $96\,\mathrm{J}$
参考答案: D
本题详细解析:
This is a perfectly inelastic collision.
py0 ;hdp+ df-l1. **Find final velocity ($v_f$)**: Use conservation of momentum (let east be positive).
$p_i = (12\,\mathrm{kg})(+4\,\mathrm{m/s}) + (6\,\mathrm{kg})(-2\,\mathrm{m/s}) = 48 - 12 = 36\,\mathrm{kg \cdot m/s}$.
$p_f = (m_1 + m_2)v_f = (12 + 6)v_f = 18v_f$.
$p_i = p_f \implies 36 = 18v_f \implies v_f = 2\,\mathrm{m/s}$.
2. **Find $KE_i$**:
$KE_i = \frac{1}{2}(12)(4)^2 + \frac{1}{2}(6)(-2)^2 = 6(16) + 3(4) = 96 + 12 = 108\,\mathrm{J}$.
3. **Find $KE_f$**:
$KE_f = \frac{1}{2}(18)(2)^2 = 9(4) = 36\,\mathrm{J}$.
4. **Find Loss**:
$KE_{loss} = KE_i - KE_f = 108\,\mathrm{J} - 36\,\mathrm{J} = 72\,\mathrm{J}$.
| 
