本题目来源于试卷:
2007美国US F=MA物理竞赛，类别为
美国F=MA物理竞赛
[单选题]
A baseball is dropped on t
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* kq;jmzijo 0i/3s n(c a basketball. The basketball hits the ground, rebou
0/zkni3c(jj ;q som*i nds with a speed of $4.0\,\mathrm{m/s}$, and collides with the baseball as it is moving downward at $4.0\,\mathrm{m/s}$. After the collision, the baseball moves upward as shown in the figure and the basketball is instantaneously at rest right after the collision. The mass of the baseball is $0.2\,\mathrm{kg}$ and the mass of the basketball is $0.5\,\mathrm{kg}$. Ignore air resistance and ignore any changes in velocities due to gravity during the very short collision times. The speed of the baseball right after colliding with the upward moving basketball is
A. $4.0\,\mathrm{m/s}$
B. $6.0\,\mathrm{m/s}$
C. $8.0\,\mathrm{m/s}$
D. $12.0\,\mathrm{m/s}$
E. $16.0\,\mathrm{m/s}$
参考答案: B
本题详细解析:
Use conservation of linear momentum for the (2 r .sf-cyp/1huhs9cpbaseball + basketball) system during their collision.
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Let $m_b = 0.2\,\mathrm{kg}$ and $m_{bk} = 0.5\,\mathrm{kg}$. Let the "upward" direction be positive.
Initial state (just before collision):
$v_{b,i} = -4.0\,\mathrm{m/s}$ (baseball moving down)
$v_{bk,i} = +4.0\,\mathrm{m/s}$ (basketball moving up)
Final state (just after collision):
$v_{b,f} = ?$ (baseball final velocity)
$v_{bk,f} = 0\,\mathrm{m/s}$ (basketball at rest)
Conservation of momentum $p_i = p_f$:
$m_b v_{b,i} + m_{bk} v_{bk,i} = m_b v_{b,f} + m_{bk} v_{bk,f}$
$(0.2\,\mathrm{kg})(-4.0\,\mathrm{m/s}) + (0.5\,\mathrm{kg})(+4.0\,\mathrm{m/s}) = (0.2\,\mathrm{kg})v_{b,f} + (0.5\,\mathrm{kg})(0)$
$-0.8 + 2.0 = 0.2 v_{b,f}$
$1.2 = 0.2 v_{b,f}$
$v_{b,f} = 1.2 / 0.2 = 6.0\,\mathrm{m/s}$.
The speed is the magnitude, $6.0\,\mathrm{m/s}$.
