题目
Problem
2.
[In this question, i \mathbf{i} i j \mathbf{j} j ]
A particle Q Q Q 0.5 k g 0.5\,\mathrm{kg} 0.5 kg Q Q Q ( 3 i + j ) m s − 1 (3\mathbf{i}+\mathbf{j})\,\mathrm{m\,s^{-1}} ( 3 i + j ) m s − 1 ( 2 i + 5 j ) N s (2\mathbf{i}+5\mathbf{j})\,\mathrm{N\,s} ( 2 i + 5 j ) N s
(a) Find the speed of Q Q Q
(4)
As a result of receiving the impulse, the direction of motion of Q Q Q θ ∘ \theta^\circ θ ∘
(b) Find the value of θ \theta θ
(2)
题目中文翻译
[ 本题中，i \mathbf{i} i j \mathbf{j} j ]
质量为 0.5 k g 0.5\,\mathrm{kg} 0.5 kg Q Q Q Q Q Q ( 3 i + j ) m s − 1 (3\mathbf{i}+\mathbf{j})\,\mathrm{m\,s^{-1}} ( 3 i + j ) m s − 1 ( 2 i + 5 j ) N s (2\mathbf{i}+5\mathbf{j})\,\mathrm{N\,s} ( 2 i + 5 j ) N s
(a) 求 Q Q Q
(4)
受到冲量使 Q Q Q θ ∘ \theta^\circ θ ∘
(b) 求 θ \theta θ
(2)
解答
(a)
解法一
思路
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冲量等于质量乘速度变化，先求末速度向量，再取模得到速率。
答题过程
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By impulse-momentum,
2 i + 5 j = 0.5 [ v − ( 3 i + j ) ] . 2\mathbf i+5\mathbf j=0.5\bigl[\mathbf v-(3\mathbf i+\mathbf j)\bigr]. 2 i + 5 j = 0.5 [ v − ( 3 i + j ) ] . Thus
v = ( 3 i + j ) + 2 ( 2 i + 5 j ) = 7 i + 11 j . \mathbf v=(3\mathbf i+\mathbf j)+2(2\mathbf i+5\mathbf j)=7\mathbf i+11\mathbf j. v = ( 3 i + j ) + 2 ( 2 i + 5 j ) = 7 i + 11 j . The speed is therefore
∣ v ∣ = 7 2 + 11 2 = 170 m s − 1 . \boxed{|\mathbf v|=\sqrt{7^2+11^2}=\sqrt{170}\text{ m s}^{-1}}. ∣ v ∣ = 7 2 + 1 1 2 = 170 m s − 1 .
(b)
解法一
思路
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初末速度都在第一象限，偏转角等于两方向角之差。注意求的是两速度之间的夹角，不是末速度对坐标轴的角。
答题过程
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Both velocity vectors lie in the first quadrant. Their angles to positive i \mathbf i i
tan − 1 ( 1 3 ) , tan − 1 ( 11 7 ) . \tan^{-1}\left(\frac13\right),\qquad \tan^{-1}\left(\frac{11}{7}\right). tan − 1 ( 3 1 ) , tan − 1 ( 7 11 ) . Hence the turning angle is
θ = tan − 1 ( 11 7 ) − tan − 1 ( 1 3 ) , \theta=\tan^{-1}\left(\frac{11}{7}\right)-\tan^{-1}\left(\frac13\right), θ = tan − 1 ( 7 11 ) − tan − 1 ( 3 1 ) , and therefore
θ = 39.1 ∘ . \boxed{\theta=39.1^\circ}. θ = 39. 1 ∘ .
解法二
思路
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也可用数量积直接求两速度的夹角，无须分别找方向角。
答题过程
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With u = 3 i + j \mathbf u=3\mathbf i+\mathbf j u = 3 i + j v = 7 i + 11 j \mathbf v=7\mathbf i+11\mathbf j v = 7 i + 11 j
u ⋅ v = 3 ( 7 ) + 1 ( 11 ) = 32. \mathbf u\cdot\mathbf v=3(7)+1(11)=32. u ⋅ v = 3 ( 7 ) + 1 ( 11 ) = 32. Therefore
cos θ = u ⋅ v ∣ u ∣ ∣ v ∣ = 32 10 170 . \cos\theta=\frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|}=\frac{32}{\sqrt{10}\sqrt{170}}. cos θ = ∣ u ∣∣ v ∣ u ⋅ v = 10 170 32 . Thus
θ = 39.1 ∘ . \boxed{\theta=39.1^\circ}. θ = 39. 1 ∘ . 