题目
Problem
1.
A particle of mass 0.5 k g 0.5\,\mathrm{kg} 0.5 kg ( 2 i + 4 j ) m s − 1 (2\mathbf{i}+4\mathbf{j})\,\mathrm{m\,s^{-1}} ( 2 i + 4 j ) m s − 1 ( − 4 i + 6 j ) N s (-4\mathbf{i}+6\mathbf{j})\,\mathrm{N\,s} ( − 4 i + 6 j ) N s
(a) Find the speed of the particle immediately after it receives the impulse.
(5)
(b) Find the size of the angle between the direction of motion of the particle immediately before it receives the impulse and the direction of motion of the particle immediately after it receives the impulse.
(3)
题目中文翻译
质量为 0.5 k g 0.5\,\mathrm{kg} 0.5 kg ( 2 i + 4 j ) m s − 1 (2\mathbf{i}+4\mathbf{j})\,\mathrm{m\,s^{-1}} ( 2 i + 4 j ) m s − 1 ( − 4 i + 6 j ) N s (-4\mathbf{i}+6\mathbf{j})\,\mathrm{N\,s} ( − 4 i + 6 j ) N s
(a) 求质点受到冲量后瞬间的速率。
(5)
(b) 求质点受到冲量前瞬间的运动方向与受到冲量后瞬间的运动方向之间的夹角。
(3)
解答
(a)
解法一
思路
展开
冲量等于动量变化，先求末速度向量，再取模得到速率。只写向量还没有回答本问。
答题过程
展开
Let the velocity immediately after the impulse be v \mathbf v v
− 4 i + 6 j = 0.5 [ v − ( 2 i + 4 j ) ] . -4\mathbf i+6\mathbf j=0.5\bigl[\mathbf v-(2\mathbf i+4\mathbf j)\bigr]. − 4 i + 6 j = 0.5 [ v − ( 2 i + 4 j ) ] . Therefore
v = ( 2 i + 4 j ) + 2 ( − 4 i + 6 j ) = − 6 i + 16 j . \mathbf v=(2\mathbf i+4\mathbf j)+2(-4\mathbf i+6\mathbf j)=-6\mathbf i+16\mathbf j. v = ( 2 i + 4 j ) + 2 ( − 4 i + 6 j ) = − 6 i + 16 j . The required speed is
∣ v ∣ = ( − 6 ) 2 + 16 2 = 2 73 m s − 1 . \boxed{|\mathbf v|=\sqrt{(-6)^2+16^2}=2\sqrt{73}\text{ m s}^{-1}}. ∣ v ∣ = ( − 6 ) 2 + 1 6 2 = 2 73 m s − 1 .
(b)
解法一
思路
展开
两方向的夹角可由初、末速度的数量积直接求得，不必分别处理它们所在的象限。
答题过程
展开
Let the angle between the directions be θ \theta θ u = 2 i + 4 j \mathbf u=2\mathbf i+4\mathbf j u = 2 i + 4 j
cos θ = u ⋅ v ∣ u ∣ ∣ v ∣ = 2 ( − 6 ) + 4 ( 16 ) 20 292 . \cos\theta=\frac{\mathbf u\cdot\mathbf v}{|\mathbf u||\mathbf v|}=\frac{2(-6)+4(16)}{\sqrt{20}\sqrt{292}}. cos θ = ∣ u ∣∣ v ∣ u ⋅ v = 20 292 2 ( − 6 ) + 4 ( 16 ) . Thus
θ = cos − 1 ( 52 20 292 ) . \theta=\cos^{-1}\left(\frac{52}{\sqrt{20}\sqrt{292}}\right). θ = cos − 1 ( 20 292 52 ) . Hence the required angle is
47.1 ∘ . \boxed{47.1^\circ}. 47. 1 ∘ .
解法二
思路
展开
也可分别求两速度相对于正水平轴的方向角。末速度在第二象限，不能直接把反正切的锐角当作方向角。
答题过程
展开
The initial velocity is in the first quadrant, making angle
ϕ = tan − 1 4 2 \phi=\tan^{-1}\frac42 ϕ = tan − 1 2 4 with the positive i \mathbf i i
ψ = 180 ∘ − tan − 1 16 6 . \psi=180^\circ-\tan^{-1}\frac{16}{6}. ψ = 18 0 ∘ − tan − 1 6 16 . The smaller angle between them is therefore
θ = ψ − ϕ = 180 ∘ − tan − 1 16 6 − tan − 1 2. \theta=\psi-\phi=180^\circ-\tan^{-1}\frac{16}{6}-\tan^{-1}2. θ = ψ − ϕ = 18 0 ∘ − tan − 1 6 16 − tan − 1 2. Hence
θ = 47.1 ∘ . \boxed{\theta=47.1^\circ}. θ = 47. 1 ∘ . 