题目
Problem
4.
A particle P P P 0.3 k g 0.3\,\mathrm{kg} 0.3 kg 5 m s − 1 5\,\mathrm{m\,s^{-1}} 5 m s − 1 J N s J\,\mathrm{N\,s} J N s P P P 8 m s − 1 8\,\mathrm{m\,s^{-1}} 8 m s − 1 P P P 60 ∘ 60^\circ 6 0 ∘
Find the value of J J J
(6)
题目中文翻译
质量为 0.3 k g 0.3\,\mathrm{kg} 0.3 kg P P P 5 m s − 1 5\,\mathrm{m\,s^{-1}} 5 m s − 1 J N s J\,\mathrm{N\,s} J N s P P P 8 m s − 1 8\,\mathrm{m\,s^{-1}} 8 m s − 1 P P P 60 ∘ 60^\circ 6 0 ∘
求 J J J
(6)
解答
解法一
思路
展开
取初速度方向为一个坐标方向，冲量与它成 60 ∘ 60^\circ 6 0 ∘
答题过程
展开
Let the magnitude of the impulse be J J J
p 0 = 0.3 ( 5 ) = 1.5 , p 1 = 0.3 ( 8 ) = 2.4. p_0=0.3(5)=1.5,\qquad p_1=0.3(8)=2.4. p 0 = 0.3 ( 5 ) = 1.5 , p 1 = 0.3 ( 8 ) = 2.4. Resolve parallel and perpendicular to the initial velocity. From impulse equals change in momentum, the final momentum components are
1.5 + J cos 60 ∘ , J sin 60 ∘ . 1.5+J\cos60^\circ,\qquad J\sin60^\circ. 1.5 + J cos 6 0 ∘ , J sin 6 0 ∘ . Hence
2.4 2 = ( 1.5 + J 2 ) 2 + ( 3 J 2 ) 2 . 2.4^2=\left(1.5+\frac J2\right)^2+\left(\frac{\sqrt3J}{2}\right)^2. 2. 4 2 = ( 1.5 + 2 J ) 2 + ( 2 3 J ) 2 . Simplifying,
J 2 + 1.5 J − 3.51 = 0. J^2+1.5J-3.51=0. J 2 + 1.5 J − 3.51 = 0. Since an impulse magnitude is positive,
J = − 1.5 + 1.5 2 + 4 ( 3.51 ) 2 . J=\frac{-1.5+\sqrt{1.5^2+4(3.51)}}2. J = 2 − 1.5 + 1. 5 2 + 4 ( 3.51 ) . Therefore
J = 1.27 N s . \boxed{J=1.27\text{ N s}}. J = 1.27 N s .
解法二
思路
展开
也可用动量三角形：初动量与冲量首尾相接，合向量是末动量。三角形的内角是 120 ∘ 120^\circ 12 0 ∘ 60 ∘ 60^\circ 6 0 ∘
答题过程
展开
The vector relation is
p 1 = p 0 + J . \mathbf p_1=\mathbf p_0+\mathbf J. p 1 = p 0 + J . Place J \mathbf J J p 0 \mathbf p_0 p 0 180 ∘ − 60 ∘ = 120 ∘ 180^\circ-60^\circ=120^\circ 18 0 ∘ − 6 0 ∘ = 12 0 ∘
The cosine rule gives
2.4 2 = 1.5 2 + J 2 − 2 ( 1.5 ) J cos 120 ∘ . 2.4^2=1.5^2+J^2-2(1.5)J\cos120^\circ. 2. 4 2 = 1. 5 2 + J 2 − 2 ( 1.5 ) J cos 12 0 ∘ . Thus
J 2 + 1.5 J − 3.51 = 0. J^2+1.5J-3.51=0. J 2 + 1.5 J − 3.51 = 0. Taking the positive root,
J = 1.27 N s . \boxed{J=1.27\text{ N s}}. J = 1.27 N s . 
