题目
Problem
4.
A particle P of mass 0.75kg is moving with velocity 4ims−1 when it receives an impulse JNs. Immediately after P receives the impulse, the speed of P is 8ms−1
Given that J=c(−i+2j), where c is a constant, find the two possible values of c.
(6)
题目中文翻译
质量为 0.75kg 的质点 P 正以速度 4ims−1 运动,随后受到冲量 JNs。受到冲量后瞬间,P 的速率为 8ms−1。
已知 J=c(−i+2j),其中 c 为常数,求 c 的两个可能值。
(6)
解答
解法一
思路
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冲量等于动量变化。先把末动量写成两个分量,再用末速率给出的模列方程。c 是有符号常数,不是冲量的大小,因此负根不能舍去。
答题过程
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Let the final velocity be v. The impulse-momentum equation gives
c(−i+2j)=43v−43(4i).
Thus the final momentum is
43v=(3−c)i+2cj.
Its magnitude is (3/4)(8)=6 N s, so
(3−c)2+(2c)2=62.
Expanding and factorising,
5c2−6c−27=0,
(c−3)(5c+9)=0.
Therefore
c=3orc=−59.
Both signs are admissible because c determines the direction as well as the magnitude of the impulse.