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M2 2021 Oct Q4

A Level / Edexcel / M2

IAL 2021 Oct Paper · Question 4

题目

Problem

4.

A particle PP of mass 0.75 kg0.75\,\mathrm{kg} is moving with velocity 4i m s−14\mathbf{i}\,\mathrm{m\,s^{-1}} when it receives an impulse J N s\mathbf{J}\,\mathrm{N\,s}. Immediately after PP receives the impulse, the speed of PP is 8 m s−18\,\mathrm{m\,s^{-1}}

Given that J=c(−i+2j)\mathbf{J}=c(-\mathbf{i}+2\mathbf{j}), where cc is a constant, find the two possible values of cc.

(6)
题目中文翻译

质量为 0.75 kg0.75\,\mathrm{kg} 的质点 PP 正以速度 4i m s−14\mathbf{i}\,\mathrm{m\,s^{-1}} 运动,随后受到冲量 J N s\mathbf{J}\,\mathrm{N\,s}。受到冲量后瞬间,PP 的速率为 8 m s−18\,\mathrm{m\,s^{-1}}。

已知 J=c(−i+2j)\mathbf{J}=c(-\mathbf{i}+2\mathbf{j}),其中 cc 为常数,求 cc 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。cc 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\mathbf v. The impulse-momentum equation gives

c(−i+2j)=34v−34(4i).c(-\mathbf i+2\mathbf j)=\frac34\mathbf v-\frac34(4\mathbf i).

Thus the final momentum is

34v=(3−c)i+2cj.\frac34\mathbf v=(3-c)\mathbf i+2c\mathbf j.

Its magnitude is (3/4)(8)=6(3/4)(8)=6 N s, so

(3−c)2+(2c)2=62.(3-c)^2+(2c)^2=6^2.

Expanding and factorising,

5c2−6c−27=0,5c^2-6c-27=0, (c−3)(5c+9)=0.(c-3)(5c+9)=0.

Therefore

c=3orc=−95.\boxed{c=3\quad\text{or}\quad c=-\frac95}.

Both signs are admissible because cc determines the direction as well as the magnitude of the impulse.