题目

Problem

Two particles, $A$ and $B$ , have masses $2m$ and $3m$ respectively. They are moving towards each other in opposite directions along the same straight line on a smooth horizontal plane when they collide directly. Immediately before they collide, the speed of $A$ is $3u$ and the speed of $B$ is $u$ . As a result of the collision, the speed of $A$ is halved and the direction of motion of each particle is reversed.

(i) Find the speed of $B$ immediately after the collision.

(ii) Find the magnitude of the impulse exerted on $A$ by $B$ in the collision.

(6)

题目中文翻译

两个质点 $A$ 和 $B$ 的质量分别为 $2m$ 和 $3m$ 。它们在光滑水平面上沿同一直线相向运动，并发生正碰。碰撞前一瞬间， $A$ 的速率为 $3u$ ， $B$ 的速率为 $u$ 。碰撞后， $A$ 的速率减半，且两个质点的运动方向均反转。

(i) 求碰撞后瞬间 $B$ 的速率。

(ii) 求碰撞过程中 $B$ 对 $A$ 施加的冲量的大小。

(6)

解答

(i)

解法一

思路

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先选定统一正方向。碰撞后两球均反向，因此 $A$ 的末速度必须带负号；水平方向无外冲量，可以用总动量守恒。

答题过程

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Take the initial direction of motion of $A$ as positive. The velocities before impact are $3u$ and $-u$ ; after impact they are $-3u/2$ and $v$ , where $v>0$ .

Conservation of momentum gives

\begin{align*} 2m(3u)+3m(-u)=&\,2m\big(-\tfrac32u\big)+3mv,\\[4mm] 3mu=&\,-3mu+3mv,\\[4mm] 6mu=&\,3mv. \end{align*}

Hence the speed of $B$ immediately after impact is $\boxed{2u}$ .

(ii)

解法一

思路

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冲量等于有方向的动量变化。反向运动时，应将末速度减去带符号的初速度，最后再取大小。

答题过程

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Using the same positive direction, the impulse on $A$ is

\begin{align*} I_A=&\,2m\big(-\tfrac32u-3u\big)\\[4mm] =&\,-9mu. \end{align*}

Thus the required magnitude is $\boxed{9mu}$ , opposite to the original motion of $A$ .

解法二

思路

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也可以计算 $B$ 的动量变化，再用牛顿第三定律：两球相互作用的冲量大小相等、方向相反。

答题过程

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From (i), the velocity of $B$ changes from $-u$ to $2u$ . Hence

I_B=3m\big(2u-(-u)\big)=9mu.

The impulse on $A$ is equal and opposite to that on $B$ , so its magnitude is $\boxed{9mu}$ .

M1 2019 Jan Q1

题目

解答

(i)

解法一

思路

答题过程

(ii)

解法一

思路

答题过程

解法二

思路

答题过程