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M1 2025 Oct Q2

A Level / Edexcel / M1

IAL 2025 Oct Paper · Question 2

题目

Problem

2.

Particle PP of mass 7m7m and particle QQ of mass 3m3m are moving in opposite directions along the same straight line on a smooth horizontal surface. The particles collide directly.

Immediately before the collision, the speed of PP is kuku and the speed of QQ is uu, as shown in Figure 2.

Immediately after the collision, the speed of PP is ww and the speed of QQ is 2w2w.

The direction of motion of QQ is reversed by the collision.

The impulse received by QQ in the collision has magnitude 72mu\frac72mu.

(a) Find ww in terms of uu.

(3)

(b) Find the two possible values of kk.

(5)
题目中文翻译

质量为 7m7m 的质点 PP 和质量为 3m3m 的质点 QQ 在光滑水平面上沿同一直线相向运动,并发生正碰。

如图 2 所示,碰撞前一瞬间,PP 的速率为 kuku,QQ 的速率为 uu。

碰撞后瞬间,PP 的速率为 ww,QQ 的速率为 2w2w。

碰撞使 QQ 的运动方向反转。

碰撞过程中 QQ 所受冲量的大小为 72mu\frac72mu。

(a) 用 uu 表示 ww。

(3)

(b) 求 kk 的两个可能值。

(5)

解答

(a)

解法一

思路

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以 PP 碰撞前的运动方向为正方向。QQ 反向运动,因此它的速度由 −u-u 变成 2w2w;冲量等于有符号的动量变化。

答题过程

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Take the initial direction of motion of PP as positive. For QQ,

I= 3m(2w−(−u)),72mu= 3m(2w+u),12u= 6w.\begin{align*} I=&\,3m\bigl(2w-(-u)\bigr),\\ \frac72mu=&\,3m(2w+u),\\ \frac12u=&\,6w. \end{align*}

Therefore

w=u12.\boxed{w=\frac{u}{12}}.

(b)

解法一

思路

展开

题目给出的是 PP 的速率,并未指定碰撞后的方向,因此分别考虑速度为 ww 与 −w-w 的情况。两球所受碰撞冲量大小相等、方向相反。

答题过程

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The impulse on PP is −72mu-\frac72mu. If its velocity after the collision is vPv_P, then

7m(vP−ku)= −72mu,vP= ku−12u.\begin{align*} 7m(v_P-ku)=&\,-\frac72mu,\\ v_P=&\,ku-\frac12u. \end{align*}

Since the speed of PP after the collision is w=u/12w=u/12, there are two cases.

If PP continues in its original direction,

u12=ku−u2⟹k=712.\frac{u}{12}=ku-\frac{u}{2} \quad\Longrightarrow\quad k=\frac7{12}.

If PP reverses direction,

−u12=ku−u2⟹k=512.-\frac{u}{12}=ku-\frac{u}{2} \quad\Longrightarrow\quad k=\frac5{12}.

Both cases allow the particles to separate, since QQ has velocity 2w2w and PP has velocity ww or −w-w. Thus

k=712 or 512.\boxed{k=\frac7{12}\ \text{or}\ \frac5{12}}.