题目

Problem

Figure 5 shows a smooth track $PQR$ which is part of a circle.

The track is fixed in a vertical plane.

The circle has centre $O$ and radius $OP=1.5\,\mathrm{m}$ .

The highest point of the track is $P$ and $PO$ is horizontal.

The lowest point of the track $Q$ is on horizontal ground.

A small ball $B$ , of mass $0.2\,\mathrm{kg}$ , is released from rest at $P$ and moves on the track until it leaves the track at $R$ , where $\text{angle }QOR=60^\circ$

The ball is modelled as a particle.

(a) Find the magnitude of the normal reaction between the ball and the track when the ball is at $Q$ .

(5)

After leaving the track, the ball moves freely under gravity and first lands on the ground at the point $S$ .

(b) Find the distance $QS$ .

(8)

题目中文翻译

图 5 所示的光滑轨道 $PQR$ 是圆周的一部分。

轨道固定于竖直平面内。

圆心为 $O$ ，半径为 $OP=1.5\,\mathrm{m}$ 。

轨道的最高点为 $P$ ，且 $PO$ 水平。

轨道的最低点 $Q$ 位于水平地面上。

将质量为 $0.2\,\mathrm{kg}$ 的小球 $B$ 从 $P$ 由静止释放，小球沿轨道运动至 $R$ 后离开轨道，其中 $\angle QOR=60^\circ$ 。

将小球视为质点。

(a) 求小球位于 $Q$ 时与轨道之间的法向反作用力大小。

(5)

离开轨道后，小球仅在重力作用下自由运动，首次落地于 $S$ 点。

(b) 求距离 $QS$ 。

(8)

解答

(a)

解法一

思路

展开

从 $P$ 到 $Q$ 的下降高度等于半径，先用能量求速率。最低点的向心方向向上，支持力需要同时克服重力并提供向心力。

答题过程

展开

The drop from $P$ to $Q$ is $1.5\ \mathrm m$ . The normal reaction does no work, so

\frac12mv_Q^2=mg(1.5) \quad\Longrightarrow\quad v_Q^2=3g.

At $Q$ , resolve upwards towards $O$ . If the normal reaction is $N$ ,

N-mg=\frac{mv_Q^2}{1.5} =\frac{m(3g)}{1.5}=2mg.

Therefore

\boxed{N=3mg=3(0.2)(9.8)=5.88\ \mathrm N.}

(b)

解法一

思路

展开

离轨点 $R$ 的速度沿圆的切线，与水平线成 $60^\circ$ 。先求离轨速率、离地高度及相对 $Q$ 的水平偏移，再做抛体运动；所求 $QS$ 还要加上这段初始偏移。

答题过程

展开

The point $R$ is above the ground by

h=1.5-1.5\cos60^\circ=0.75\ \mathrm m,

and its horizontal distance to the right of $Q$ is

x_R=1.5\sin60^\circ.

The drop from $P$ to $R$ is $1.5\cos60^\circ=0.75\ \mathrm m$ . Energy conservation gives the departure speed $u$ :

\frac12mu^2=mg(0.75) \quad\Longrightarrow\quad u=\sqrt{1.5g}.

At $R$ the velocity is tangential, directed upwards and to the right at $60^\circ$ to the horizontal. Thus

u_x=\sqrt{1.5g}\cos60^\circ,\qquad u_y=\sqrt{1.5g}\sin60^\circ.

Let $t$ be the flight time from $R$ to the first ground contact. With upward positive,

-0.75=u_y t-\frac12gt^2.

The positive root is

t=\frac{u_y+\sqrt{u_y^2+2g(0.75)}}{g} =0.85636\ldots\ \mathrm s.

The other root is negative and is outside the flight interval. Horizontal motion gives

QS=x_R+u_x t =1.5\sin60^\circ+\sqrt{1.5g}\cos60^\circ(0.85636\ldots).

Using $g=9.8$ without rounding intermediate values,

\boxed{QS=2.94\ \mathrm m\quad\text{(3 s.f.)}.}

M3 2026 June Q6

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程