题目

Problem

A small ball is released from rest from a point that is $40\,\mathrm{m}$ above horizontal ground. The ball bounces on the ground and rebounds vertically. Each time the ball bounces on the ground, the speed of the ball is instantaneously reduced by $50\%$ . The ball is modelled as a particle moving freely under gravity, from the instant when it is released until it first hits the ground, and between each successive bounce.

(a) Find the time from the instant when the ball is released from rest to the instant when it hits the ground for the second time.

(5)

(b) Find the maximum height reached by the ball above the ground after the ball’s third bounce.

(4)

题目中文翻译

一个小球从水平地面上方 $40\,\mathrm{m}$ 处由静止释放。小球撞击地面后竖直反弹。每次撞击地面时，小球的速率瞬间减小 $50\%$ 。从释放到首次撞击地面的过程中，以及相邻两次碰撞之间，将小球视为仅在重力作用下运动的质点。

(a) 求从小球由静止释放的瞬间，到第二次撞击地面的瞬间所经过的时间。

(5)

(b) 求小球第三次反弹后离地面的最大高度。

(4)

解答

(a)

解法一

思路

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先求首次下落的时间与落地速率。首次反弹的初速减半，回到同一地面的上升、下降时间相等；把这次完整弹跳时间加到首次下落时间上。

答题过程

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Let $t_0$ be the time to the first impact. For the initial fall,

40=\frac12gt_0^2, \qquad t_0=\sqrt{\frac{80}{9.8}}=\frac{20}{7}\,\mathrm{s}.

The speed immediately before this impact is

v_0=gt_0=9.8\frac{20}{7}=28\,\mathrm{m\,s^{-1}}.

The rebound speed is $14\,\mathrm{m\,s^{-1}}$ . Taking upwards as positive during this flight, the time $\tau>0$ to return to the ground satisfies

0=14\tau-\frac12g\tau^2, \qquad \tau=\frac{28}{g}=\frac{20}{7}\,\mathrm{s}.

Therefore the time to the second impact is

t_0+\tau=\frac{40}{7}\,\mathrm{s} =\boxed{5.71\,\mathrm{s}}\quad\text{(3 s.f.)}.

(b)

解法一

思路

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每次碰地前的速率等于上一次弹起速率；每次碰撞又把速率减半。第三次反弹的初速是首次落地速率的八分之一，而不是四分之一。

答题过程

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The successive speeds immediately after the first, second and third bounces are

14,\quad 7,\quad 3.5\quad\mathrm{m\,s^{-1}}.

For the upward flight after the third bounce, use $v^2=u^2+2as$ , with $v=0$ at maximum height and $a=-g$ :

0=3.5^2-2gh.

Hence

h=\frac{3.5^2}{2(9.8)} =\boxed{0.625\,\mathrm{m}}.

M1 2019 Oct Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程