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M1 2024 June Q5

A Level / Edexcel / M1

IAL 2024 June Paper · Question 5

题目

Problem

5.

A parachute is used to deliver a box of supplies. The parachute is attached to the box.

  • the parachute and box are dropped from rest from a helicopter that is hovering at a height of 520 m520\,\mathrm{m} above the ground
  • the parachute and box fall vertically and freely under gravity for 55 seconds, then the parachute opens
  • from the instant the parachute opens, it provides a resistance to motion of magnitude 3200 N3200\,\mathrm{N}
  • the parachute and box continue to fall vertically downwards after the parachute opens
  • the parachute and box are modelled throughout the motion as a particle PP of mass 250 kg250\,\mathrm{kg}

(a) Find the distance fallen by PP in the first 55 seconds.

(2)

(b) Find the speed with which PP lands on the ground.

(7)

(c) Find the total time from the instant when PP is dropped from the helicopter to the instant when PP lands on the ground.

(3)

(d) Sketch a speed-time graph for the motion of PP from the instant when PP is dropped from the helicopter to the instant when PP lands on the ground.

(2)
题目中文翻译

用降落伞投送一箱物资,降落伞连接在箱子上。

  • 降落伞和箱子从悬停于离地 520 m520\,\mathrm{m} 高处的直升机上由静止投下。
  • 降落伞和箱子在重力作用下竖直自由下落 55 秒后,降落伞打开。
  • 从降落伞打开的瞬间起,它提供大小为 3200 N3200\,\mathrm{N} 的运动阻力。
  • 降落伞打开后,降落伞和箱子继续竖直向下运动。
  • 在整个运动过程中,将降落伞和箱子视为质量为 250 kg250\,\mathrm{kg} 的质点 PP。

(a) 求 PP 在前 55 秒内下落的距离。

(2)

(b) 求 PP 落地时的速率。

(7)

(c) 求从 PP 被投下直升机的瞬间到落地瞬间的总时间。

(3)

(d) 画出从 PP 被投下直升机的瞬间到落地瞬间的速率—时间图像草图。

(2)

解答

(a)

解法一

思路

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第一阶段自由落体,从静止出发,加速度为 gg。最终答案按有效数字写,但后续计算保留未舍入距离。

答题过程

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Taking downwards as positive, the distance fallen in 5 s is

s1=12gt2=12(9.8)(52)=122.5 m.s_1=\frac12gt^2=\frac12(9.8)(5^2)=122.5\ \mathrm m.

Thus the distance is 123 m\boxed{123\ \mathrm m} (3 s.f.). Retain s1=122.5s_1=122.5 m for subsequent calculations.

(b)

解法一

思路

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打开降落伞前已有速率 49,不能把第二阶段当成从静止开始。用重力与阻力求新的加速度,再用剩余距离求落地速率。

答题过程

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Immediately before the parachute opens,

u2=0+g(5)=49 m s−1u_2=0+g(5)=49\ \mathrm{m\,s^{-1}}

downwards. The remaining distance is

s2=520−122.5=397.5 m.s_2=520-122.5=397.5\ \mathrm m.

After opening, taking downwards as positive,

250g−3200=250a,a=−3 m s−2.250g-3200=250a,\qquad a=-3\ \mathrm{m\,s^{-2}}.

Using v2=u2+2asv^2=u^2+2as for this second stage,

v2= 492+2(−3)(397.5),= 2401−2385=16.\begin{align*} v^2=&\,49^2+2(-3)(397.5),\\ =&\,2401-2385=16. \end{align*}

The box is still moving downwards, so its landing speed is 4.00 m s−1\boxed{4.00\ \mathrm{m\,s^{-1}}} (3 s.f.).

(c)

解法一

思路

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先求打开伞后至落地的时间,再加上自由落体的 5 秒。整个过程不能共用同一个加速度。

答题过程

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Let τ\tau be the time from parachute opening to landing. For the second stage,

4=49−3τ,τ=15 s.4=49-3\tau,\qquad \tau=15\ \mathrm s.

Therefore the total time is

5+15=20.0 s(3 s.f.).5+15=\boxed{20.0\ \mathrm s}\quad\text{(3 s.f.)}.

(d)

解法一

思路

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速率先线性增加,再以较小斜率绝对值线性下降。落地前仍有速率 4,终点不能落在时间轴上,也不画竖直线接到零。

答题过程

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Draw time (s) horizontally and speed (m s−1\mathrm{m\,s^{-1}}) vertically. Join

(0,0)⟶(5,49)⟶(20,4)(0,0)\longrightarrow(5,49)\longrightarrow(20,4)

with two straight line segments. The first rises steeply and the second falls more gently. Label speeds 49 and 4 and the final time 20.

End the graph at (20,4)(20,4), above the time axis; do not extend it to zero speed or add a vertical drop.