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M1 2023 Oct Q3

A Level / Edexcel / M1

IAL 2023 Oct Paper · Question 3

题目

Problem

3.

A hammer is used to hit a tent peg into soft ground.

The hammer has mass 1.8 kg1.8\,\mathrm{kg} and the tent peg has mass 0.2 kg0.2\,\mathrm{kg}.

The hammer and tent peg are both modelled as particles and the impact is modelled as a direct collision.

Immediately before the impact, the tent peg is stationary and the hammer is moving vertically downwards with speed 10 m s−110\,\mathrm{m\,s^{-1}}

Immediately after the impact, the hammer and tent peg move together, vertically downwards, with the same speed v m s−1v\,\mathrm{m\,s^{-1}}

(a) Find the value of vv

(2)

(b) Find the magnitude of the impulse exerted on the tent peg by the hammer, stating the units of your answer.

(3)

The ground exerts a constant vertical resistive force of magnitude RR newtons, bringing the hammer and tent peg to rest after they travel a distance of 12 cm12\,\mathrm{cm}.

(c) Find the value of RR.

(5)
题目中文翻译

用锤子将帐篷钉打入松软的地面。

锤子的质量为 1.8 kg1.8\,\mathrm{kg},帐篷钉的质量为 0.2 kg0.2\,\mathrm{kg}。

将锤子和帐篷钉均视为质点,将撞击视为正碰。

撞击前一瞬间,帐篷钉静止,锤子以 10 m s−110\,\mathrm{m\,s^{-1}} 的速率竖直向下运动。

撞击后瞬间,锤子与帐篷钉以相同速率 v m s−1v\,\mathrm{m\,s^{-1}} 一起竖直向下运动。

(a) 求 vv 的值。

(2)

(b) 求锤子对帐篷钉施加的冲量大小,并写出答案的单位。

(3)

地面施加大小为 RR 牛顿的恒定竖直阻力,使锤子和帐篷钉运动 12 cm12\,\mathrm{cm} 后静止。

(c) 求 RR 的值。

(5)

解答

(a)

解法一

思路

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碰撞前后总动量守恒;碰后两物体一起运动,质量合计为 2 kg。取向下为正。

答题过程

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Taking downwards as positive, conservation of momentum gives

1.8(10)+0.2(0)=(1.8+0.2)v.1.8(10)+0.2(0)=(1.8+0.2)v.

Thus 2v=182v=18, so v=9\boxed{v=9}.

(b)

解法一

思路

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帐篷钉由静止变为向下运动,锤子对它的冲量等于其动量变化;答案必须有冲量单位。

答题过程

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The impulse on the peg is

I=0.2(9−0)=1.8 N s.I=0.2(9-0)=1.8\ \mathrm{N\,s}.

Its magnitude is 1.8 N s\boxed{1.8\ \mathrm{N\,s}}.

解法二

思路

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也可先求锤子的动量变化,再利用两个相互作用冲量等大反向。

答题过程

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The impulse on the hammer is

IH=1.8(9−10)=−1.8 N sI_H=1.8(9-10)=-1.8\ \mathrm{N\,s}

when downwards is positive. The impulse exerted on the peg by the hammer is equal and opposite, so its magnitude is 1.8 N s\boxed{1.8\ \mathrm{N\,s}}.

(c)

解法一

思路

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先把 12 cm 换成 0.12 m,用碰后速率 9 求减速加速度,再对锤子与钉的整体列牛顿第二定律。阻力必须同时克服重力并使物体减速。

答题过程

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The combined mass is 2 kg. Taking downwards as positive, use v2=u2+2asv^2=u^2+2as during penetration:

0=92+2a(0.12),a=−337.5 m s−2.0=9^2+2a(0.12),\qquad a=-337.5\ \mathrm{m\,s^{-2}}.

The weight acts downwards and the resistance upwards, so

2g−R= 2a,R= 2(9.8)+2(337.5),= 694.6 N.\begin{align*} 2g-R=&\,2a,\\ R=&\,2(9.8)+2(337.5),\\ =&\,694.6\ \mathrm N. \end{align*}

Therefore R=695 N\boxed{R=695\ \mathrm N} (3 s.f.).

解法二

思路

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恒定阻力意味着加速度恒定。用平均速度求停止时间,再把整体的合冲量等同于动量变化。

答题过程

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Since the acceleration is constant, the stopping time τ\tau satisfies

0.12=9+02τ,τ=275 s.0.12=\frac{9+0}{2}\tau,\qquad \tau=\frac2{75}\ \mathrm s.

Taking upwards as positive, the resultant force is R−2gR-2g and the velocity changes from −9-9 to 0. The impulse-momentum equation is

(R−2g)275= 2(0−(−9)),R= 19.6+675,= 694.6 N.\begin{align*} (R-2g)\frac2{75}=&\,2(0-(-9)),\\ R=&\,19.6+675,\\ =&\,694.6\ \mathrm N. \end{align*}

Hence R=695 N\boxed{R=695\ \mathrm N} (3 s.f.).

解法三

思路

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也可用功能关系:下降时重力做正功,阻力做负功,最终动能为零。地面阻力所做负功的大小等于初动能加重力释放的能量。

答题过程

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During the downward displacement, gravity does positive work and the resistance does negative work. By the work-energy principle,

2g(0.12)−R(0.12)= 0−12(2)(92),0.12R= 81+2(9.8)(0.12),R= 694.6 N.\begin{align*} 2g(0.12)-R(0.12)=&\,0-\frac12(2)(9^2),\\ 0.12R=&\,81+2(9.8)(0.12),\\ R=&\,694.6\ \mathrm N. \end{align*}

Thus R=695 N\boxed{R=695\ \mathrm N} (3 s.f.).