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M2 2020 Oct Q7

A Level / Edexcel / M2

IAL 2020 Oct Paper · Question 7

题目

Problem

7.

Particle AA of mass 3m3m is moving in a straight line with speed 2u2u on a smooth horizontal surface. Particle AA collides directly with particle BB of mass mm, which is moving along the same straight line and in the same direction as AA.

Immediately before the collision, the speed of BB is uu.

As a result of the collision, the direction of motion of BB is unchanged and the kinetic energy gained by BB is 4825mu2\frac{48}{25}mu^2

(a) Find the coefficient of restitution between AA and BB.

(8)

After the collision, BB hits a smooth fixed vertical wall that is perpendicular to the direction of motion of BB. The coefficient of restitution between BB and the wall is ff.

Given that the speed of BB immediately after first hitting the wall is equal to the speed of AA immediately after its first collision with BB,

(b) find the value of ff.

(2)
题目中文翻译

质量为 3m3m 的质点 AA 在光滑水平面上以速率 2u2u 作直线运动。质点 AA 与质量为 mm 的质点 BB 发生正碰,BB 沿同一直线向与 AA 相同的方向运动。

碰撞前一瞬间,BB 的速率为 uu。

碰撞后 BB 的运动方向不变,且 BB 增加的动能为 4825mu2\frac{48}{25}mu^2。

(a) 求 AA 和 BB 之间的恢复系数。

(8)

碰撞后,BB 撞击一面垂直于其运动方向的光滑固定竖直墙。BB 与墙之间的恢复系数为 ff。

已知 BB 第一次撞墙后瞬间的速率,等于 AA 与 BB 第一次碰撞后瞬间 AA 的速率,

(b) 求 ff 的值。

(2)

解答

(a)

解法一

思路

展开

由 BB 增加的动能先求其末速度,再由动量守恒求 AA 的末速度,最后取分离速率与接近速率之比。

答题过程

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Take the initial direction of motion as positive. Let the velocities of AA and BB after their collision be vv and ww.

The kinetic energy gained by BB is

12mw2−12mu2=4825mu2.\frac12mw^2-\frac12mu^2=\frac{48}{25}mu^2.

Therefore

w2=12125u2.w^2=\frac{121}{25}u^2.

Since BB continues in its original direction,

w=115u.w=\frac{11}{5}u.

Conservation of momentum gives

3m(2u)+mu= 3mv+mw,7mu= 3mv+115mu,v= 85u.\begin{align*} 3m(2u)+mu=&\,3mv+mw,\\ 7mu=&\,3mv+\frac{11}{5}mu,\\ v=&\,\frac85u. \end{align*}

The coefficient of restitution is

e=speed of separationspeed of approach=w−v2u−u=35.e=\frac{\text{speed of separation}}{\text{speed of approach}} =\frac{w-v}{2u-u} =\boxed{\frac35}.

(b)

解法一

思路

展开

固定墙的速度为零,恢复系数等于反弹速率除以入射速率。题给反弹速率等于 AA 碰后的速率。

答题过程

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The speed of BB before hitting the wall is 11u/511u/5, and its rebound speed is 8u/58u/5. Thus

f=8u/511u/5=811.f=\frac{8u/5}{11u/5}=\boxed{\frac8{11}}.