Reviewed CSCA Physics question · Standard

在光滑水平面上，质量为2 kg的小球A以速度3 m/s向右运动，与质量为1 kg、静止的小球B发生正碰。碰撞后A球速度为1 m/s，方向仍向右。求碰撞后B球的速度大小和方向。

4 m/s，向右
4 m/s，向左
2 m/s，向右
2 m/s，向左

Show the answer and explanation

Correct answer

A. 4 m/s，向右

Principle or equation

动量守恒定律：系统不受外力或合外力为零时，碰撞前后总动量守恒。m1v1 = m1v1' + m2v2'。

Why this answer is correct

取向右为正方向。碰撞前总动量 p = 2×3 = 6 kg·m/s。碰撞后A动量 pA' = 2×1 = 2 kg·m/s。B动量 pB' = p - pA' = 6 - 2 = 4 kg·m/s。B速度 vB' = 4/1 = 4 m/s，方向向右。

Example

若A碰后速度为0.5 m/s，则B速度为5 m/s向右。

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