Space Time & Motion A.2 Forces & Momentum (id: 8ca1c1cf2)

本题目来源于试卷: Space Time & Motion A.2 Forces & Momentum，类别为 IB物理学

[单选题]
A ball of mass m strikes a vertical wall with a speed of v. It bounces back with the same speed. The angle between the wall and the trajectory of the incoming ball is $\theta_2$​. The angle between the wall and the trajectory of the ball after the bounce is $\theta_2$. During the collision, the only force that acts on the ball is normal to the wall.


What is the magnitude of the change in momentum of the ball?

A. $2mv(\cos \theta_1)$
B. $2mv(\sin \theta_1)$
C. $2mv(\sin (\theta_1+\theta_2))$
D. $2mv(\cos (\theta_1+\theta_2))$


参考答案:  B


本题详细解析:

$\underline{\text{xplanation}}$:

The question states that the only force from the wall on the ball during the collision is perpendicular to the surface. Due to the absence of a vertical force, there will be no change in the vertical component of the momentum of the ball, and all of the momentum change is in the horizontal direction.

Because the collision is elastic, the speed of the ball is unchanged in the collision. As the vertical component of the velocity is unchanged, it follows that the magnitude of the horizontal component of the velocity is also unchanged.

Given that the magnitudes of the components are both unchanged, the angles ${\theta }_1$and ${\theta }_2$ must be equal.

It has been determined that the only change in momentum is in the horizontal direction.

Therefore,

• $\text{∆}{p}_h=p_{fh}-p_{ih}$

• $p_{ih}=mv(sin{\theta }_1)$

and

• $p_{fh}=-mv(sin{\theta }_2)$

• $\mathrm{\Delta }{p}_h=p_{fh}-p_{ih}=-mvsin{\theta }_2-mvsin{\theta }_1$

But

• ${\theta }_1={\theta }_2$

Therefore

• $\mathrm{\Delta }{p}_h=-2mv(sin{\theta }_1)$

Thus

• $\mathrm{\Delta }p=-2mv(sin{\theta }_1)$

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