In case of an elastic impact of a body against a fixed wall, its velocity v changes only in the direction. Determine the change after the impact of the velocity of this body, if the wall is moving: a) at a speed u towards the body; b) at a speed w < v in the direction of movement of the body.

Solutions of Savchenko Problems in Physics

Aliaksandr Melnichenka
October 2023

    <h3 id="back-link"><a href="/#1.4">$\leftarrow$Back</a></h3>

    <h3> Statement </h3>
    <p>
        $1.4.8.$ In case of an elastic impact of a body against a fixed wall, its velocity $v$ changes only in the direction. Determine the change after the impact of the velocity of this body, if the wall is moving: a) at a speed $u$ towards the body; b) at a speed $w < v$ in the direction of movement of the body.

    <h3>Solution</h3>
    <p>
        <p>The whole point of the problem is to move to an inertial system associated with the wall, correctly describing the relative velocity.</p>

$a)$ In the case when the wall "catches up" with the ball, it is necessary to move to a frame of reference in which the wall is at rest.

In this frame of reference, the relative velocity of the ball before the impact is

$$v_{0} + u$$

After the elastic impact, the velocity projection changes sign: it becomes equal to

$$(- v_{0} - u)$$

If we then return to the original (laboratory) frame of reference, where the wall moves with a velocity of $u$, then the velocity of the ball here is

$${v}' = - v_{0} - 2u$$

The change in velocity in this case is

$$\fbox{$\Delta v = {v}' - v_{0}$ $= - 2(v_{0} + u)$}$$

$b)$ Similarly, we need to go to the frame of reference in which the wall is at rest.

In this frame of reference, the relative velocity of the ball before the impact is $v_{0} - w$. After the elastic impact, the projection of the velocity changes sign: it becomes equal to

$$(- v_{0} + w)$$

If we then return to the original (laboratory) frame of reference, where the wall moves with the velocity $w$, then the velocity of the ball is

$${v}' = - v_{0} + 2w$$

The change in velocity in this case is

$$\fbox{$\Delta v = {v}' - v_{0}$ $= - 2(v_{0} - w)$}$$

    <h4>Answer</h4>
    <p>
        $\begin{aligned}&{a) }\Delta v = -2(v + u).\quad б) \Delta v = -2(v - w). (\text{Projection on the direction of the initial}\\&\text{speeds are considered positive.)}\end{aligned}$
    </p>


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