Irodov 1.31: Elastic Bounces Walk Down the Slope – Amit & Deepali Sinha Foundation

Problem Irodov 1.31

A ball is released from rest directly above a point $A$ on a smooth (frictionless) inclined plane that makes angle $\alpha$ with the horizontal. It falls freely through a vertical distance $h$ before striking the incline at $A$ , and rebounds perfectly elastically. Find the distance, measured along the incline, from $A$ to the point $B$ where the ball strikes the incline a second time.

The setup: a smooth plane inclined at α, and the ball falling from rest through the vertical height h to its first impact at A, arriving with the downward velocity v_1. — The setup: a smooth plane inclined at $\alpha$ , and the ball falling from rest through the vertical height $h$ to its first impact at $A$ , arriving with the downward velocity $\vect v_1$ .

Given

$h$
$\alpha$
$g$

Find

the distance

l=AB

along the incline

Solution

Step 1Speed and direction at the first impact

Free fall from rest through a height $h$ under gravity $g$ gives the squared speed at the bottom,

v_1^2 = 2gh

and the impact speed is its positive root.

v_1 = \sqrt{2gh}

(1)

Step 2Resolve

\vect v_1

along and normal to the incline

Using the down-slope direction and the inward normal as the local axes, the vertical vector $\vect v_1$ projects onto each by the angle $\alpha$ between the vertical and the incline’s normal.

\begin{align*}v_{1\parallel} &= v_1\sin\alpha \quad\text{(down slope)} \\ v_{1\perp} &= v_1\cos\alpha \quad\text{(into the surface)}\end{align*}

Step 3Apply the elastic, frictionless rebound

The rebound velocity v_2, drawn beside A for clarity, resolved along the incline into v_2∥ down slope and v_2⊥ along the outward normal. — The rebound velocity $\vect v_2$ , drawn beside $A$ for clarity, resolved along the incline into $v_{2\parallel}$ down slope and $v_{2\perp}$ along the outward normal.

Immediately after the bounce, in the same local axes (down slope, outward normal),

v_{2\parallel} = v_1\sin\alpha, \qquad v_{2\perp} = v_1\cos\alpha \quad\text{(now outward)}

Step 4Motion after the bounce, in the incline’s own axes

With $x'$ measured down slope and $y'$ measured along the outward normal, both from $A$ ,

x'(t) = v_1\sin\alpha\,t+\frac{g\sin\alpha}{2}t^2, \qquad y'(t) = v_1\cos\alpha\,t-\frac{g\cos\alpha}{2}t^2 \tag{2}

Step 5Time of the second impact

The ball returns to the surface when the normal coordinate of (2) vanishes again; factoring out $t$ and $\cos\alpha$ isolates the nonzero root.

\begin{align*}0 &= v_1\cos\alpha\,t-\frac{g\cos\alpha}{2}t^2 \\ 0 &= t\cos\alpha\left(v_1-\frac{g}{2}t\right)\end{align*}

The factor $\cos\alpha$ never vanishes for a genuine incline and $t=0$ is the bounce itself, so the remaining bracket fixes the flight time.

t_2 = \dfrac{2v_1}{g}

(3)

Step 6The along-slope distance to the second impact

What this step measures: the flight from the first impact at A over the slope to the second impact at B, and the along-slope separation l=AB evaluated below. — What this step measures: the flight from the first impact at $A$ over the slope to the second impact at $B$ , and the along-slope separation $l=AB$ evaluated below.

Substituting (3) into the down-slope coordinate of (2) gives the distance $AB$ directly.

\begin{align*} l &= x'(t_2) = v_1\sin\alpha\cdot\frac{2v_1}{g}+\frac{g\sin\alpha}{2}\left(\frac{2v_1}{g}\right)^2 \\ &= \frac{2v_1^2\sin\alpha}{g}+\frac{2v_1^2\sin\alpha}{g} \tag{4} \\ &= \frac{4v_1^2\sin\alpha}{g} \end{align*}

Putting the impact speed (1) in as $v_1^2=2gh$ ,

l = \frac{4(2gh)\sin\alpha}{g}

l = 8h\sin\alpha

(5)

Check

As $\alpha\to0$ (a horizontal surface), $l\to0$ : a ball dropped onto a flat floor bounces straight back up and comes straight back down on the same spot, exactly as expected. Dimensionally $h\sin\alpha$ is a length, matching $l$ .

Remark

The two terms of (4) are equal by no coincidence: $\vect v_1$ and $\vect g$ are both purely vertical, so resolving them onto the incline’s down-slope and normal axes multiplies each by the same trigonometric factor, and the ratio of resolved speed to resolved deceleration along either axis is always $v_1/g$ , whatever the tilt $\alpha$ . That is why the normal motion returns to the surface at exactly $t_2=2v_1/g$ , the same flight time as a ball thrown straight up and caught at its launch height; and at that same instant, along the down-slope axis, the initial-speed term $v_1\sin\alpha\,t_2$ and the acceleration term $(g\sin\alpha/2)t_2^2$ are forced equal by the identical doubling that makes a vertical throw’s two halves match.