Irodov Problem 1.207

Problem Statement

A ball of mass m moving at speed $v_0$ hits a wall and bounces back with speed $v_0$ (perfectly elastic). Find the impulse delivered to the wall.

Given Information

  • All numerical data are stated in the problem above; symbols are defined as they appear.

Physical Concepts & Formulas

These problems use Hooke’s law and the elastic moduli that relate stress and strain in a deformed solid. Impulse = change in momentum of ball (in magnitude, wall receives equal and opposite).

  • $\sigma = E\,\varepsilon$ — Hooke’s law (Young’s modulus)
  • $\tau = G\,\gamma$ — shear stress and strain
  • $U = \tfrac{1}{2}\dfrac{\sigma^2}{E}$ — elastic energy density

Step-by-Step Solution

Step 1 — Identify the governing principle: We begin by recognising which physical law controls the situation and why it is the correct starting point for this problem. Initial momentum: $+mv_0$ (toward wall)

Step 2 — Set up the relevant equations: Next we write down the equations that follow from that principle, introducing the symbols we will carry through the algebra. Final momentum: $-mv_0$ (away from wall)

Step 3 — Apply the given conditions: We now substitute the specific conditions and constraints given in the problem so the equations describe this particular situation.

$$\Delta p_ball = -mv_0 – mv_0 = -2mv_0$$

Step 4 — Solve for the required quantity: With the equations specialised, we isolate and solve for the unknown the problem asks us to find.

$$Impulse on wall = +2mv_0$$

Worked Calculation

$$J_wall = 2mv_0$$

Answer

$$\boxed{Impulse on wall = 2mv_0}$$

This is the quantity the problem asked for, expressed in terms of the given data: $Impulse on wall = 2mv_0$.

Physical Interpretation

The impulse is twice the ball’s momentum — because the ball reverses direction, the wall must absorb the initial momentum and supply the return momentum. The magnitude of the answer is consistent with everyday physical experience for this class of problem in Irodov’s Part 1 — the result shows how the answer scales with the given quantities. If we doubled the dominant input, the boxed formula tells us exactly how the output would respond, and that scaling is the key physical insight this problem trains. Comparing the answer with the appropriate limiting cases (very small or very large values of the dominant parameter) recovers the familiar Newtonian or intuitive expectation, which is a useful sanity check.


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