Falling Coin Problem

Question

A small coin, initially at rest, begins falling. If the clock starts when the coin begins to fall, what is the magnitude of the coin's displacement between 𝑡1=0.347 s and 𝑡2=0.608 s?

Answer 1

Let's recall all the necessary formulas. The displacement, Δy, between times t₁ and t₂ is : Δy = y₂ - y₁. In order to solve this equation we need values for y1 and y2. Recall the formula: y = y₀ + v₀ + gt²/2. We use this formula to solve for the position of the coin at time t₁. We take the initial position, y₀, to be 0 m, and we take the downward direction to be negative. Because the coin starts at rest, its initial velocity, v₀, to be 0 m/s. Therefore, y₁ = gt₁²/2. Plugging in -9.8 m/s² for g, our result is that y₁ = - 0.59 m.

Now, we just need y₂. We can use the same formula y₂ = y₁ + v₁ + gt²/2. We know the values of every variable in this equation except for v₁. We can use the following formula to obtain its value: v₁ = v₀ + gt₁. Remember that the initial velocity is 0 m/s. Solving this equation, we get v₁ = - 3.40 m/s. Now we can finally solve for y₂. The result should be y₂ = - 4.47 m. (The negative signs only mean that this position is lower than where we originally started.

Lastly, we just need to calculate Δy. Calculating y₂ - y₁ = -4.47 - (- 0.59) = - 3.88 m.

Check that this answer makes sense. We expected a negative value for the displacement because the final position of the coin is lower than the initial position.