Question

A 25 g rubber ball is dropped from a height of 3.0 m above the center of a horizontal, concave mirror. The ball and its image coincide 0.65 s after the ball is released. What is the mirror’s radius of curvature?

Short answer

Radius of curvature of the mirror is 0.93 m

Step-by-step solution

Step 1. Given information is :Mass of the rubber ball, m = 25g = 0.025 kgHeight above the center of the wall from where the ball is dropped, ∆y=3.0 mInitial velocity of the ball, uyi=0Time after which the image coincides with the ball, ∆t=0.65 s

We need to find out the radius of curvature of the mirror.

Step 3. Using the equation of motion.

The ball is under a constant acceleration that is gravitational acceleration.

$$\begin{gathered} ∆y=v_{yi}∆t+\frac{1}{2}g{t}^2 \\ ∆y=\left(0\right)(0.65s)+\frac{1}{2}(9.80m/s^2){(0.65s)}^2∆y=2.07s \end{gathered}$$

Using figure, R can be calculated as :

$$\begin{gathered} R=3m-2.07m \\ R=0.93m \end{gathered}$$

Radius of curvature of the mirror is 0.93 m