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Chapter 34: Q. 67 (page 993)

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

Expert verified

Radius of curvature of the mirror is 0.93 m

Step by step solution

01

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.

02

Step 3. Using the equation of motion.

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

∆y=vyi∆t+12gt2∆y=(0)(0.65s)+12(9.80m/s2)(0.65s)2∆y=2.07s

Using figure, R can be calculated as :

R=3m-2.07mR=0.93m

Radius of curvature of the mirror is 0.93 m

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