An egg is dropped from a third-floor window and lands on a foam-rubber pad without breaking. The acceleration of gravity is 9.81 m/s 2 . If a 56.4 g egg falls 14.0 m from rest and the 7.85 cm thick foam pad stops it in 6.05 ms, by how much is the pad compressed? Assume constant upward acceleration as the egg compresses the foam-rubber pad. (Assume that the potential energy that the egg gains while the pad is being compressed is negligible.) Answer in units of m.

The pad compresses for about 5 cm.

Explanation:

Given:

- The mass of the egg m = 56.4 g

- The height at which egg drops h = 14.0 m

- The total thickness of foam d = 7.85 cm

- Total time taken to stop the egg t = 6.05 ms

Find:

By how much is the pad compressed? s

Solution:

- Apply conservation of energy from release point to point just before egg contacts the pad. To compute the speed of egg at impact

                                  ΔP.E = ΔK.E

                                 m*g*( h - t ) = 0.5*m*v^2

                                 v = sqrt( 2*g*( h - d ) )

                                 v = sqrt ( 2*9.81*(14 - 0.0785))

                                 v = 16.53 m/s

- Now to determine the the acceleration (Assuming constant ) at which pad compresses we will use first equation of motion till the point the final velocity is zero.

                                 vf = vi + a*t

                                 0 = 16.53+ a*(0.00605)

                                 a = 16.53 / 0.00605

                                 a = -2731.73 m/s^2

- To determine the total amount of pad compressed (s), we will use the third equation of motion as follows:

                                 vf^2 = vi^2 + 2*a*s

                                  0 = 16.53^2 + 2*(-2731.73)*s

                                  s = 16.53^2 / (2*2731.73)

                                  s = 0.05 m = 5 cm

- The pad compresses for about 5 cm.

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