Two identical coins are initially held at height h = 11.8 m. Coin 1 is dropped a

Question

Two identical coins are initially held at height h = 11.8 m. Coin 1 is dropped at time t = 0 and then lands on a muddy field where it sticks. Coin 2 is dropped at t = 0.500 s and then lands on the field. What is the magnitude of the acceleration of the center of mass (com) of the two-coin system (a) between r = 0 and t = 0.500 s, (b) between t = 0.500 s and time when coin 1 hits and sticks, and (c) between and time^when coin 2 hits and sticks? What is the speed of the center of mass when t is (d) 0.250 s, (e) 0.750 s, and (f) 1.75 s?

Explanation / Answer

a) accelaration of center of mass of first coin is 9.8 m.s2 , second coin is 0 so net accelaration is

9.8/2 = 4.9 m/s2

b) Both will accelarate at 9.8 m/s2 , so accelaration is 9.8m/s2

C) accelaration of center of mass of first coin is 0 m.s2 , second coin is 9.8 so net accelaration is

9.8/2 = 4.9 m/s2

d) v = 9.8 x .25 = 2.45 m/s

so centre of mass is V/2 = 1.225 m/s

e) v1 = 9.8 x .75 = 7.35 m/s

v2= 9.8 x .25 = 2.45 m/s

f) v1 = 0 , beacuse after 1.25 seconds it had come to rest

v2 = 12.25 m/s

Velocity of centre of mass is 6.125 m/s

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