Question

A .4 kilogram ball is rolled off a 1 meter high level table. If the ball lands in line with a spring of 50 N/M and the spring compresses exactly .5 meters, how much time did our ball spend in the air..what horizontal displacement did it undergo.. what was the kinetic energy of the ball before it reached the spring.. and what was the initial velocity of the spring?

Answer 1

time spent in the air is found by the displacement formula:

h = 1/2 g t^2

=> 1 = 1/2 g t^2

so t = sqrt (2/g) = 0.452 s

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now find v of the ball when it just lefts the table :

=> (KE + PE)1 = 1/2 k x^2

=> 1/2 (0.4) v^2 + 0.4(9.8)(1.5) = 25 (0.5)^2

=> v = 1.36 m/s

so the horizontal distance is :

dx = v * t = 1.36 * 0.452 = 0.615 m

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the kinetic energy of the ball before it reached the spring is :

KE = 1/2 k x^2 = 25 (0.25) = 6.25 J

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the initial velocity of the spring :

=> KE = 1/2 m v^2

=> 6.25 = 0.2 v^2

=> v = 5.59 m/s