a ball of mass .240 kg is shot straight up into the air where it collides with a

Question

a ball of mass .240 kg is shot straight up into the air where it collides with an ideal spring with a spring constant K of 220 N/m. the lowest part of the spring is initially a distance of 0.530 meters above the ball. the ball collides with the spring and continues to move upward compressing the spring a distance of 0.270 meters before stopping and coming back down. ignore air resistance and let g= 9.80 m/s^2....a) what was the initial speed of the (when it was a distance d below the spring)? b) later, the ball has left the spring and it on the was back toward the ground. it is a distance of 26 cm below the lowest part of the spring (now back in its equilibrium position). what is the speed of the ball at this time?

Explanation / Answer

initial height of ball is zero

final height of ball is 0.270 + 0.530 = 0.800 m

so

initial K = final U

(1/2) m v^2 = (1/2) kx^2 + mgh

(1/2) 0.240 v^2 = (1/2) * 220 * 0.270^2 + 0.240 * 9.8 * 0.800

solve for v

v = 9.083 m/s was the initial speed

now part b

26 cm below the lowest point of spring is 0.53 - 0.26 = 0.27 m above the ground

The spring was initially relaxed, when the ball was shot, and it is now relaxed again. So

initial K = final K + final potential energy

(1/2) m v^2 = (1/2) m v^2 + mgh

(1/2) * 9.083^2 = (1/2) * v^2 + 9.8 * 0.27

solve for v

v = 8.787 m/s now for part b

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