Question

My Notes

A toy cannon uses a spring to project a 5.20-g soft rubber ball. The spring is originally compressed by 5.04 cm and has a force constant of 8.03 N/m. When the cannon is fired, the ball moves 14.1 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.033 0 N on the ball.

(a) With what speed does the projectile leave the barrel of the cannon?
m/s

(b) At what point does the ball have maximum speed?
cm (from its original position)

(c) What is this maximum speed?
m/s

Answer 1

- tiwn ball man m= 5.20g = 5.2xlorlog d = 14.1cm=0.141 mm - Xo = 5,04ca k=8.03N/M 19-633N 0.0504 e spring eacray E= 1 kr = 1 (8.03) (0.0 504) = 0.010198 I Energy but by friction in the ballel E = fx distance tutelled in the barrel = 0.03 X 0.141 = 4.653x103 I Mzorert work done = change in ke E, E = m(www) 0.010198 - -4.653x103 = £ (5.2x103) (0-0) Tu=1.4603 m/s • osmis velouty I velouty when leaving the barrel. mayinline speed occurs at O Energy gain th (Eg) forest (8:03) (0.0504 gring force = friction force K(x-x) = f 80% (0.0504-) = 0.03] [x= 0.0462 m/ this is the point when having Max speed. speed of ball of y = (4.62 cm Energy gain from Spring = kr - t ka-n't 2010198 - (8.02) (0.0504-0.0462) = 0.01013019 J Energy cost due to friction = txan (ED) 0.033 x 0.0462 = 1.5246810 So, Eg - Ex at man .. 0.01013019-115246x13?= {(5,2x133), u = 3.309842 TU,= 1.819 mys this in max speed