A toy cannon uses a spring to project a 5.30 g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.30 g soft rubber ball. The spring is originally compressed by 5.00 cm and has a force constant of 9.00 N/m. When it is fired, the ball moves 13.0 cm through the horizontal barrel of the cannon, and a constant frictional force of 0.0320 N exists between barrel and 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 the starting position) (c) What is this maximum speed?

Explanation / Answer

Using spring and kinematics concepts to solve this is possible. [All values here are in SI quantities] Energy supplied by spring E = 1/2 kx^2 = (1/2)(8.6)(0.05^2) = 0.01075 J Energy goes to giving initial kinetic energy to ball 0.01075 = 1/2 mv^2 (kinetic energy) initial (launch) v = 1.943 m/s Retarding frictional force F = ma Deceleration due to friction a = F/m = 0.032/0.0053 = 6.03 m/s^2 Using v^2 = u^2 + 2as, Final v after leaving barrel: v^2 = 1.943^2 - 2(6.03)(0.15) (a is negative as it is a deceleration) v final (after leaving barrel) = 1.40 m/s Therefore a) 1.40 m/s b) When it was first fired (because it was fired horizontally through the 'horizontal barrel' so no vertical component involved here). c) 1.94 m/s

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