A toy cannon uses a spring to project a 5.39-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.39-g soft rubber ball. The spring is originally compressed by 5.04 cm and has a force constant of 8.10 N/m. When the cannon is fired, the ball moves 15.6 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 5 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

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)(8.1)(0.054^2)
= 0.01 J

Energy goes to giving initial kinetic energy to ball
0.01 = 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
= 0.0325/0.00539
= 6.03 m/s^2

Using v^2 = u^2 + 2as,
Final v after leaving barrel:
v^2 = 1.943^2 - 2(6.03)(0.156)
(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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