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

Question

A toy cannon uses a spring to project 5.39-g soft rubber ball. The spring is originally compressed by 4.99 cm and has a force constant of 7.96 N/m. when the cannon is fired, the ball moves 41.9 cm through the horizontal barrel of cannon, and the barrel exerts a constant friction force 0.031.8N on the ball. With speed does the projectile leave the barrel of the cannon? If you calculated the speed ignoring the force of friction of you would find a smaller value than this. Does that make sense? m/s At what point does the ball have maximum speed? Cm (from its original position) What is this maximum speed? m/s

Explanation / Answer

given that

Ff = friction force on the ball = 0.0318 N

m = mass of ball = 5.39 g = 0.0053 kg

k = 7.96 N/m

we know that spring force is
Fs = k*x

Fs = 7.96 * 0.0499 = 0.397 N

and combined force acting on ball is

F = m*a

F = Fs - Ff

F = 0.397 - 0.0318 = 0.365 N

a = F/m = 0.365 / 0.00539 = 67.7m/s^2

Now using third equation of motion

a = 0.5 * ( ( v^2-u^2 ) / (x-x0))
u = 0
x0 = 0
a = 0.5 * (v^2/x)

v = sqrt (2*a*x)

for x = 0.0499 m therefore at the end of the spring push

v = sqrt (2*67.7*0.0499)

v = sqrt (6.75) = 2.59 m/s
This is the maximum speed and again it happens at 4.99 cm.

part (a)

the decceleration during the rest of the barrel length using

F = m*a
a = -(0.0318 /0.00539) = -5.89 m/s^2

again we use third equation of motion
a = 0.5 *[ (v^2 - u^2 ) / (x-x0) ]

now we take x0 = 0.0499 m & x = 0.149 m

x-x0 = 0.149 - 0.0499 = 0.099 m/s^2

u = 2.59 m/s (we have calculated this which is maximum speed v)

v = sqrt ( 2a (x-x0) + u^2 )

v = sqrt ((2*(-5.89)*(0.099 )+(2.59)^2)

v =sqrt(5.54) m/s

v = 2.89 m/s

this is the speed the ball leaves the barrel .

answer

(a) the speed the ball leaves the barrel = 2.89 m/s

(b) the point where speed is maximum = 4.99 cm

(c) maximum speed = 2.59 m/s

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