A toy cannon uses a spring to project a 5.22-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.22-g soft rubber ball. The spring is originally compressed by 4.98 cm and has a force constant of 8.02 N/m. When the cannon is fired, the ball moves 14.6 cm through the horizontal barrel of the cannon, and the barrel exerts constant friction force of 0.031 9 N on the ball. (a) With what speed does the projectile leave the barrel of the cannon? m/s ) At what point does the bal have maximum speed? cm (from its original position) e) What is this maximum speed? m/s

Explanation / Answer

(a)Break the barrel up into two parts. For the first 4.98 cm of the barrel the ball will accelerate from the spring force. For the last 9.62 cm of the barrel the ball will decelerate from frictional force.

First part
Let F = combined forces acting on the ball
Let Fs = Spring force on the ball
Let Ff = friction force on the ball
Let m = mass of ball (5.22 g)
Let a = acceleration


Since the spring is linear you can just use its average force on the ball which is the force when it is halfway compressed.
Fs = kx = 8.02 N/m * (0.0498m/2) = 0.199 N

Use F = ma to calculate acceleration
F = Fs - Ff = 0.199 N - 0.0319 N = 0.167 N
solve for a:
a = F/m = 0.167 N / .00522 kg = 31.9 m/s^2

Now use:
a = 0.5 *[(v^2-v0^2)/(x-x0)]
v0 = 0
x0 = 0
so: a = 0.5 * (v^2/x)
solve for v
v = squareroot of (2ax)
we solved for a above and x = 0.0498 m therefore at the end of the spring push
v = sqaureroot of (2*31.9 m/s^2*0.0498 m) = 1.78 m/s
This is the maximum speed and again it happens at 4.98 cm.

Now to answer question (a) calculate the decceleration during the rest of the barrel length using F = ma
a = -1*(0.0319 N/.00522 kg = -6.11 m/s^2

again use:
a = 0.5 *[(v^2-v0^2)/(x-x0)]
x-x0 = 0.146 m - 0.0499 m = 0.0961 m
v0 = 1.78 m/s (calculated above as v, note that this is not the same v0 as used above )
solve equation for v
v = sqrt [2a(x-x0)+v0^2]
v = sqrt [2*(-6.11m/s^2)*(0.0961 m)+(1.78 m/s)^2]
v = 2.27 m/s this is the answer to a the speed the ball leaves the barrel

(b)

Because friction applies a constant backward (let's assume leftward) force, the ball will only accelerate to the right if the spring force exceeds the force of friction. So let's set the two equal to see when this ceases to happen.
F_spring = F_friction
kx = 0.0319 N
(8.02 N/m)(x) = 0.0319 N
And we find that x = 0.0040 m or 0.40 cm. Of course, this is how much the spring is compressed. Since the spring originally was compressed by 4.98 cm, the ball has now traveled a distance of (4.98 -0.40) = 4.58 cm.

(c)
First, let's consider how much work is done on the ball as it travels this distance by the two forces we need to consider: namely, the spring force and friction. We can use W = 1/2 k x^2 to calculate the spring's work, but keep in mind that the spring hasn't finished releasing, so it will actually look like this:
W_spring = 1/2 * ( 8.02 N/m) * [(0.0498 m)^2 - (0.0040 m)^2]
W_spring = 0.00988 J

The work done by friction is easy because it is a constant force, so
W_fric = Fd
W_fric = (-0.0319 N) (0.0458 m)
W_fric = -.00146 J

I have called the frictional force negative because it is pushing the ball left, and since it is opposite the displacement it makes negative work. So adding the works, we get(-.00146 J + 0.00988 J) = 0.00842 J. Of course, we need to know the velocity after this net work has been applied, so we use the work-energy theorem to find that
K = W (since K_initial was zero)
1/2 m v^2 = 0.00842 J
1/2 (0.00498 kg) (v^2) = 0.0084 J
And solving for v,

we get v = 1.83 m/s,

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