A toy cannon uses a spring to project a 5.31-g soft rubber ball. The spring is o

Question

A toy cannon uses a spring to project a 5.31-g soft rubber ball. The spring is originally compressed by 4.97 cm and has a force constant of 8.09 N/m. When the cannon is fired, the ball moves 15.1 cm through the horizontal barrel of the cannon, and the barrel exerts a constant friction force of 0.032 2 N on the ball.
(a) With what speed does the projectile leave the barrel of the cannon?

(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

Break the barrel up into two parts. For the first 4.96 cm of the barrel the ball will accelerate from the spring force. For the last 10.04 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 (5.34 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 = 7.93 N/m * (0.0496 m/2) = 0.197 N Use F = ma to calculate acceleration F = Fs - Ff = 0.197 N - 0.0319 N = 0.165 N solve for a: a = F/m = 0.165 N / .00534 kg = 30.9 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 = 0.0496 m therefore at the end of the spring push v = sqaureroot of (2*30.9 m/s^2*0.0496 m) = 1.75 m/s This is the maximum speed and again it happens at 4.96 cm. Now to answer question (a) calculate the decceleration during the rest of the barrel length using F = ma a = -1*(0.0319 N/.00534 kg = -5.97 m/s^2 again use: a = 0.5 *[(v^2-v0^2)/(x-x0)] x-x0 = 0.15 m - 0.0496 m = 0.1004 m v0 = 1.75 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*(-5.97m/s^2)*(0.1004 m)+(1.75 m/s)^2] v = 1.37 m/s this is the answer to a the speed the ball leaves the barrel Source(s): Sorry, I found the error. I had forgotten to square v (1.75 m/s) in the last equation. The new answer IS correct - I've triple checked it.

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