Question

A toy cannon uses a spring to project a 5.38-g soft rubber ball. The spring is originally compressed by 4.91 cm and has a force constant of 8.08 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 6 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

Answer 1

Part A.

Using Energy conservation:

KEi + PEi + Wf = KEf + PEf

KEi = 0 & PEf = 0

PEi = (1/2)*k*x^2

where k = 8.08 N/m & x = 4.91 cm = 0.0491 m

KEf = (1/2)*m*V^2

where m = mass of ball = 5.38 gm = 5.38*10^-3 kg

V = Speed of ball when it leaves the cannon

Wf = Ff*d

Ff = Force of friction = -0.0326 N (-ve sign because friction force is always in opposite direction)

d = length of barrel = 15.6 cm = 0.156 m

Using these values:

0 + (1/2)*k*x^2 - Ff*d = (1/2)*m*V^2 + 0

V = sqrt [(k*x^2 - 2*Ff*d)/m]

V = sqrt [(8.08*0.0491^2 - 2*0.0326*0.156)/(5.38*10^-3)]

V = 1.31 m/sec

Part B.

The ball will have maximum speed when spring force will be equal to friction force, because after this moment ball will stop speeding up.

Fs = Ff

k*xf = Ff

xf = Ff/k = 0.0326/8.08

xf = 0.004035 = 0.0040 m = 0.40 cm

Now since spring is originally compressed to 4.91 cm,

So, the point where spring have maximum speed, d1 = 4.91 cm - 0.40 cm

d1 = 4.51 cm

Part C.

At this point speed will be:

Using Energy conservation:

KEi + PEi + Wf = KEf + PEf

0 + (1/2)*k*xi^2 - Ff*d1 = (1/2)*m*V^2 + (1/2)*k*xf^2

0 + k*xi^2 - k*xf^2 - 2*Ff*d1 = m*V^2

V = sqrt [(k*(xi^2 - xf^2) - 2*Ff*d1)/m]

V = sqrt [(8.08*(0.0491^2 - 0.004^2) - 2*0.0326*0.0451)/(5.38*10^-3)]

V = 1.75 m/sec = Max speed of projectile

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