Question

A toy cannon uses a spring to project a 5.20-g soft rubber ball. The spring is originally compressed by 5.03 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 1 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?

(c) What is this maximum speed?

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 = 5.03 cm = 0.0503 m

KEf = (1/2)*m*V^2, where m = mass of ball = 5.20 gm = 5.20*10^-3 kg

V = Speed of ball when it leaves the cannon

Wf = Ff*d

Ff = Force of friction = -0.0321 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*0.0503^2 - 2*0.0321*0.156)/(5.20*10^-3)]

V = 1.42 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 = 0.0321/8.08

xf = 0.003972 = 0.0040 m = 0.40 cm

Now since spring is originally compressed to 5.03 cm, so

Spring is originally compressed by 5.03 cm, So

the point where spring have maximum speed, d1 = 5.03 cm - 0.40 cm = 4.63 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*(0.0503^2 - 0.004^2) - 2*0.0321*0.0463)/(5.20*10^-3)]

V = 1.83 m/sec = Max speed of projectile

Please Upvote.