Chapter 08, Problem 028 Figure (a) applies to the spring in a cork gun (Figure (

Question

Chapter 08, Problem 028 Figure (a) applies to the spring in a cork gun (Figure (b)); it shows the spring force as a function of the stretch or compression of the spring. The spring is compressed by 5.00 cm and used to propel a 3.50 g cork from the gun. (a) What is the speed of the cork if it is released as the spring passes through its relaxed position? (b) Suppose, instead, that the cork sticks to the spring and stretches it 1.50 cm before separation occurs. What now is the speed of the cork at the time of release? Force (N) Compressed pring 0.2 Cork x (cm -4 -0.2 0 -0.4 (a) Number Units (b) Number Units

Explanation / Answer

a) it is clear from the graph, the force at 5cm is clearly 0.5N.  
we have
a = f / m

v = sqrt {2aS} (where S is 5cm)

average force on spring = 0.5N / 2 = 0.25 N

m = 3.50 g = 0.0035 kg

a = 0.25 / 0.0035 = 71.43 m/s2 = average acceleration over the 5 cm

S = 5cm = 0.05 m
therefore
v = sqrt{2aS} = sqrt{2*71.43*0.05} = 2.67 m/s

b) when the cork stretches the spring 1.5cm at the point of release, we see from the graph,the spring is applying a braking force of 0.15N
The average braking force is 0.15/2 = 0.075N

decceleration -a = f / m = 0.075 / 0.0035 = -21.43

1.5 cm = 0.015m

v = sqrt{2aS}

the reduction in speed = root {2 x 21.43 x 0.015} = 0.802 m/s

therefore new speed = 2.67 - 0.802 = 1.87 m/s

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