1. Sep 9, 2007

### EvanQ

1. The problem statement, all variables and given/known data
A spring-loaded toy gun is used to shoot a ball of mass m= 1.50kg straight up in the air, as shown in the figure. The spring has spring constant k = 667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height h_max (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis.

Find v_m the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position y=0).

2. Relevant equations

F = kx
F = ma
v^2 = u^2 + 2as

3. The attempt at a solution

I think I am using some formula's that don't really belong. Not sure which steps I go wrong in though.

F=kx
=667x0.25
=166.75

F=ma
166.75=1.5a
a=111.16m/s

v^2 = u^2 + 2as
= 0+(2x111.16x0.25)
= 55.58
v = 7.455m/s

2. Sep 9, 2007

### Staff: Mentor

Several problems:
(1) F = kx gives the spring force at spring extension x, which changes as the ball moves. Plugging in x = 0.25 m only gives the force at one point.
(2) The spring tension is not the only force acting on the ball.

Hint: Use conservation of energy.

3. Sep 9, 2007

### EvanQ

So potential energy = kinetic energy?
like:
kx = .5mv^2
667x0.25=.5(1.5)v^2
166.75=.75v^2
v^2=222.33
v=14.91m/s

4. Sep 9, 2007

### learningphysics

elastic potential energy in the spring is (1/2)kx^2. You also need to consider gravitational potential energy.

Analyze the system at two points. At the bottom when the spring is compressed, and at the muzzle...

5. Sep 10, 2007

### EvanQ

hmm ok.

Initially Ei = Ki + Ui = 0 + (Ugravity + Uspring)
= 0 + (0 + 1/2kx^2)
= (1/2)667 x .25^2
= 20.84375J

Final Ef = Kf + Uf = (1/2)mv^2 + (Ugravity + Uspring)
= (1/2)mv^2 + (mgh + 0)
= (1/2)(1.5)v^2 + (1.5x-9.8x0.25)
= .75v^2 - 3.675

Through conservation of energy:
Ei = Ef
20.84375 = .75v^2 - 3.675
24.51875 = .75v^2
v^2 = 32.69
v = 5.72m/s

??
bit confused as to what to use for h in the gravitational potential formula.

6. Sep 10, 2007

### learningphysics

This part should be: = (1/2)(1.5)v^2 + (1.5x9.8x0.25)

so that will change your results.

other than that it all looks good to me.

You used h = 0 at the bottom where the spring was compressed. So the place where the spring is uncompressed (the equilibrium position y = 0)... h = 0.25, just as you did.

You could also have used h = -0.25 at the bottom, and h = 0 at the top. It doesn't matter as long as the difference in the heights is 0.25.

7. Sep 10, 2007

### EvanQ

ok great thanks heaps.
got it correct, and finished off the next 3 questions which required that answer also.

8. Sep 10, 2007

### learningphysics

cool. no prob.

9. Sep 10, 2007

### EvanQ

similar question...

A toy gun uses a spring to project a 5.3 g soft rubber sphere. The spring constant k = 8.0 N m-1. The barrel of the gun is 15 cm long and there is a constant frictional force of 0.032 N between the barrel and the projectile. If the spring is compressed 5 cm what is the speed of the projectile from the barrel of the gun?

can do it using the same principles as above, just unsure on how to incorporate the friction force in instead of the potential gravity.

i tried using F=ma using the friction and mass of the projectile to get a deceleration, but then couldn't see how to put that in.

10. Sep 10, 2007

### learningphysics

Basically the same idea except this time mechanical energy isn't conserved:

work done by friction = Final mechanical energy - initial mechanical energy

By mechanical energy I mean kinetic energy, gravitational potential energy and elastic potential energy...

remember that friction does negative work.

Last edited: Sep 10, 2007