# Spring gun and muzzle speed

1. Dec 4, 2004

### Heart

Here's the question:
"A spring-loaded toy gun is used to shoot a ball of mass m straight up in the air. The spring has spring constant k. If the spring is compressed a distance x_0 from its equilibrium position 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 x_0<<<h_max."

"Find , the muzzle velocity of the ball (the velocity of the ball at height above the equilibrium position). Express your answer in terms of given quantities."

I tried v = sqrt(2g(h_max-h)); but it didn't work.
I'm not sure if I should try v = sqrt[((kx_0^2)-(2mgh))/m], since the question suggested me after my first answer that g, h_max, and h are not relevant variables - in the past, for some weird reasons, the generic variables that are not given in the question worked.

Another question is finding h_max; if h_max is not related to the previous question then how am I suppose to find it.

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2. Dec 4, 2004

### Heart

BTW, these were the conservation of energy equation I made
(1/2)kx_0^2 = (1/2)mv^2 + mgh = mgh_max

The first end was only (1/2)kx_0^2 because x_0 <<<<< h_max

Where did I go wrong?

3. Dec 4, 2004

### CartoonKid

Since the spring does work against gravitational force to bring the ball a distance $$x_0$$ higher, you can consider the spring has done $$mgx_0$$ of work. Therefore, by using the method of conservation of energy, your equation should be:
$$\frac{kx_0^2}{2}=\frac{mv^2}{2}+mgx_0$$

4. Dec 4, 2004

### CartoonKid

You can approach the question by using the integration method as well. First,
$$F=-(kx_0-mg)$$
$$ma=-kx_0+mg$$
$$v\frac{dv}{dx}=-\frac{k}{m}x+g$$
$$\int(v)dv=\int(\frac{k}{m}x+g)dx$$
This method also yields the same answer.

Last edited: Dec 4, 2004
5. Dec 4, 2004

### CartoonKid

To find $$h_{max}$$, you have to make use of the velocity you found in the first question. Equalling the kinetic energy at the equilibrium point to the potential energy at the highest point. There will be no kinetic energy at the highest point, the velocity of the ball is instantaneously zero at that point.

6. Dec 4, 2004

### Staff: Mentor

Two problems:
(1) The muzzle velocity is the speed of the ball at a height = 0 above the equilibrium position, not h.
(2) h_max is not given, it's just the label for the maximum height. So you can't use it to represent your answer.

To find the muzzle velocity, use the energy conservation method that CartoonKid gave:
$$\frac{kx_0^2}{2}=\frac{mv^2}{2}+mgx_0$$
Since x_0 << h_max, the mgx_0 factor may be ignored:
$$\frac{kx_0^2}{2}=\frac{mv^2}{2}$$
Use the exact same method, only now have all the spring PE go to gravitational PE (set KE = 0):
$$\frac{kx_0^2}{2}= mg(x_0 + h_{max})$$
Again, since x_0 << h_max, the mgx_0 factor may be ignored:
$$\frac{kx_0^2}{2}= mg(h_{max})$$

7. Dec 4, 2004

### Heart

Hi CartoonKid,

Thanks for your help. It didn't work though, I actually tried that twice so now I've got only one more chance left. Do you think I might have put in the wrong answr

what I put in was
v = sqrt((k((x_0)^2)/m)-(2*gx_0))
the first time I put g(x_0) and it told me that g() function is not part of the answer
I then tried put it as gx_0 and it told me that gx_0 variable is not part of the answer