# AP Physics: Momentum & Springs

1. Nov 14, 2015

### Thomas Brown

1. A mass M is connected to a wall by an ideal spring. The mass is on a frictionless surface. The mass is pushed toward the wall, compressing the spring by a distance X. Use the impulse-momentum theorem to demonstrate that the mass will reach a maximum velocity of X * (K / M)^1/2.

2. J = delta P, P = MV

3. J = delta P
Fn * t = MV - 0
-kx * t = MV What is t supposed to be? How can I finish this problem?

Last edited: Nov 14, 2015
2. Nov 14, 2015

### Student100

What is $\frac{\Delta v}{\Delta t}$ and what would the question wanting the maximum imply?

3. Nov 14, 2015

### Thomas Brown

I see that delta V / delta T would be the acceleration and that the velocity would be greatest when acceleration is zero (maximum) but if I divide both sides by delta T I end up with:
-kx = m dV/dT
-kx = ma
Fn = ma
And I'm not really sure what I could do next...

4. Nov 14, 2015

### Student100

So you have $a=\frac{-kx}{m}$ or $a={w^2}{x}$, is this looking familiar?

5. Nov 15, 2015

### Thomas Brown

I think I got it but can you please verify? I know that a = w^2 x is the acceleration in SHM, so I integrated the A(t) formula to get V(t) = xw * sin(wt) and velocity is greatest when sin(wt) = 1 so the maximum velocity is xw, or v = x * (k / m)^1/2.

6. Nov 15, 2015

### Student100

Yes, $v(t)=- \omega xsin(\omega t)$ when you take the phase constant to be zero. So what you said originally was true, the maximum speed in this motion occurs when the acceleration is zero, as well as when the displacement from equilibrium is the same. (Also applies for the min!) $V_{max/min}= \pm \omega A$ depending on the direction of oscillation $V_{max}= \sqrt{\frac{k}{m}} x$ depending on your coordinate system choosen.

Is this the best way to show what you wanted to show? I can't think of another way from the impulse-momentum theorem off the top of my head, but looks good to me.

7. Nov 15, 2015

### Thomas Brown

Yup, this makes sense! Thanks for your help!