# A question on Newton's 3rd Law

1. Aug 2, 2008

### DiamondGeezer

I don't have a diagram for this but I hope the description is vivid enough.

Imagine a mass M on a horizontal smooth surface (ie no friction). The mass M is between two bars equidistant from M and connected to the mass via two springs (if you imagine a square mass with springs attached to the corners which connect to the fixed bars). The springs are ideal springs of spring constant $$k$$ and the mass M is sitting at the equilibrium position

At some time $$t=0$$ a small mass m is ejected from the larger mass M and embeds itself in putty in one of the fixed bars.

The mass (M-m) therefore will react to the ejection of m by moving in the opposite direction and will exhibit SHM about the equilibrium position.

I guess that if the mass m leaves the larger mass M at speed u, then the law of conservation of momentum would mean that the remain mass will recoil at velocity v such that

$$mu + (M-m)v=0$$

Is this correct?

What would be the equation of motion of the mass (M-m)?

2. Aug 3, 2008

### NERV

At this very moment(t=0),the mass(M-m) will recoil at the speed v0=mu/(M-m).Then from now on,the remain mass can be regarded as a spring oscillator,whose initial speed is v0.

If you want to know the motion equation of the remain mass(Let's call it M').You need the knowledge of trigonometric functions.

For M',the energy conservation equation is:
0.5(M-m)v02=0.5×2k×A2(A:amplitude)---------(1)
while ω=(2k/m)0.5--------------(2)
and x=Asin(ωt)----------(3)
Now you can get an function x(t) by (1)-(3).I'm sorry that I cannot write the result here for it's a little complicated.

As you have got x=x(t),you can write down v=v(t) and even a=a(t) easily by derivation.

Last edited: Aug 3, 2008
3. Aug 3, 2008

### aniketp

$$\xi$$=Asin(kx-wt) is the general eqn.
To get A equate initial K.Eof (M-m) to the spring energy when the K.E of (M-m) becomes zero. "w" is the angular frequency and "k" is the wave no of the wavw describing the motion of (M-m).Hope I helped...