A question on Newton's 3rd Law

[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)?

 

[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.

 

[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...