- #1
DiamondGeezer
- 126
- 0
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 [tex]k[/tex] and the mass M is sitting at the equilibrium position
At some time [tex]t=0[/tex] 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
[tex] mu + (M-m)v=0[/tex]
Is this correct?
What would be the equation of motion of the mass (M-m)?
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 [tex]k[/tex] and the mass M is sitting at the equilibrium position
At some time [tex]t=0[/tex] 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
[tex] mu + (M-m)v=0[/tex]
Is this correct?
What would be the equation of motion of the mass (M-m)?