Conservation of linear momentum elastic string

[Kurokari]

Homework Statement

Two bodies P and Q of different masses on a smooth table are connected by a light elastic string which is stretched. P and Q are then released.

Which of the following is of the same magnitude for both P and Q

Speed, Acceleration, Momentum, Distance moved.

Answer: Momentum

Homework Equations

m1v1 + m2v2 = m1u1 + m2u2

The Attempt at a Solution

Since both bodies are at rest first, so initial velocity for both are 0.

This makes m1v1 + m2v2 = 0, m1v1 = - m2v2

My concern is that, since the question stated that both the bodies P and Q have different masses, so this must mean that in the equation m1v1 = - m2v2, the velocity v1 and v2 is different. The question does not mention anything about the length of the string extension, so I assume that it's the same length for both bodies P and Q.

Ok here's the deal, does this mean that velocity is somehow influence by the mass, like a ratio thing seeing how the equation can be expressed in a ratio form only with a negative sign.

If so, does this apply in a general sense, or only towards this case which is conservation of momentum.

or am I thinking far far away and has gotten it all wrong? :D

 

[gneill]

Ok here's the deal, does this mean that velocity is somehow influence by the mass, like a ratio thing seeing how the equation can be expressed in a ratio form only with a negative sign.

If so, does this apply in a general sense, or only towards this case which is conservation of momentum.

or am I thinking far far away and has gotten it all wrong? :D

The light elastic string applies forces to the bodies that must be equal and opposite (since the tension in a stretched string is everywhere the same). The resulting accelerations, and hence velocities, depend upon the individual masses of the bodies. F = MA. So, yes, you could say it's a 'ratio thing'.

 

[chaoseverlasting]

Velocity is not related to mass. As the momentum is conserved in the horizontal direction, the velocities of the bodies are related to mass. This particular equation is only applicaple where momentum is conserved. In other cases, you'd have to work out the equation fully for the exact expression.

You've got it right though.