Question

a 1 - kg object moving at 6 m/s collides with a 3 - kg object moving at 4 m/s in the same direction. data for the momentum p as a function of time t for both objects is shown in the graph. which of the following statements best describes this collision?
a the momentum of the two - object system is constant, and the magnitude of the relative velocity between the objects is the same before and after the collision.
b the momentum of the two - object system is constant, and the magnitude of the relative velocity between the objects is different before and after the collision.
c the momentum of the 1 - kg object is constant, and the magnitude of the relative velocity between the objects is the same before and after the collision.
d the momentum of the 1 - kg object is not constant, and the magnitude of the relative velocity between the objects is different before and after the collision.

Explanation:

Step1: Check momentum conservation

According to the law of conservation of momentum, for a closed - system (no external forces), the total momentum of the system is constant. In a collision, if we consider the two - object system, there are no external forces acting on the system during the collision (assuming internal forces only, like the force between the two colliding objects). So, the momentum of the two - object system is constant.

Step2: Calculate relative velocity before and after

The relative velocity before the collision \(v_{rel - before}=v_1 - v_2\), where \(v_1 = 6m/s\) and \(v_2=4m/s\), so \(v_{rel - before}=6 - 4=2m/s\).
Let the velocities after the collision be \(v_1'\) and \(v_2'\). From the graph (since momentum \(p = mv\), and for the two - object system \(m_1v_1 + m_2v_2=m_1v_1'+m_2v_2'\), \(1\times6+3\times4=1\times v_1'+3\times v_2'\), \(6 + 12=v_1'+3v_2'\), \(v_1'=18 - 3v_2'\)).
If we assume elastic collision (from the fact that relative velocity magnitude is same, which is a property of elastic collision), using \(v_{rel - before}=v_{rel - after}\), \(v_{rel - after}=v_2'-v_1'\) (if we assume the direction is such). Substituting \(v_1'=18 - 3v_2'\) into \(v_{rel - after}=v_2'-v_1'\), we get \(v_{rel - after}=v_2'-(18 - 3v_2')=4v_2'-18\). Setting \(4v_2'-18 = 2\) (since \(v_{rel - before} = 2m/s\)), we find \(v_2'=5m/s\) and \(v_1'=3m/s\), and \(v_{rel - after}=5 - 3=2m/s\)

Answer:

A. The momentum of the two - object system is constant, and the magnitude of the relative velocity between the objects is the same before and after the collision.