Center of Mass Movement in a perfectly inelastic collision

[libelec]

1. There are two 50 kg skaters in a friction-less horizontal surface. Both have velocities of |V|=10 m/s, in opossite directions (that is, skater 1's velocity is 10 m/s i, while skater 2's velocity is -10 m/s i). They both skate parallel to each other with a distance of 1,5m. Skater 1 has a stick, whose mass is [tex]\approx[/tex]0kg and 1,5m in lenght. In one point, Skater 2 will grab the stick.

The question is to describe the position and velocity of the center of mass just before and just after the collision.



2. Because the only external forces of the skater 1-skater 2 system are each skater's weight P and normal N, I found that [tex]\sum[/tex]F[tex]_{external}[/tex]=0. Therefore, [tex]\vec{P}[/tex] (momentum) is constant and therefore conserves. Because after one of the skaters holds the stick, both skaters will have the same speed, I assumed this is a perfectly inelastic collision. So, if skater 1's mass is m1, and skater 2's mass is m2[tex]\Rightarrow[/tex] m1[tex]\vec{V1i}[/tex] + m2[tex]\vec{V2i}[/tex] = (m1+m2)[tex]\vec{Vf}[/tex].

I can find the position of the center of mass using [tex]\vec{r}[/tex][tex]_{CM}[/tex]= (m1[tex]\vec{r1}[/tex] + m2[tex]\vec{r2}[/tex])/(m1+m2).

And its velocity with [tex]\vec{V}[/tex][tex]_{CM}[/tex]= (m1[tex]\vec{V1}[/tex] + m2[tex]\vec{V2}[/tex])/(m1+m2)

The Attempt at a Solution

Here's the thing. The problem asks to give the position of the center of mass and its velocity before and after the collision. In any case, using the conservation of momentum and the equations to find [tex]\vec{r}[/tex][tex]_{CM}[/tex] and [tex]\vec{V}[/tex][tex]_{CM}[/tex], I find that the position and velocity of the center of mass are the same just before and just after the collision, and particularly, the velocity is 0 m/s.

Am I doing something wrong? Because, intuitively, in a situation like this I expect the skaters to begin a circular motion with center in the center of mass. What am I missing?

 

[Redbelly98]

You're correct that the CM velocity is 0 m/s. The two skaters have opposite momenta of ±500 kg*m/s, making v_CM zero. And yes, the skaters will move in a circle after the collision.

 

[nlightner]

I would first like to commend you for approaching this problem with a critical and analytical mindset. Your approach using the conservation of momentum and equations for finding the position and velocity of the center of mass is correct. However, your intuition about the skaters beginning a circular motion with the center of mass is not entirely accurate in this scenario.

In a perfectly inelastic collision, the two objects stick together after the collision and move with a common velocity. In this case, the two skaters will indeed move with the same velocity after the collision, but their motion will not necessarily be circular. The stick, which is initially held by skater 1, will now be held by both skaters and will continue to move in a straight line with them.

The center of mass, being the average position of the two objects, will not experience any change in its position or velocity during the collision. This is because the external forces acting on the skaters, namely their weight and normal force, do not change during the collision. Therefore, the center of mass will continue to move with a constant velocity of 0 m/s.

Your intuition about circular motion may come from the idea of rotational motion, where an object can rotate around its center of mass. However, in this scenario, the skaters are moving in a straight line and do not experience any rotational motion.

In conclusion, your analysis of the position and velocity of the center of mass is correct, and the center of mass will not experience any change in its motion during the perfectly inelastic collision. I hope this explanation helps clarify any confusion and further develops your understanding of center of mass movement in collisions.