2D Collision b/w two objects, How do I find the speed of the centre of mass?

AI Thread Summary
To find the center of mass in a 2D collision involving two objects, the correct formula is x_CM = (Σ(x_i * m_i)) / Σm_i for both x and y coordinates. In this case, with two 500g masses, the center of mass can be calculated by multiplying each position by its mass and dividing by the total mass. For momentum, the center of mass velocity is the average of the velocities of both masses, weighted by their respective masses. It is essential to consider the velocities before and after the collision, but only the masses of the two objects should be included in the calculation. Understanding these principles will clarify how to approach the problem effectively.
TaintedLove
Messages
9
Reaction score
0
Hi, I was attempting to work through this question.
Anyways, I was working from a graph, so the radius 2.5 cm, the two of them collided, it was an inelastic collision as the two objects did part ways, so how do I find the centre of gravity?
I attempted to calculate it... but if someone could help me out? The masses were 500g and 500g.

So for x and y, would I do:
Xcg = (500)(0.025) + (500)(1.5)(0.025) / 1000 ?
Ycg = (500)(0.025) + (500)(0.025) / 100 ?

That's what I was thinking but I'm not entirely sure.

After this, I also have to find the momentum of the centre of gravity... so for that, I need the masses and the speed.

Even though the objects didn't stick together, would I add the masses when solving for the momentum at the centre of mass?

I was reading on the internet and it said: "The center of mass velocity of a system of particles is the average velocity of all the particles weighted relative to their mass"

but what I'm confused about is, do I add all the velocities... like the velocities of both the masses before and after the collision? So I'd have 4 velocities? But I'd only add the masses twice right? So it would be the velocities / 1000 g

Could someone really help me. Please.
 
Physics news on Phys.org
It'd be rather easier for all involved if you use the homework question template. Or at least post the exact wording of the problem, followed by your attempt at solving it.

The formula for center of mass is
x_\text{CM} = \frac{\sum_i x_i m_i}{\sum_i m_i}
and similarly for y_\text{CM}. That is, for each particle involved, you multiply its position by its mass, then add up all those products, and finally divide by the sum of all the masses. That would be the average position of all the particles weighted by their masses. You can do the same thing with the velocities of the particles to find the velocity of the center of mass.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top