Conservation of Momentum of masses

AI Thread Summary
The discussion revolves around a physics problem involving the conservation of momentum following an explosion of a mess kit into two parts. The original speed of the mess kit is calculated using momentum equations, resulting in a velocity of 3.5 m/s. Participants clarify that while the masses are equal, their momenta can differ due to the directions in which they are emitted. The conversation highlights the distinction between momentum and kinetic energy, emphasizing that kinetic energy is not conserved during the explosion. Overall, the participants confirm the calculations and explore the implications of momentum conservation in non-symmetric explosions.
MostlyHarmless
Messages
344
Reaction score
15

Homework Statement


A 4.0kg mess kit, ##m_1## sliding on a frictionless surface explodes in two 2.0kg parts, ##m_2## and ##m_3##: ##m_2##@3.0m/s due north and##m_3##@ 5.0m/s, 30 degrees north of east. What is the original speed of the mess kit?


Homework Equations


Momentum=mv


The Attempt at a Solution


After thinking about it for a bit, I thought that I could think about the problem in reverse, i.e. the two 2kg masses colliding in sticking. I broke the momentum of ##m_3## into x and y components and got ##P_{3x}=8.66N*s## and ##P_{3y}=5N*s## ##m_2## has only a y component of 6N*s. So the total momentum in x, after the collision, would be 8.66N*s, and in y would be 11N*s. Using these two values I found the magnitude of the momentum of the resulting mass,##m_1##, to be 13.99N*s, divided by its mass, 4kg, and found its velocity to be 3.5 m/s.

And to verify that this seemed right, it would have been moving at 51 degrees below the negative x axis.

It all seems within reason, but I'm not sure.

This was an even numbered problem and again I'm unsure if what I'm doing is right. Does this look right? If not, could someone steer me in the right direction.
 
Physics news on Phys.org
Your approach looks fine, and the result looks good as well.
 
Awesome, thank you. I have a follow up question though.

How is it possible for a moving object to break into two equal mass pieces, but those pieces have different magnitudes for their momentum? I'm having a hard time visualizing that in my head.
 
The two pieces are not emitted in a symmetric way relative to the initial motion of the object. As extreme example, one object could be emitted in the original direction of motion (->quicker than the initial object), and the other object could be emitted in the opposite direction (can be slower than the original object, or even be at rest afterwards).
 
Oh ok, I was picturinf a literal explosion, I guess it could be 2 springs directed at different directions. I think I keep getting hung up on the idea kinectic energy not alway being conserved.
 
Kinetic energy is not conserved in this process, right. The mass needs some other type of energy to "explode".
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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