Understanding the implications of a momentum equation

  • Thread starter Thread starter gibberingmouther
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
The discussion centers on understanding why a zero rate of change in total momentum implies that momentum is constant. The equation dP/dt = 0 indicates that the total momentum does not change over time, leading to the conclusion that momentum is a constant value. The integration of this equation confirms that the total momentum, represented as P, equals a constant C. One participant emphasizes that there is no alternative explanation, as the mathematical principle directly reflects the concept of constancy. Thus, the relationship between the derivative and the constancy of momentum is fundamentally straightforward.
gibberingmouther
Messages
120
Reaction score
15

Homework Statement



the rate of change of the total momentum in a system with respect to time is zero.
thus, my textbook says, the total momentum of that system is constant.

i'd like to understand how one implies the other if anyone could give me a conceptual explanation? i understand the steps to get the derivative of total momentum equation. i just don't have a good grasp on how that implies that the total momentum of the system is constant.

Homework Equations



dP/dt = 0 (where P is a vector quantity and is the total momentum of a system)

The Attempt at a Solution



integrate dP/dt = 0 and you get P = C, a constant. that is one way to get the answer but i am not satisfied. i want a better explanation.
 
Physics news on Phys.org
gibberingmouther said:
integrate dP/dt = 0 and you get P = C, a constant. that is one way to get the answer but i am not satisfied. i want a better explanation.
There is no "better" explanation than the explanation. The primitive function of zero is a constant. There is absolutely nothing more to it. The derivative with respect to time equal to zero literally tells you that the function does not change with time.
 
  • Like
Likes gibberingmouther
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