Kinetic energy and momentum of two objects

AI Thread Summary
Two objects can have the same kinetic energy but different momentum, as demonstrated by a 5 kg mass moving at 50 m/s and a 125 kg mass moving at 10 m/s, both possessing 6.25 kJ of kinetic energy but differing in momentum (250 kg*m/s vs. 1250 kg*m/s). If both objects have zero momentum, it implies they are at rest, resulting in zero kinetic energy as well. The relationship between kinetic energy and momentum is defined by their formulas: kinetic energy (KE = 1/2 mv^2) and momentum (p = mv). Therefore, while kinetic energy can be the same for different masses and velocities, momentum is dependent on both mass and velocity. This discussion highlights the distinct nature of kinetic energy and momentum in physics.
wilmerena
Messages
11
Reaction score
0
do 2 objects that have the same kinetic energy necessarily have the same momentum? I can't think of a simple example
 
Physics news on Phys.org
wilmerena said:
do 2 objects that have the same kinetic energy necessarily have the same momentum? I can't think of a simple example
Consider a 5 kg mass going at 50 m/s, thus having 6.25 kJ of kinetic energy. Now, a mass of 125 kg going 10 m/s also has 6.25 kJ of kinetic energy. However, the first object has a momentum of 250 kg*m/s, and the second has 1250 kg*m/s of momentum.
 
thanks so much =o) !
 
what if they both have 0 momentum, so does it follow that the kinetic energy of the system doesn't have to be zero as well? or does it?
 
wilmerena said:
what if they both have 0 momentum, so does it follow that the kinetic energy of the system doesn't have to be zero as well? or does it?
Think about it. For a simple object:
KE = 1/2 mv^2
momentum = mv
The only way an object can have zero momentum is if its speed (v) is what? Then what is its KE?
 
thanks again, i think i just spaced out on that one ;oP
 
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 '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 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