About perfectly inelastic collision

AI Thread Summary
In a perfectly inelastic collision, two objects collide and stick together, moving as one mass with a final velocity. This type of collision does not conserve kinetic energy, as demonstrated by the comparison of initial and final kinetic energy calculations. The conservation of momentum applies, but the total kinetic energy decreases because some energy is transformed into other forms, such as heat or sound. The mathematical analysis shows that the initial kinetic energy is always greater than the final kinetic energy after the collision. Therefore, this scenario cannot be classified as an elastic collision.
jack1234
Messages
132
Reaction score
0
[SOLVED] About perfectly inelastic collision

I have one problem about prefectly inelastic condition. When I read the reference book, it mentioned that when one object moving with an initial speed u1, and another object is in rest, after collision, if both object moved together as one object with a final speed v, this is perfectly inelastic condition.

My question is, what is the reason this cannot be an elastic collision? ie the total kinetic energy of system before =the total kinetic energy of the system after?
 
Physics news on Phys.org
jack1234 said:
I have one problem about prefectly inelastic condition. When I read the reference book, it mentioned that when one object moving with an initial speed u1, and another object is in rest, after collision, if both object moved together as one object with a final speed v, this is perfectly inelastic condition.

My question is, what is the reason this cannot be an elastic collision? ie the total kinetic energy of system before =the total kinetic energy of the system after?

Because, if the objects stick together and move with the same velocity after collision, energy is not conserved.

For a simple example, let's say you have an object of mass m1 and velocity v1 colliding with another object of mass m2 which is at rest. The conservation of momentum will give you,

m_1 v_1 = (m_1 + m_2)v_f

where, v_f is the final velocity of the combined objects which are now stuck together.

If you take the ratio of the KE before and after collision, you have

\frac{KE_{initial}}{KE_{final}} = \frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} (m1+m2) v_f^2}

Substituting for the value of v_f from the conservation of momentum, you get

\frac{KE_{initial}}{KE_{final}} = \frac{m1+m2}{m1}

This shows that the intial KE is always greater than the final KE and so energy is not conserved.
 
Last edited:
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