7:16 is FALSE. (Although he is correct that Initial energy > Final energy). Here is the correction: Firstly, I will relabel: m1 as m, m2 as m' and u1 as u. Secondly, bearing in mind that V is clearly less than u & m is less than M: The energy change is _defined as_ : ∆KE= Final energy - Initial energy (This is from energy momentum theorem.) ∆KE=½(MV²-mu²), M=m+m' =½(M(mu)²/M² -mu²) =½mu²(m/M -1)< 0, as m/M=1/(1+m'/m)
Great explained sir
Love you sir♥️♥️♥️
What you have is the _energy loss_ ,
which can be thought of as a _negative_ energy gain!!
The final energy is initial _plus_ that gain.
Super❤❤❤
How we can explain the existance of reduced mass in the expression of the the difference between final kinetic energy and initial kinetic energy?
Yes. The reduced mass is m1m2/m1+m2
7:16 is FALSE.
(Although he is correct that
Initial energy > Final energy).
Here is the correction:
Firstly, I will relabel:
m1 as m, m2 as m' and u1 as u.
Secondly, bearing in mind that
V is clearly less than u & m is less than M:
The energy change is _defined as_ :
∆KE= Final energy - Initial energy
(This is from energy momentum theorem.)
∆KE=½(MV²-mu²), M=m+m'
=½(M(mu)²/M² -mu²)
=½mu²(m/M -1)< 0, as m/M=1/(1+m'/m)