My physics book says, "A firecracker sliding on ice has the same total momentum before and after it explodes." I understand this part. This is because of Newton's 3rd law, and no external forces. This is what I really don't get. "The same, however, is not true of a system's kinetic energy. Energetically, that firecracker is very different after it explodes; internal potential energy has become kinetic energy of fragments." It goes on to say, "Nevertheless, the centre-of-mass concept remains useful in categorizing the kinetic energy associated with a system of particles."
How is it that the kinetic energy increased but momentum stayed the same?
My problem lies within the equation $K = \frac{1}{2}mv^2$ and $p= mv$.
If kinetic energy increased doesn't that mean that the velocities increased as well? How else would $K$ become more positive? And since $K$ increased $v$ increased and thus $p=mv$ must increase?
Best Answer
The formula for momentum is not $p=mv$ but it is $\vec p=m\vec v$. This being said, after an explosion, the velocities of the fragments have increased and so is the kinetic energy of the entire system (chemical energy $\rightarrow$ kinetic energy in the explosion) but the net momentum does not change. If the system had zero net momentum before the explosion, after the explosion it will have the same (zero) momentum.
Consider a system of $N$ fragments of masses $m_1,m_2,m_3....m_N$ which are released after the explosion with different velocities $v_1,v_2,v_3,....v_N$ in different direction. What conservation of momentum says is:-
$$\vec P_{net}\text{before explosion} =\vec P_{net} \text{after explosion}$$ $$0=\vec P_{net} \text{after explosion}=m_1\vec v_1+m_2\vec v_2+....m_N\vec v_N=\Sigma_{i=1}^{N} m_i\vec v_i$$ $$\sum_{i=1}^{N}m_i \vec v_i=0$$ That implies, that the mass weighted vectorial sum of all fragment velocities must be zero for the system's net momentum to be zero, they can have individual non-zero magnitudes, and hence non-zero individual momentums.
If the initial momentum is not zero but $\vec {p_f}$ then $$\vec {p_f}=\sum_{i=1}^{N}m_i\vec v_i$$