# Kinetic energy and momentum conservation in an explosion?

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?

-
momentum is a vector and the contributions of parts of the fire cracker blasted in opposite directions will cancel – Christoph Oct 29 '13 at 10:14

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 $$\vec {p_f}=\sum_{i=1}^{N}m_i\vec v_i$$

-

This might be easier to understand if you think of a simpler case where the "firecracker" is only made up of two masses. Let's say the two masses are 40 g each, and when the device explodes it sends each mass going the opposite way at 130 m/s.

Before the explosion, the combined 80 g mass is moving to the right at 1 m/s. Its momentum is therefore 80 g m / s to the right. It's kinetic energy is ½ * 80g * (1m/s)2 = 40 mJ.

After the explosion, 40 g is moving to right at 131 m/s, and the other 40 g to the left at 129 m/s. The total momentum is (-129m/s * 40g)+(+131m/s * 40g), which is still the same 80 g m / s as before. However, the kinetic energy is now quite different:

½ * 40g * (129m/s)2 + ½ * 40g * (131m/s)2 = 676 J

-
@Satwik: Thanks for trying to make this post look nicer, but I'm rolling back the edit because you introduced several serious errors into the equations. – Olin Lathrop Oct 29 '13 at 15:48

And since K increased v increased and thus p=mv must increase?

For an individual particle, yes. For a system of particles, no.

Consider two identical particles, co-located and initially at rest. At initial rest, the total kinetic energy is zero and the total momentum is zero.

Now, due to some mechanism, the particles are sent in opposite directions with the same speed so the momenta are equal and opposite thus, the total momentum is zero even though the individual momenta are not.

$$p = mv + m (-v) = m(v-v) = 0$$

However, the kinetic energy of each particle are equal since KE depends on speed, not velocity. Thus, the total KE is twice the individual KE.

$$KE = \frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mv^2$$

-