Momentum is conserved in an inelastic collision

Momentum is conserved in an inelastic collision. Let us consider the case of two masses, \(m_1\) and \(m_2\), with velocities \(v_1\) and \(v_2\), respectively. The initial momentum is given by \[\begin{equation} p_0 = m_1 v_1 + m_2 v_2\ \text{initial momentum}. \end{equation}\] In a perfectly inelastic collision, the two masses stick together, forming a single mass equal to \(m_1+m_2\). The final velocity is then obtained from momentum conservation \[\begin{equation} p_f = (m_1 + m_2) v_f = p_0. \end{equation}\] We can substitute for \(p_0\) and solve for \(v_f\): \[\begin{align} (m_1 + m_2) v_f &= m_1 v_1 + m_2 v_2 \\ \dfrac{1}{\cancel{m_1 + m_2}} \cancel{(m_1 + m_2)} v_f &= \dfrac{1}{(m_1 + m_2)} (m_1 v_1 + m_2 v_2) \\ v_f &= \dfrac{m_1 v_1 + m_2 v_2}{m_1 + m_2} . \end{align}\]

Energy decreases in an inelastic collision

Energy decreases during an inelastic collision. For our perfectly inelastic collision, the kinetic energy before the collision is \[\begin{equation} E_0 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \end{equation}\] Some energy is lost when the two bodies stick together. The kinetic energy after the collision is \[\begin{align} E_f &= \frac{1}{2} (m_1 + m_2) v_f^2 \\ &= \frac{1}{2} \dfrac{(m_1 v_1 + m_2 v_2)^2}{m_1 + m_2} \end{align}\]

Note that these results are for a perfectly inelastic collision. A collision in which the objects do not stick, but bounce off at velocities that show the final system kinetic energy is less than the initial system kinetic energy is an inelastic collision too, just not a perfect one.

Momentum is conserved in an explosion

Momentum is conserved in an explosion. First, let us consider the case of two masses, \(m_1\) and \(m_2\), initially at rest. The initial momentum is given by \[\begin{equation} p_0 = m_1 0 + m_2 0 = \qty{0}{\kilo\gram\meter\per\second}\ \text{initial momentum}. \end{equation}\]

The two masses explode apart and head off with velocities \(v_{1,f}\) and \(v_{2,f}\), respectively so that the final momentum is \[\begin{equation} p_f = m_1 v_{1,f} + m_2 v_{2,f} = p_0. \end{equation}\]

Typically in a problem like this, one of the velocities is given and we must solve for the other velocity. Let’s assume \(v_{2,f}\) is given, then: \[\begin{align} m_1 v_{1,f} + m_2 v_{2,f} &= 0 \\ m_1 v_{1,f} &= -m_2 v_{2,f} \\ v_{1,f} &= - \dfrac{m_2}{m_1} v_{2,f}. \end{align}\] It’s worthwhile to consider some limiting cases.

If \(m_1 \gg m_2\), which might be the case where \(m_1\) is a cannon and \(m_2\) is a cannon ball, then \(v_{1,f}\ll v_{2,f}\). This makes sense when firing a cannon; hopefully the recoil velocity of the cannon is much less than the velocity of the cannonball.

If \(m_1 = m_2\), then both will be shot off at equal velocities. On physics tests this might be the case when two astronauts of equal mass push off each other during a spacewalk.

Finally, if \(m_1 \ll m_2\), it is like the opposite of the cannon/cannonball case. This might be encountered if, say, an astronaut tried to push off a very heavy satellite, or if a chihuahua jumped off a Mack truck.

Energy increases in an explosion

Energy increases during an explosion. The kinetic energy before the explosion (if starting from rest), by inspection, is \[\begin{equation} E_0 = 0. \end{equation}\]

Energy has to be supplied to cause the explosion: \[\begin{equation} E_f = \frac{1}{2} m_1 v_{1,f}^2 + m_2 v_{2,f}^2, \end{equation}\] noting that this must be positive due to the squared terms.

Momentum and energy are conserved in an elastic collision

Momentum and energy are conserved in an elastic collision. First, let us consider the case of two equal masses, each mass \(m\), moving towards one another, each at velocity \(v\). Before they collide: \[\begin{align} p_0 &= m v - m v = \qty{0}{\kilo\gram\meter\per\second}\ \text{initial momentum}\\ E_0 &= \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2\ \text{initial energy} \end{align}\]

The two masses collide and bounce so that each is now heading apart at velocity \(v\). It is clear that after the collision, the momentum and energy of the system are both the same \[\begin{align} p_f &= -m v + m v = 0 = p_0, \\ E_f &= \dfrac{1}{2}mv^2 + \dfrac{1}{2}mv^2 = E_0. \end{align}\]

More general case of elastic collision

Let’s consider the more general case of an elastic collision in a system with two masses \(m_1\) and \(m_2\). Momentum conservation gives \[\begin{align} p_0 &= m_1 v_{1,0} + m_2 v_{2,0} \\ p_f &= m_1 v_{1,f} + m_2 v_{2,f} \\ p_0 &= p_f. \end{align}\] If \(v_{1,0}\) and \(v_{2,0}\) were known, momentum conservation gives me one equation and two unknowns, so I need a way to get a second equation in order to solve for both \(v_{1,f}\) and \(v_{2,f}\). We can get a second equation from energy conservation: \[\begin{align} E_0 &= \dfrac{1}{2} m_1 v_{1,0}^2 + \dfrac{1}{2} m_2 v_{2,0}^2 \\ E_f &= \dfrac{1}{2} m_1 v_{1,f}^2 + \dfrac{1}{2} m_2 v_{2,f}^2 \\ E_0 &= E_f, \end{align}\] but now we are a little stuck as to how to proceed.

If we were to rewrite the velocities relative to the velocity of the center of mass, we could pull a trick like what we did in solving for momentum conservation during an explosion with the system initially moving. Let’s try it. \[\begin{align} V_{com} &= \dfrac{m_1 v_{1,0} + m_2 v_{2,0}}{m_1 + m_2} \\ V_1 &= v_{1,0} - V_{com} \\ V_2 &= v_{2,0} - V_{com} \end{align}\]

Working with these, we can rewrite momentum and energy conservation: \[\begin{align} p_0 &= m_1 (V_1 + V_{com}) + m_2 (V_2 + V_{com}) \\ &= m_1 V_1 + m_2 V_2 + (m_1 + m_2) V_{com} \\ E_0 &= \dfrac{1}{2} m_1 (V_1 + V_{com})^2 + \dfrac{1}{2} m_2 (V_2 + V_{com})^2 \\ \end{align}\]

We note that \(m_1 V_1 + m_2 V_2=0\). The solution is that \(m_1\) is reflected back at \(-V_1\), and likewise, \(m_2\) is reflected back at \(-V_2\), so that: \[\begin{align} p_f &= m_1 (-V_1 + V_{com}) + m_2 (-V_2 + V_{com}) \\ &= - m_1 V_1 - m_2 V_2 + (m_1 + m_2) V_{com} \\ E_f &= \dfrac{1}{2} m_1 (-V_1 + V_{com})^2 + \dfrac{1}{2} m_2 (-V_2 + V_{com})^2 \end{align}\]