Analyzing inelastic collisions

An inelastic collision is a collision in which energy is not conserved. A perfectly inelastic collision is a special type of inelastic collision in which two or more objects "stick together" to form one final object. (One object splitting into more than one object can also be treated as a perfectly inelastic collision.) As with any collision, conservation of momentum can always be used.

Conservation of momentum:

\[m_1v_{1_0} + m_2v_{2_0} = m_1v_{1_f}+m_2v_{2_f}\]

Analysis of inelastic collisions permits investigation into very common interactions in physics that are inaccessible with conservation of energy considerations. As a result, the more robust conservation of momentum forms a foundational bedrock for all of physics.

Consider two particles of mass \(m_{1}\) and \(m_{2}\) moving at velocities \(\vec{v}_{1}\) and \(\vec{v}_{2}\), respectively. Before they collide, they have a combined energy of \(E_{\text{init}} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{1} v_{2}^{2}\) and a combined momentum of \(\vec{p}_{\text{init}} = m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}\).

Since the collision is perfectly inelastic, after the collision there is a single combined object of mass \(m_{1} + m_{2}\). Since momentum is conserved, this object has momentum equal to the total intitial momentum, \(\vec{p} = (m_{1} + m_{2}) \vec{v}_{f}\). The velocity of the combined object \(\vec{v}_f\) is thus

\[\begin{align} (m_{1} + m_{2}) \vec{v}_{f} &= m_{1} \vec{v}_{1} + m_{2} \vec{v}_{2}\\ \vec{v}_{f} &= \frac{m_1}{m_1 + m_2} \vec{v}_1 + \frac{m_{2}}{m_1 + m_2} \vec{v}_2. \end{align}\]

The energy depends on the squared magnitude of \(\vec{v}_f\), which is the dot product of \(\vec{v}_f\) with itself. If the angle between \(\vec{v}_1\) and \(\vec{v}_2\) is \(\theta\), then this equals

\[ \| \vec{v}_f \|^2 = \frac{m_1^2}{(m_1 + m_2)^2} v_1^2 + \frac{m_2^2}{(m_1 + m_2)^2} v_2^2 + \frac{m_1 m_2}{(m_1 + m_2)^2} v_1 v_2 \cos \theta.\]

The final energy \(E_f\) is

\[\begin{align} E_f &= \frac{1}{2} (m_1 + m_2) \| \vec{v}_f \|^2\\ &= \frac{1}{2} \left[\frac{m_1^2}{(m_1 + m_2)} v_1^2 + \frac{m_2^2}{(m_1 + m_2)} v_2^2 + 2 \frac{m_1 m_2}{(m_1 + m_2)} v_1 v_2 \cos \theta\right]. \end{align}\]

This equation is the general solution for perfectly inelastic collisions. It's somewhat ugly, but exploring how it works in particular simplified cases can help build intuition for what it says.

What is the energy difference \(\Delta E = E_ f - E_ i\) if \(m_2\) is much much smaller than \(m_1?\) (Physicists express this with symbols as \(m_2 \ll m_1\).)

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In this case, \(m_1 + m_2 \approx m_1\), which simplifies the equation to

\[\begin{align} E_f &= \frac{1}{2} \left(\frac{m_1^2}{m_1} v_1^2 + \frac{m_2}{m_1} v_2^{2} + 2 \frac{m_1 m_2}{m_1} v_1 v_2 \cos \theta\right) \\ &= \frac{1}{2} \left(m_1 v_1^2 + 2 m_2 v_1 v_2 \cos \theta + \frac{m_2}{m_1} m_1 v_2^2\right). \end{align}\]

Since \(m_2 \ll m_1\), \(\frac{m_2}{m_1} \ll 1\), so the last term is small if in addition \(v_2\) is smaller than or not much larger than \(v_1\). These combined assumptions allow \(E_f\) to be further simplified to

\[\begin{align} E_f &= \frac{1}{2} (m_1 v_1^2 + 2 m_2 v_1 v_2 \cos \theta)\\ \\ \Rightarrow \Delta E &= E_f - E_i \\&= m_2 v_1 v_2 \cos \theta - \frac{1}{2} m_2 v_2^2. \end{align}\]

This equation gives a nice interpretation for this limiting case. The second term "eliminates" the energy of the original particle, while the first term "creates" a particle of mass \(m_2\) with velocity projected in the direction of the more massive \(m_1\), because it's stuck to \(m_1\). The energy of the mass \(m_1\) is left unchanged.

Take special care that this simplification required that the velocity of the smaller particle was not too high. If it were, then the smaller party would have sufficiently high energy to change the momentum and energy of the larger particle significantly, and this approximation would be invalid.

Perfectly inelastic collisions are only one type of inelastic collision. Many collisions--like a basketball bouncing on the floor--involve collisions that seem elastic, but still lose energy. Basketballs don't bounce up and down at the same height forever. They lose energy and momentum every time they bounce. Without guaranteeing any conserved quantity, solving this problem in the general case is very difficult.

One solvable case is the case in which the relative velocities of the outgoing particles have a magnitude that is some fixed fraction of the relative velocities of the incoming particles. This fraction is called the coefficient of restitution. Mathematically, the coefficient of restitution \(e\) for a collision between two particles is given by

\[e = \frac{\| \vec{v}_{2f} - \vec{v}_{1f} \|}{\| \vec{v}_{2i} - \vec{v}_{1i} \|},\]

where \(\vec{v}_{1i}\) and \(\vec{v}_{2i}\) are the initial velocities and \(\vec{v}_{1f}\) and \(\vec{v}_{2f}\) are the final velocities.

Unless latent energy is released in a collision (for instance, a bound spring inside of a ball that is released), then \(e \leq 1\) since kinetic energy is lost to the environment. In the ideal case of a perfectly elastic collision, \(e = 1\).

Consider the diagram shown below. It shows an inelastic collision between two moving objects on a frictionless ground.

Here, \({u_1}\) and \({u_2}\) are initial velocities of block \({m_1}\) and \({m_2}\), respectively, and \({v_1}\) and \({v_2}\) are their respective final velocities after the collision.

The two blocks travel along a frictionless line and hit each other. The collision is inelastic with coefficient of restitution \(e\).

As there is no net external force in the horizontal direction, linear momentum is conserved:

\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}. \qquad (1)\]

From the equation for coefficient of restitution,

\[e = \frac{{{v_2} - {v_1}}}{{{u_1} - {u_2}}}. \qquad (2)\]

From equations (1) and (2), we get

\[\begin{align} {v_1} &= \left( {\frac{{{m_1} - e{m_2}}}{{{m_1} + {m_2}}}} \right)\,{u_1} + \left[ {\frac{{(1 + e)\,{m_2}}}{{{m_1} + {m_2}}}} \right]\,{u_2}\\ {v_2} &= \left[ {\frac{{(1 + e)\,{m_1}}}{{{m_1} + {m_2}}}} \right]\,{u_1} + \left( {\frac{{{m_2} - e\,{m_1}}}{{{m_1} + {m_2}}}} \right)\,{u_2}. \end{align}\]

Note that by setting \(e = 1,\) we can get \({v_1}\) and \({v_2}\) for a perfectly elastic collision in one dimension.

A ball is dropped from a height of \(h\) above the ground, with initial velocity 0. It bounces up and down many times before eventually coming to rest. If the coefficient of restitution for the bounces is \(e\), what are the total time taken and the total vertical distance traveled, before it comes to rest?

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First, determine the velocity \(v_1\) at which the ball first hits the ground. This can be derived in several ways, but the simplest is considering conservation of energy. The potential energy \(U\) at a height \(h\) is \(U = mgh\). By conservation, the change in kinetic energy is

\[\begin{align} \Delta K &= - \Delta U\\ \frac{1}{2}mv_1^2 - 0 &= - (0 - mgh)\\ v_1^2 &= 2gh\\ \Rightarrow v_1 &= \sqrt{2gh}. \end{align}\]

Next, determine the time \(t\) it takes to reach the ground:

\[\begin{align} s &= ut + \frac{1}{2}a{t^2}\\ h &= \frac{1}{2}g{t^2}\\ \Rightarrow t &= \sqrt {\frac{{2h}}{g}}. \end{align} \]

Since the coefficient of restitution is \(e\), each time the ball rebounds its next velocity is reduced by the factor \(e\). Specifically, \(v_{n+1} = e v_{n}\). Since \(v_1 = \sqrt{2gh}\), we have \(v_{n} = e^{n-1} \sqrt{2gh}\).

Now, the ball will reach a height less than \(h,\) and let the height reached be \({h_1}\). Then taking the upward direction as positive, one can write \[{v^2} = {u^2} + 2aS.\] At the highest point, the speed is zero, so \[\begin{align} {0^2} &= \big(e\sqrt {2gh} \big)^2 + 2( - g){h_1}\\ \Rightarrow {h_1} &= {e^2}h. \end{align}\]

Also, after the rebound, the time taken to come back to ground can be calculated from
\[s = ut + \frac{1}{2}a{t^2}.\]
The displacement is zero as the particle reaches back to the ground, and the acceleration is \(-g\) taking the upward direction as positive: \[\begin{array}{l} 0 = e\sqrt {2gh} \,t + \frac{1}{2}( - g){t^2}\\ \Rightarrow t = 2e\sqrt {\frac{{2h}}{g}}. \end{array}\]

This process of bouncing will continue and the total number of collisions will be infinite. However, the time taken and the distance traveled during each bounce shrink quickly enough that the total sums are finite. These series are called convergent series.

The total distance traveled \(H\) will be

\[\begin{align} H &= {h_0} + 2{e^2}{h_0} + 2{e^4}{h_0} + 2{e^6}{h_0} + \cdots \\
&= {h_0}\left[1 + 2{e^2}\left(1 + {e^2} + {e^4} + {e^6}+\cdots\right)\right]\\ &= {h_0}\left[ {1 + 2{e^2}\left( {\frac{1}{{1 - {e^2}}}} \right)} \right] \qquad \left(\text{since } 1 + {e^2} + {e^4} + \cdots = \frac{1}{{1 - {e^2}}}\right)\\
&= {h_0}\left[ {\frac{{1 + {e^2}}}{{1 - {e^2}}}} \right]. \end{align}\]

Similarly, the total time \(T\) taken will be

\[\begin{align} T &= \sqrt {\frac{{2{h_0}}}{g}} + 2\sqrt {\frac{{2{e^2}{h_0}}}{g}} + 2\sqrt {\frac{{2{e^4}{h_0}}}{g}} +\cdots \\ &= \sqrt {\frac{{2{h_0}}}{g}} \cdot \left[1 + 2e\left(1 + e + {e^2} + {e^3} + \cdots \right)\right]\\ &= \sqrt {\frac{{2{h_0}}}{g}} \cdot \left[ {1 + 2e\left( {\frac{1}{{1 - e}}} \right)} \right]\\ &= \left( {\frac{{1 + e}}{{1 - e}}} \right)\,\sqrt {\frac{{2{h_0}}}{g}}. \end{align}\]