Between the two extremes
The last two lessons solved the two extreme collisions. Bodies that stick need only momentum. Elastic bodies need momentum and the rule that separation equals approach. A real collision lies between them: the bodies separate, but more slowly than they approached. One number places a collision on that range.
The coefficient of restitution \(e\) is the velocity of separation divided by the velocity of approach. Both are measured along the line of centres: \(e = (v_2 - v_1)/(u_1 - u_2)\).
The coefficient of restitution depends mainly on the materials of the two bodies. Two billiard balls have \(e\) near \(0.95\), a cricket ball on a wooden bat about \(0.5\), and wet clay on anything nearly \(0\). Its value fixes the type of collision.
- \(e = 1\): separation equals approach, so the collision is elastic and no kinetic energy is lost.
- \(0 < e < 1\): the bodies separate more slowly than they approached. The collision is inelastic, and some kinetic energy is lost.
- \(e = 0\): there is no separation. The bodies continue together with the common velocity, a completely inelastic collision.
Newton first defined \(e\) through the impulses. The contact force delivers one impulse while the bodies are squeezed together, and a second while they push apart. The ratio of the second to the first is \(e\). It can be shown that this ratio equals separation over approach. A value above \(1\) needs extra energy, such as a firecracker between the bodies.
Two bodies collide head-on. The coefficient of restitution compares
Two equations for every collision
With \(e\) known, two equations hold for any head-on collision. The first is momentum. The second is the definition of \(e\), turned around.
\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2, \qquad v_2 - v_1 = e\,(u_1 - u_2) \]
Two equations fix the two unknowns \(v_1\) and \(v_2\). So every head-on collision can be solved, not only the two extremes. Solving the pair once and for all gives the results below.
In practice, most people write the two equations and solve them by hand. That is quicker than remembering the formulas, and it keeps the signs in view.
A \(2\,\text{kg}\) ball moving at \(4\,\text{m/s}\) strikes a \(1\,\text{kg}\) ball at rest, head-on, with \(e = 0.5\). Find both velocities after the collision and the kinetic energy lost.
Take the direction of the moving ball as positive, so \(u_1 = 4\,\text{m/s}\) and \(u_2 = 0\).
Momentum: \(2 \times 4 = 2v_1 + v_2\). Restitution: \(v_2 - v_1 = 0.5 \times 4 = 2\).
\(3v_1 = 6\), so \(v_1 = 2\,\text{m/s}\) and \(v_2 = 4\,\text{m/s}\). Kinetic energy: \(16\,\text{J}\) before, \(4 + 8 = 12\,\text{J}\) after.
The \(2\,\text{kg}\) ball continues at \(2\,\text{m/s}\), the \(1\,\text{kg}\) ball at \(4\,\text{m/s}\), and \(4\,\text{J}\) of kinetic energy is lost.
The loss of kinetic energy also has a compact form. It is the loss for sticking, from Lesson 2, multiplied by \((1 - e^2)\).
\[ \Delta K = \dfrac{m_1m_2}{2(m_1 + m_2)}\,(1 - e^2)\,(u_1 - u_2)^2 \]
With \(e = 1\) nothing is lost, and with \(e = 0\) the loss is the sticking loss. So a measured loss can give \(e\), as the next example shows.
A ball moving at speed \(v\) strikes an identical ball at rest, head-on. After the collision the kinetic energy of the two balls is three-quarters of the original. Find the coefficient of restitution.
(a) Momentum with equal masses: \(v_1 + v_2 = v\). Restitution: \(v_2 - v_1 = ev\). So \(v_1 = \tfrac12(1 - e)v\) and \(v_2 = \tfrac12(1 + e)v\).
(b) The kinetic energy after, as a fraction of \(\tfrac12 mv^2\), is \(\tfrac14\left[(1 - e)^2 + (1 + e)^2\right] = \tfrac12(1 + e^2)\).
(c) Set \(\tfrac12(1 + e^2) = \tfrac34\). Then \(e^2 = \tfrac12\), and \(e = 1/\sqrt2 \approx 0.71\).
A ball and a moving wall
A wall is a body of enormous mass. Its velocity does not change in the collision, so only the ball's velocity is unknown, and the restitution equation alone gives it. Suppose a ball moves at \(2\,\text{m/s}\) towards a massive wall that moves towards it at \(1\,\text{m/s}\). The collision is elastic.
The gap closes at \(2 + 1 = 3\,\text{m/s}\). With \(e = 1\), it must open at \(3\,\text{m/s}\) too. The wall still moves at \(1\,\text{m/s}\) towards where the ball came from, so the ball must leave at \(1 + 3 = 4\,\text{m/s}\). A cricket bat meeting the ball is a moving wall of this kind. So a well-timed shot sends the ball away faster than it arrived.
A \(3\,\text{kg}\) trolley moving at \(4\,\text{m/s}\) strikes a \(1\,\text{kg}\) trolley at rest, head-on, with \(e = 0.5\). Find the velocity of the \(1\,\text{kg}\) trolley after the collision, in m/s.
A collision has \(e = 0\). Which statement must be true?
The bouncing ball
The floor is a wall that never moves. A ball dropped from a height \(h_0\) reaches it at \(v = \sqrt{2gh_0}\), from the chapter on Potential Energy. It leaves at \(ev\), because the floor's velocity stays zero. Rising with \(ev\), the ball reaches the height \(h_1 = (ev)^2/2g = e^2h_0\).
Each bounce repeats the story with the new height, so the heights are \(h_0\), \(e^2h_0\), \(e^4h_0\), and so on. In the sim, \(e = 0.8\) gives heights of \(1.25\), \(0.80\) and \(0.51\,\text{m}\). A small \(e\) makes the ball come to rest after a few low bounces. The two heights also give \(e\) directly: \(e = \sqrt{h_1/h_0}\).
A ball is dropped from \(5\,\text{m}\) onto a floor with \(e = 0.5\). Take \(g = 10\,\text{m/s}^2\). Find the height of the first bounce, the total distance the ball travels before it comes to rest, and the total time.
(a) \(h_1 = e^2h_0 = 0.25 \times 5 = 1.25\,\text{m}\).
(b) The ball falls \(5\,\text{m}\), then goes up and down \(1.25\,\text{m}\), then \(0.3125\,\text{m}\), and so on. The distance is \(5 + 2(1.25 + 0.3125 + \dots) = 5 + 2 \times \dfrac{1.25}{1 - 0.25} = 5 + 3.33 = 8.33\,\text{m}\).
(c) The first fall takes \(\sqrt{2h_0/g} = 1\,\text{s}\). Each flight after that takes \(e\) times the one before, so the total time is \(1 + 2(0.5 + 0.25 + \dots) = 1 + 2 = 3\,\text{s}\).
A ball dropped from a height of \(1\,\text{m}\) on a floor rebounds to \(0.64\,\text{m}\). Find the coefficient of restitution between the ball and the floor.
The ballistic pendulum
How fast is a bullet? A ballistic pendulum answers with a ruler. A wooden block hangs from strings. The bullet enters the block and stays in it, and the block swings up through a height \(h\). The event has two parts, and each part keeps a different quantity.
The collision itself is completely inelastic and over in a moment, so momentum is conserved: \(mv = (m + M)V\). The swing afterwards is slow, and the string does no work, so energy is conserved: \(\tfrac12 (m + M)V^2 = (m + M)gh\). The swing gives \(V = \sqrt{2gh}\), and the collision then gives \(v\).
A \(20\,\text{g}\) bullet is fired into a \(1.98\,\text{kg}\) wooden block hanging from light strings. The bullet stays in the block, which swings up through \(20\,\text{cm}\). Take \(g = 10\,\text{m/s}^2\). Find the speed of the bullet, and the fraction of its kinetic energy that is lost.
Two stages: the collision, then the swing. The block and bullet together have mass \(2.0\,\text{kg}\).
Swing: \(V = \sqrt{2gh}\). Collision: \(0.02\,v = 2.0\,V\).
\(V = 2\,\text{m/s}\), so \(v = 2.0 \times 2/0.02 = 200\,\text{m/s}\). Kinetic energy: \(\tfrac12 \times 0.02 \times 200^2 = 400\,\text{J}\) before, \(\tfrac12 \times 2.0 \times 2^2 = 4\,\text{J}\) after.
The bullet was moving at \(200\,\text{m/s}\), and \(99\,\%\) of its kinetic energy went into heat and splintered wood.
Energy conservation across the impact gives \(\tfrac12 mv^2 = (m + M)gh\), and a speed about ten times too small here. The impact is completely inelastic. Use momentum for the impact and energy only for the swing.
A \(10\,\text{g}\) bullet is fired into a \(990\,\text{g}\) block hanging from a string, and stays in it. The block rises through \(5\,\text{cm}\). Take \(g = 10\,\text{m/s}^2\). Find the speed of the bullet, in m/s.
In the Principia of 1687, Newton described collisions between balls hung as pendulums. The height of a swing told him each speed just before and after the impact. Balls of tightly wound wool separated at about five-ninths of the speed with which they met. Glass balls separated at about fifteen-sixteenths. The ratio was the same at every speed for the same pair, and it is the \(e\) of this lesson.
- Taking \(e\) as one body's speed after divided by its speed before.\(e\) compares relative velocities: separation over approach, along the line of centres.
- Using \(e = h_1/h_0\) for a bouncing ball.Height goes as the square of speed, so \(e = \sqrt{h_1/h_0}\).
- Conserving kinetic energy across the impact of a ballistic pendulum.Momentum across the impact, energy across the swing.