Lesson 6 · 40 min
Direct Central Impact
Two bodies collide head-on. Momentum is conserved through the collision, which gives one equation, but the collision leaves two velocities to find. The second equation comes from the materials: how much of the squeeze they give back. That is the coefficient of restitution, and with it every head-on collision, from a perfectly elastic bounce to two railcars locking together, is a two-line calculation.
Learning objectives
- Identify the line of impact, and tell central from oblique and direct from oblique impact.
- Describe the deformation and restitution phases, and define the coefficient of restitution \(e\) both as a ratio of impulses and as a ratio of relative velocities.
- Find the velocities after a direct central impact from conservation of momentum and \(e\).
- Calculate the kinetic energy lost, recognize perfectly elastic and perfectly plastic impact, and measure \(e\) with a drop test.
The line of impact
When two bodies collide, the line of impact is the common normal to their surfaces at the point of contact. The impact is central when the line of impact passes through both mass centers, as for two spheres or two blocks striking face to face. In a direct central impact, both velocities lie along the line of impact; in an oblique impact, one or both do not. This lesson treats direct central impact; Lesson 7 adds the oblique case.
Take the line of impact as the \(x\) axis, with one positive direction for all four velocities: \((v_A)_1\) and \((v_B)_1\) just before the impact, \((v_A)_2\) and \((v_B)_2\) just after. For an impact to happen, \(A\) must be catching up with \(B\): \((v_A)_1 > (v_B)_1\).
Deformation and restitution
No real body is rigid. In the brief contact, the two bodies squeeze each other in two phases:
- Deformation. The contact force \(P\) grows as the bodies flatten each other, slowing \(A\) and speeding up \(B\), until the instant of maximum deformation, when both move with the same velocity \(v\).
- Restitution. The bodies spring back, partly or completely, and the contact force \(R\) pushes them apart until they separate.
Apply impulse and momentum to body \(A\) alone, over each phase:
\[ m_A(v_A)_1 - \int P\,dt = m_A v \qquad\qquad m_A v - \int R\,dt = m_A(v_A)_2 \]The coefficient of restitution is the ratio of the restitution impulse to the deformation impulse. From the two equations,
\[ e = \frac{\int R\,dt}{\int P\,dt} = \frac{v - (v_A)_2}{(v_A)_1 - v} \]The same steps for \(B\) give \(e = \left[(v_B)_2 - v\right]/\left[v - (v_B)_1\right]\). Eliminate the common velocity \(v\) between the two expressions:
Coefficient of restitution
\[ e = \frac{(v_B)_2 - (v_A)_2}{(v_A)_1 - (v_B)_1} \]The relative velocity of separation divided by the relative velocity of approach, with \(0 \le e \le 1\).
\(e\) depends on the two materials, and also on the impact speed and the shapes of the bodies, so a tabulated value is an approximation. Steel on steel gives about \(0.5\) to \(0.8\), a golf ball on a hard floor about \(0.8\), clay or putty nearly \(0\).
Solving a direct central impact
During the impact, the contact forces are internal to the pair and impulsive; weights, springs and ordinary friction are nonimpulsive. So the momentum of the two bodies is conserved, and \(e\) gives the second equation:
Two equations, two unknowns
\[ m_A(v_A)_1 + m_B(v_B)_1 = m_A(v_A)_2 + m_B(v_B)_2 \] \[ (v_B)_2 - (v_A)_2 = e\left[(v_A)_1 - (v_B)_1\right] \]One positive direction for all four velocities. A velocity that comes out negative points the other way.
Example 6.1 — A partly elastic impact
Block \(A\) (\(2\ \text{kg}\)) slides at \(6\ \text{m/s}\) on a smooth surface and strikes block \(B\) (\(3\ \text{kg}\)), which is at rest. The coefficient of restitution is \(0.5\). Find the velocities just after the impact and the kinetic energy lost.
Show solution
Positive to the right.
\[ 2(6) + 3(0) = 2(v_A)_2 + 3(v_B)_2 \qquad (v_B)_2 - (v_A)_2 = 0.5(6 - 0) = 3 \]Substitute \((v_B)_2 = (v_A)_2 + 3\) into the first: \(12 = 5(v_A)_2 + 9\), so
\[ (v_A)_2 = 0.6\ \text{m/s},\qquad (v_B)_2 = 3.6\ \text{m/s} \]Both blocks move to the right. \(T_1 = \tfrac12(2)(6)^2 = 36\ \text{J}\) and \(T_2 = \tfrac12(2)(0.6)^2 + \tfrac12(3)(3.6)^2 = 19.8\ \text{J}\), so \(16.2\ \text{J}\), or \(45\%\), is lost in the impact.
Example 6.2 — A head-on collision
Ball \(A\) (\(1\ \text{kg}\)) moving right at \(5\ \text{m/s}\) collides head-on with ball \(B\) (\(2\ \text{kg}\)) moving left at \(2\ \text{m/s}\). With \(e = 0.6\), find the velocities after the impact and the kinetic energy lost.
Show solution
Positive to the right, so \((v_B)_1 = -2\ \text{m/s}\). The approach speed is \(5 - (-2) = 7\ \text{m/s}\).
\[ 1(5) + 2(-2) = (v_A)_2 + 2(v_B)_2 \qquad (v_B)_2 - (v_A)_2 = 0.6\left[5 - (-2)\right] = 4.2 \] \[ 1 = 3(v_A)_2 + 8.4 \quad\Rightarrow\quad (v_A)_2 = -2.47\ \text{m/s},\qquad (v_B)_2 = 1.73\ \text{m/s} \]Both balls bounce back: \(A\) moves left at \(2.47\ \text{m/s}\) and \(B\) right at \(1.73\ \text{m/s}\). \(T_1 = 12.5 + 4 = 16.5\ \text{J}\), \(T_2 = 3.04 + 3.00 = 6.05\ \text{J}\), so \(10.5\ \text{J}\) is lost.
Elastic, plastic, and in between
| Impact | \(e\) | Afterward | Kinetic energy |
|---|---|---|---|
| Perfectly elastic | \(1\) | separate as fast as they approached | conserved |
| Partly elastic | \(0 < e < 1\) | separate more slowly | partly lost |
| Perfectly plastic | \(0\) | move on together, at \(v\) | the largest possible loss |
The kinetic energy lost in a direct central impact depends only on the masses, \(e\), and the approach speed:
\[ \Delta T = \frac{m_A m_B}{2(m_A + m_B)}\left(1 - e^2\right)\left[(v_A)_1 - (v_B)_1\right]^2 \]It is the energy of the relative motion, scaled by \(1 - e^2\). Momentum is conserved in every case; only the energy depends on \(e\). And a plastic impact does not stop the bodies: they move on together, carrying all the momentum they had.
Example 6.3 — Newton's cradle
Two identical balls collide with \(e = 1\): \(A\) strikes \(B\), which is at rest. What happens?
Show solution
With \(m_A = m_B = m\) and \((v_B)_1 = 0\): \(m(v_A)_1 = m(v_A)_2 + m(v_B)_2\) and \((v_B)_2 - (v_A)_2 = (v_A)_1\). Adding them gives \((v_B)_2 = (v_A)_1\), and then \((v_A)_2 = 0\).
The balls exchange velocities: \(A\) stops dead and \(B\) leaves with \(A\)'s speed. In a row of steel balls, the impulse passes down the line and the last ball swings out. Since steel's \(e\) is a little less than \(1\), the swing dies away after a while.
Measuring \(e\): the drop test
Drop a ball from rest from height \(h_1\) onto a floor and measure how high it bounces, \(h_2\). The floor does not move, so with \((v_B)_1 = (v_B)_2 = 0\) the definition becomes \(e = -(v_A)_2/(v_A)_1\): the rebound speed over the landing speed. Each speed comes from its height, \(v = \sqrt{2gh}\), so
\[ e = \sqrt{\frac{h_2}{h_1}} \qquad\text{and each bounce reaches } e^2 \text{ times the height of the one before.} \]Example 6.4 — A drop test
A ball dropped from \(1.5\ \text{m}\) rebounds to \(0.96\ \text{m}\). Find \(e\), and the height of the second bounce.
Show solution
The ball loses \(36\%\) of its energy at each bounce, since \(1 - e^2 = 0.36\).
Check your understanding
Key takeaways
- An impact has a deformation phase, ending at a common velocity, and a restitution phase. \(e = \int R\,dt / \int P\,dt\).
- \(e = \left[(v_B)_2 - (v_A)_2\right]/\left[(v_A)_1 - (v_B)_1\right]\): separation over approach, \(0 \le e \le 1\).
- Conservation of momentum plus the restitution equation give the two velocities after a direct central impact.
- \(e = 1\): no energy lost; \(e = 0\): the bodies move on together and the loss is largest. Momentum is conserved either way.
- A drop test gives \(e = \sqrt{h_2/h_1}\).
- Next, Lesson 7 handles impacts at an angle.