Lesson 7 · 30 min

Oblique Impact

A ball skids off the floor at a flatter angle than it arrived. A cue ball glances off another and the two roll away at nearly a right angle. When the velocities are not along the line of impact, split them into components along it and across it. Across it, on smooth surfaces, nothing changes. Along it, the impact is exactly the direct impact of Lesson 6.

Learning objectives

The \(n\) and \(t\) axes

At the instant of contact, put the \(n\) axis along the line of impact, normal to the contact surfaces, and the \(t\) axis along their common tangent. Resolve each velocity into its \(n\) and \(t\) components. With smooth surfaces, the impulsive contact force is entirely along \(n\), and that settles everything:

Oblique central impact on smooth surfaces

  • Along \(t\), each body separately: no impulse, so each body keeps its own tangential velocity: \((v_{A})_{t2} = (v_{A})_{t1}\) and \((v_{B})_{t2} = (v_{B})_{t1}\).
  • Along \(n\), the two bodies together: momentum is conserved, \(m_A(v_A)_{n1} + m_B(v_B)_{n1} = m_A(v_A)_{n2} + m_B(v_B)_{n2}\).
  • Along \(n\), restitution: \((v_B)_{n2} - (v_A)_{n2} = e\left[(v_A)_{n1} - (v_B)_{n1}\right]\).

Four equations for the four components after the impact. The two along \(n\) are those of Lesson 6.

The coefficient of restitution applies only along the line of impact. Applying it to the whole velocity is the most common mistake in this topic.

A ball on a fixed surface

When a ball strikes a fixed smooth surface, the surface does not move, so the restitution equation gives \((v_n)_2 = e\,(v_n)_1\) for the size of the normal component, now pointing away from the surface, and the tangential component is unchanged. Measure the angles from the surface:

\[ \tan\theta_1 = \frac{(v_n)_1}{(v_t)_1} \qquad \tan\theta_2 = \frac{(v_n)_2}{(v_t)_2} = \frac{e\,(v_n)_1}{(v_t)_1} \qquad\Rightarrow\qquad \tan\theta_2 = e\tan\theta_1 \]

Unless \(e = 1\), the ball leaves at a flatter angle than it arrived, and more slowly.

Figure 7.1 A ball strikes a smooth floor. After the bounce, the figure shows the velocity just before the impact (dashed, ending at the contact point) and just after (solid), each split into its tangential and normal components: the tangential component \(v_t\) is the same, the normal component \(v_n\) has turned around and shrunk by the factor \(e\). Change the speed, the angle and \(e\).

Example 7.1 — A bounce off a smooth floor

A ball strikes a smooth floor at \(12\ \text{m/s}\), at \(50^\circ\) to the floor. The coefficient of restitution is \(0.6\). Find the rebound velocity.

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Tangential, along the floor: no impulse.

\[ (v_t)_2 = (v_t)_1 = 12\cos 50^\circ = 7.71\ \text{m/s} \]

Normal: the floor is fixed, so

\[ (v_n)_2 = e\,(v_n)_1 = 0.6\,(12\sin 50^\circ) = 5.52\ \text{m/s},\ \text{upward} \] \[ v_2 = \sqrt{7.71^2 + 5.52^2} = 9.48\ \text{m/s} \qquad \theta_2 = \tan^{-1}\frac{5.52}{7.71} = 35.6^\circ \]

Check: \(\tan 35.6^\circ = 0.716 = 0.6\tan 50^\circ\). The ball keeps \(62\%\) of its kinetic energy.

Two bodies at an angle

For two moving bodies, the \(n\) axis runs through both mass centers at the instant of contact: for two disks or spheres, along the line joining their centers. Work out each body's components along that line and across it, then apply the four equations.

Example 7.2 — Two pucks

Puck \(A\) (\(0.2\ \text{kg}\)) slides at \(4\ \text{m/s}\) and strikes puck \(B\) (\(0.2\ \text{kg}\)), which is at rest on smooth ice. At contact, the line joining their centers is at \(30^\circ\) to \(A\)'s velocity. The coefficient of restitution is \(0.8\). Find the velocities of both pucks after the impact.

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Components of \(A\)'s velocity before: \((v_A)_{n1} = 4\cos 30^\circ = 3.464\ \text{m/s}\) and \((v_A)_{t1} = 4\sin 30^\circ = 2\ \text{m/s}\). \(B\) is at rest.

Along \(t\): \((v_A)_{t2} = 2\ \text{m/s}\) and \((v_B)_{t2} = 0\).

Along \(n\):

\[ 0.2(3.464) + 0 = 0.2(v_A)_{n2} + 0.2(v_B)_{n2} \qquad (v_B)_{n2} - (v_A)_{n2} = 0.8(3.464 - 0) = 2.771 \] \[ (v_A)_{n2} = 0.346\ \text{m/s},\qquad (v_B)_{n2} = 3.118\ \text{m/s} \]

Result. \(B\) moves off along the line of impact at \(3.12\ \text{m/s}\), \(30^\circ\) to one side of \(A\)'s original path. \(A\) moves at \(\sqrt{0.346^2 + 2^2} = 2.03\ \text{m/s}\), at \(\tan^{-1}(2/0.346) = 80.2^\circ\) from the line of impact, which is \(50.2^\circ\) to the other side of its original path. The kinetic energy drops from \(1.6\ \text{J}\) to \(1.384\ \text{J}\).

Figure 7.2 Two pucks on smooth ice, seen from above. Puck \(A\) slides toward puck \(B\), at rest, offset sideways by \(b\). At the moment of contact the figure draws the line of impact \(n\), through both centers, and the tangent \(t\); the impulse acts along \(n\) only, so \(B\) always leaves along \(n\). With equal masses and \(e = 1\) the two paths afterward are at right angles, whatever the offset.

Example 7.3 — The billiards right angle

Show that when a ball strikes an identical ball at rest, off-center, with \(e = 1\), the two move off at right angles.

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Along \(n\), equal masses with \(e = 1\) exchange velocities, as in Newton's cradle: \((v_A)_{n2} = 0\) and \((v_B)_{n2} = (v_A)_{n1}\). Along \(t\), \(A\) keeps \((v_A)_{t1}\) and \(B\) gets nothing. So afterward \(A\) moves purely along \(t\) and \(B\) purely along \(n\): at right angles. (For a real billiard ball, spin and \(e\) a little under \(1\) bend the rule slightly.)

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Key takeaways