Mi13–Mi15 - Oblique impact with a smooth plane

Find the line of impulse

Can a ball hit a wall with coefficient of restitution zero and still move immediately afterwards? A direct impact would leave it at rest. A slanting impact has another component of motion to consider; we will return to the question after resolving a bounce.

The new decision is which direction restitution controls. The line of impulse is the direction of the collision impulse. For a smooth fixed plane it is perpendicular to the plane, so it is the normal direction.

The bodies' shape matters in the second definition: two particles have no preferred line of centres to determine an oblique collision impulse. A particle can, however, strike a plane obliquely, because the plane supplies a normal direction.

For this model, mobile bodies are uniform, with spherical or circular symmetry. Contact is smooth, and objects do not rotate before or after impact. The collision impulse passes along the line of centres for two such bodies, or perpendicular to a plane. It acts through each body's centre and produces no rotation. Rotating objects are outside this model.

Smooth contact means no impulse parallel to the plane. That is why the object's tangential momentum, and hence its tangential velocity, stays constant during impact. Its normal velocity changes.

The direction perpendicular to the impulse is the direction parallel to the plane. Keeping those two descriptions linked prevents a common reversal of the rules.

Resolve a smooth-plane bounce

Choose a tangential unit vector $\mathbf{t}$ along the plane and an outward normal unit vector $\mathbf{n}$. Write the incoming velocity as $\mathbf{u}=u_t\mathbf{t}+u_n\mathbf{n}$. A body approaching the plane has $u_n<0$.

Take a horizontal plane, with $\mathbf{i}$ rightwards along it and $\mathbf{j}$ vertically upwards. In this case $\mathbf{t}=\mathbf{i}$ and $\mathbf{n}=\mathbf{j}$.

[DIAGRAM: asset_name: Components of a smooth-plane bounce; asset_slug: oblique_plane_components; caption: Smooth contact preserves the rightwards component and reverses only the normal component; description: An exact vector diagram above a horizontal plane shows an incoming velocity with components 6 right and 8 down, and an outgoing velocity with components 6 right and 4 up for e equal to one half. Incoming and outgoing arrows meet the same contact point in their respective panels. The positive tangent is rightwards and the outward normal is upwards.]

Change only $e$ to $0.75$. The horizontal component still has no reason to change.

For a tilted plane, choose axes along and perpendicular to that plane first. Horizontal and vertical components are useful only if they coincide with these directions or you resolve them again.

Use different incoming components in the next bounce.

Find rebound angles and deflection

Label angles to a named line. Let the incoming velocity make an acute angle $\alpha$ with the plane, and the outgoing velocity an angle $\beta$ with the plane. Take the tangential motion to be positive in both cases.

An incoming speed $U$ has tangential component $U\cos\alpha$ and normal component $-U\sin\alpha$. The outgoing components are $U\cos\alpha$ and $eU\sin\alpha$.

The sum occurs because the incoming velocity points below the positive tangent and the outgoing velocity points above it. If the angles are measured to the normal, use their complements or resolve directly; do not insert them unchanged into this formula.