MB1, MB2 - Oblique impacts

Normal and tangential components

A ball reaches a smooth horizontal wall with velocity $(3\mathbf i-4\mathbf j)\,\mathrm{m\,s}^{-1}$. If restitution is $1/2$, is its outgoing speed simply half its incoming speed? The wall changes the velocity perpendicular to itself; the component along it behaves differently.

The normal direction is perpendicular to the surface at contact. The tangential direction is parallel to it. For smooth spheres, the normal is the line joining their centres at contact, or line of impact. Choose perpendicular unit vectors $\mathbf n,\mathbf t$ and write

These are signed components: $u_n=\mathbf u\mathbin{\cdot}\mathbf n$ and $u_t=\mathbf u\mathbin{\cdot}\mathbf t$. This uses the A-level vector rule for projection onto a unit vector. For a horizontal wall with accessible region above it, take $\mathbf t=\mathbf i$ rightwards and $\mathbf n=\mathbf j$ away from the wall. An incoming velocity then has $u_n<0$.

[DIAGRAM: asset_name: A smooth wall changes only the normal component; asset_slug: aqa7367-oblique-components; caption: Incoming velocity (3, minus 4) becomes (3, 2) when a horizontal smooth wall has restitution one half; description: A horizontal wall has a normal arrow vertically upwards and tangent arrow to the right. A blue incoming velocity arrow runs down and right to the contact point with components right 3 and down 4. A sage outgoing arrow runs up and right from contact with components right 3 and up 2. A note states tangential component unchanged and normal component reverses and is multiplied by one half. The arrows and component triangles use equal scales.]

When an angle $\alpha$ is measured to the wall, the tangential magnitude is $u\cos\alpha$ and the normal magnitude is $u\sin\alpha$. If the angle is to the normal, those sine and cosine roles exchange. Add signs from the actual directions after finding magnitudes.

Keep the speed and numerical angle, but change the reference from the wall to the normal. That changes which component is adjacent to the angle.

Resolve relative to the contact geometry even when the original velocity is given using different horizontal and vertical axes. The impact laws belong to normal and tangential components.

Try the normal-angle convention with a new speed.

Smooth walls and two-body impacts

A smooth contact exerts no tangential impulse. The component of each body's momentum parallel to the contact surface is therefore unchanged during the impact, provided other impulses in that direction are negligible. With constant mass, each body's tangential velocity component stays unchanged individually.

For a fixed wall, restitution acts on the normal component and reverses it. With $\mathbf n$ away from the wall, $u_n<0$ gives $v_n=-eu_n\ge0$, while $v_t=u_t$. The opening velocity $3\mathbf i-4\mathbf j$ therefore becomes $3\mathbf i+2\mathbf j$ when $e=1/2$. The entire velocity vector is not multiplied by $-e$.

For two smooth bodies, use a common normal $\mathbf n$ from $A$ towards $B$ at contact. Their normal velocities obey the same two equations as a direct collision. Their tangential velocities remain their respective incoming values.

For two bodies, normal momentum is conserved for the pair; normal momentum need not stay constant for either body separately. The unchanged tangential components follow from smoothness, a stronger and more specific fact than total momentum conservation alone.