Momentum, impulse and collisions requiring resolvingAQA A-Level Further Maths: Revision notes

In many collisions the velocities are not along the line of the force. For a smooth surface or smooth spheres the only impulse acts along the normal to the surface, or along the line of centres of the spheres at the instant of impact. So we resolve each velocity into a component along that line and a component perpendicular to it.

• Perpendicular to the impulse: the component of velocity is unchanged for each body.
• Along the impulse: use conservation of momentum (two bodies) and Newton's experimental law, $\text{separation}=e\times\text{approach}$. Finally recombine the components with Pythagoras and trigonometry to get the speed and direction after the collision.

A particle hits a fixed smooth wall with speed $u$ at angle $\theta$ to the wall. Resolve parallel and perpendicular to the wall:

• Parallel: $u\cos\theta$ before and after (unchanged, as the wall is smooth).
• Perpendicular: $u\sin\theta$ before; after the impact it is $e\,u\sin\theta$ in the opposite direction. Speed after $=u\sqrt{\cos^2\theta+e^2\sin^2\theta}$. The angle after, $\phi$ with the wall, satisfies $\tan\phi=e\tan\theta$. Impulse on the particle (perpendicular to the wall) $=m(1+e)u\sin\theta$. Example: $m=0.2$, $u=10$, $\theta=60^\circ$, $e=0.5$: parallel $5$; perpendicular $5\sqrt3\to2.5\sqrt3$; speed $=\sqrt{43.75}=6.61$ m s⁻¹; impulse $=0.2(1.5)(5\sqrt3)=1.5\sqrt3$ N s.

Read the angle carefully. If the angle $\theta$ is to the wall, the parallel component is $u\cos\theta$ and the perpendicular is $u\sin\theta$. If the angle is to the normal, the roles are swapped. When $\tan\theta$ is given as a ratio (e.g. $\frac43$), build a right-angled triangle to read off $\sin$ and $\cos$ exactly rather than using a calculator. Example: $\tan\alpha=\frac43$ gives $\sin\alpha=\frac45$, $\cos\alpha=\frac35$. For $u=20$: parallel $12$, perpendicular $16$. The change in direction is the sum of the angles before and after, measured on opposite sides of the normal.

When smooth spheres collide, the impulse acts along the line of centres. For each sphere:

1. Resolve the velocity along the line of centres and perpendicular to it.
2. The perpendicular components do not change for either sphere.
3. Along the line of centres, apply conservation of momentum and Newton's experimental law to the components.
4. Combine the components for each sphere to find its speed and direction. Example: $A$ at $10$ m s⁻¹, $\cos\alpha=\frac45$, hits $B$ at rest, $e=\frac12$, equal masses. Along the line: $8=v_A+v_B$ and $v_B-v_A=4$, so $v_B=6$, $v_A=2$. $A$'s perpendicular component stays $6$. $A$'s speed is $\sqrt{4+36}=\sqrt{40}$ m s⁻¹, and $B$ moves along the line of centres at $6$ m s⁻¹.

The impulse on a body from a smooth surface or sphere acts along the normal or line of centres. Its magnitude is the change in momentum in that direction only: $I=m(v_n-u_n)$, using signed components. For an impulse not along the original velocity (for example from a bat), use $\mathbf{I}=m\mathbf{v}-m\mathbf{u}$ and resolve into components along two perpendicular directions. Kinetic energy lost $=$ KE before $-$ KE after, using the full speeds. For the sphere collision above: before $50m$, after $\frac12m(40)+\frac12m(36)=38m$, so $12m$ is lost. The loss equals the loss in kinetic energy of the components along the line of centres: $32m-20m=12m$, as perpendicular components do not change. A good check: kinetic energy after should never exceed the kinetic energy before.