Worked Examples

proton collision

Let the initial proton speed be

    \[ w=8.2\times10^5\ \text{m s}^{-1}. \]

After the collision:

  • the first proton moves at speed u, 60^\circ above the horizontal;
  • the second proton moves at speed v, at angle \phi below the horizontal.

1. Conservation of momentum

Since both particles have the same mass m, the mass cancels from the momentum equations.

Horizontal direction:

    \[ mw=mu\cos60^\circ+mv\cos\phi \]

(1)   \[ w=u\cos60^\circ+v\cos\phi. \]

Vertical direction:

    \[ 0=mu\sin60^\circ-mv\sin\phi \]

(2)   \[ u\sin60^\circ=v\sin\phi. \]

In vector form, conservation of momentum gives

    \[ \vec w=\vec u+\vec v. \]

Squaring the magnitudes:

(3)   \[ w^2=u^2+v^2+2uv\cos(60^\circ+\phi). \]

The angle between the two final velocity vectors is 60^\circ+\phi.

2. Use conservation of kinetic energy

The collision is elastic, so kinetic energy is conserved:

    \[ \frac12mw^2=\frac12mu^2+\frac12mv^2. \]

Cancelling \frac12m:

(4)   \[ w^2=u^2+v^2. \]

Comparing equations (3) and (4):

    \[ u^2+v^2+2uv\cos(60^\circ+\phi)=u^2+v^2, \]

so

    \[ 2uv\cos(60^\circ+\phi)=0. \]

Neither proton is stationary after the collision, so u\neq0 and v\neq0. Therefore,

    \[ \cos(60^\circ+\phi)=0. \]

For the physical angles shown,

    \[ 60^\circ+\phi=90^\circ, \]

giving

    \[ \boxed{\phi=30^\circ}. \]

Thus, the two protons move at right angles after the collision.

3. Calculate u and v

Using vertical momentum conservation, equation (2):

    \[ u\sin60^\circ=v\sin30^\circ. \]

Substitute

    \[ \sin60^\circ=\frac{\sqrt3}{2}, \qquad \sin30^\circ=\frac12: \]

    \[ u\left(\frac{\sqrt3}{2}\right)=v\left(\frac12\right). \]

Hence

(5)   \[ v=\sqrt3\,u. \]

Now use conservation of kinetic energy:

    \[ w^2=u^2+v^2. \]

Substituting v=\sqrt3u:

    \[ w^2=u^2+(\sqrt3u)^2 \]

    \[ w^2=u^2+3u^2=4u^2. \]

Therefore,

    \[ u=\frac{w}{2}. \]

With w=8.2\times10^5\ \text{m s}^{-1},

    \[ u=\frac{8.2\times10^5}{2} =4.1\times10^5\ \text{m s}^{-1}. \]

Thus,

    \[ \boxed{u=4.1\times10^5\ \text{m s}^{-1}}. \]

Finally,

    \[ v=\sqrt{3}\left(4.1\times10^{5}\right) \approx 7.10\times10^{5}\ \mathrm{m\,s^{-1}}. \]

To two significant figures,

    \[ \boxed{v=7.1\times10^5\ \text{m s}^{-1}}. \]

Therefore, the answers are

    \[ \boxed{\phi=30^\circ,\quad u=4.1\times10^5\ \text{m s}^{-1},\quad v=7.1\times10^5\ \text{m s}^{-1}}. \]