Category Archives: Worked Examples

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

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

Worked Examples

Lens Equation

A converging lens is used to magnify a virtual image to four times it original size. Given then focal length of the converging lens is 20.0 cm, what is the distance of the object from the lens?

m=v/u

4=v/u   =>  v=4u

But image is virtual, so v must be negative (ie. v = -4u)

1/f = 1/u +1/v

1/f = 1/u + 1/(-4u)

1/20 = (4-1)/4u

u = 15 cm

In this question it is important that you recognise the image is a virtual image and thus v has to be a negative number.

 

2022, Worked Examples

Charge moving through magnetic and electric fields

The magnitude of the force created by the electric field is given by

FE = Eq

FE = EQ (using the values in this question)

The magnitude of the force created by the magnetic field is given by

FM = Bqv

FM = BQv (using the values in this question)

 

Q could be +ve or -ve.

If Q is +ve, then:

    • the electric field will exert a downward force on Q
    • the magnetic field will exert an upwards force on Q (from Fleming’s Left Hand Rule)

This could still create the motion indicated in the diagram, however FM would need to be smaller than FE.

FM < FE

BQv < EQ

v < E/Q

If Q is –ve, then:

    • the electric field will exert an upwards force on Q
    • the magnetic field will exert a downwards force on Q (from Fleming’s Left Hand Rule)

This could still create the motion indicated in the diagram, however FM would need to be larger than FE.

FM > FE

BQv > EQ

v > E/Q

Which is answer D