A student sets up an experiment to investigate the interaction between two trolleys on a smooth, horizontal track - Scottish Highers Physics - Question 3 - 2022

To calculate the time, we can use the impulse-momentum principle:

$F \Delta t = \Delta p$

Given that the force is 6.25 N and we've calculated the change in momentum to be 0.35 kg m s⁻¹, we can rearrange this to find time:

$\Delta t = \frac{\Delta p}{F}$

Substituting in the values:

$\Delta t = \frac{0.35}{6.25}$
$\Delta t = 0.056 \, \text{s}$

So, the time during which the plunger exerts a force on trolley Y is approximately 0.056 seconds.

Using the principle of conservation of momentum, we set the total momentum before equal to the total momentum after:

Before:
$m_X v_{X_i} + m_Y v_{Y_i} = (0.50 \text{ kg} \times 0.40 \, \text{m s}^{-1}) + (0.25 \text{ kg} \times 0.40 \, \text{m s}^{-1})$
$= 0.50 \times 0.40 + 0.25 \times 0.40$
$= 0.40 + 0.10 = 0.50 \, \text{kg m s}^{-1}$

After:
Using the final velocity of trolley Y as 1.80 m s⁻¹: $m_X v_{X_f} + m_Y v_{Y_f} = 0.50 \, \text{kg m s}^{-1}$
Substituting the known values: $0.50 v_{X_f} + 0.25 \times 1.80 = 0.50$
$0.50 v_{X_f} + 0.45 = 0.50$
By rearranging, we get: $0.50 v_{X_f} = 0.50 - 0.45$
$= 0.05$
Now, solving for $v_{X_f}$ gives: $v_{X_f} = \frac{0.05}{0.50} = -0.30 \, \text{m s}^{-1}$

Thus, the velocity of trolley X immediately after the trolleys separate is -0.30 m s⁻¹ (indicating the direction to the left).

The student would compare the total kinetic energy before and after the interaction.

First, they would calculate the total initial kinetic energy using: $KE_{initial} = \frac{1}{2} m_X v_{X_i}^2 + \frac{1}{2} m_Y v_{Y_i}^2$

Then, after the interaction, they would calculate the total kinetic energy: $KE_{final} = \frac{1}{2} m_X v_{X_f}^2 + \frac{1}{2} m_Y v_{Y_f}^2$

If (KE_{initial} = KE_{final}), then the interaction is elastic. If not, it is inelastic.

The effect is known as the photovoltaic effect.

When photons of light strike the photodiode, they provide energy to the electrons within it. This energy allows some electrons to gain enough energy to move from the valence band to the conduction band, creating free electrons. As these free electrons move towards the n-type region, they create a potential difference across the junction, allowing for current flow. This process is fundamental to the operation of photodiodes.