In an experiment to verify the principle of conservation of momentum, body A was set in motion with a velocity u - Leaving Cert Physics - Question 1 - 2018

1. Gravitational force - This was minimised by ensuring that the apparatus is level and positioned on a flat surface.
2. Frictional force - This was minimised by using a low-friction track and ensuring that all surfaces in contact were lubricated or designed to reduce friction.

The velocity of body A before the collision is calculated using the distance travelled:

$u = \frac{distance\; travelled}{time} = \frac{161 \; \text{mm}}{0.12 \; \text{s}} = 1.342 \; \text{m/s}$

After the collision, the common velocity is:

$v = \frac{distance \; travelled}{time} = \frac{83 \; \text{mm}}{0.12 \; \text{s}} = 0.692 \; \text{m/s}$

Initial momentum (before collision) is:

$p = mv$

For body A: $m_1 = 0.3607 \; \text{kg}, \; u = 1.342 \; \text{m/s} \rightarrow p_1 = m_1 u = 0.484 \; \text{kg m/s}$

For body B before collision, $p_2 = 0 \; \text{kg m/s}$ since it’s at rest.

Total momentum before collision: $p_{initial} = p_1 + p_2 = 0.484 \; \text{kg m/s}$

After the collision, for body A and B:

The initial kinetic energy (KE) of body A before collision is:

$E_{initial} = \frac{1}{2} m_1 u^2 = \frac{1}{2} \times 0.3607 \; \text{kg} \times (1.342)^2 = 0.325\; \text{J}$

The final kinetic energy after collision is:

$E_{final} = \frac{1}{2} (m_1 + m_2) v^2 = \frac{1}{2} \times (0.3607 + 0.3409) \times (0.692)^2 = 0.168\; \text{J}$

The loss of kinetic energy: $\Delta E = E_{initial} - E_{final} = 0.325 - 0.168 = 0.157 \; \text{J}$

The loss of kinetic energy during the collision could be accounted for as sound energy or thermal energy due to frictional forces and deformation of the bodies during the impact.