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 - 2019

1. Gravitational Force:

   • Minimised by using a horizontal setup which reduces the effect of gravity on the bodies, ensuring they remain on the same level.

2. Frictional Force:

   • Minimised by utilizing a smooth air track. An air cushion reduces contact with the track, minimizing frictional effects.

Given:

• Distance travelled by A before collision = 161 mm = 0.161 m
• Distance travelled by A and B after collision = 83 mm = 0.083 m
• Time = 0.12 s

Calculating velocities:

• For body A: $u = \frac{\text{Distance}}{\text{Time}} = \frac{0.161 \text{ m}}{0.12 \text{ s}} = 1.342 \text{ m/s}$

• For bodies A and B after the collision: $v = \frac{0.083 \text{ m}}{0.12 \text{ s}} = 0.692 \text{ m/s}$

Using the formula for momentum:

• Momentum before collision (p1): $p_1 = m_A \cdot u + m_B \cdot 0 = 0.3607 \text{ kg} \cdot 1.342 \text{ m/s} = 0.484 \text{ kg m/s}$

• Momentum after collision (p2): $p_2 = (m_A + m_B) \cdot v = (0.3607 + 0.3409) \text{ kg} \cdot 0.692 \text{ m/s} = 0.485 \text{ kg m/s}$

Since p1 ≈ p2, this demonstrates the principle of conservation of momentum.

The kinetic energy (E) can be calculated using the formula:

$E = \frac{1}{2} mv^2$

1. Kinetic energy before collision for body A: $E_1 = \frac{1}{2} m_A u^2 = \frac{1}{2} \times 0.3607 \text{ kg} \times (1.342)^2 \approx 0.325 ext{ J}$

2. Kinetic energy after collision for body A and B: $E_2 = \frac{1}{2} (m_A + m_B) v^2 = \frac{1}{2} \times (0.3607 + 0.3409) \text{ kg} \times (0.692)^2 \approx 0.168 ext{ J}$

3. Loss of kinetic energy: $\Delta E = E_1 - E_2 = 0.325 - 0.168 = 0.157 ext{ J}$

The loss of kinetic energy during the collision can be accounted for by the transformation into:

• Sound energy: Energy is dissipated as sound when the bodies collide.
• Heat energy: Friction generated during the collision may also lead to increased thermal energy in the bodies.