The Bagger 293 excavator is the world's largest land vehicle, with a mass of 14200 tonnes - Leaving Cert Physics - Question 7 - 2021

The principle of conservation of momentum states that in a closed system with no external forces acting, the total momentum before an interaction is equal to the total momentum after the interaction. This can be expressed as:

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

where:

• $m_1$ and $m_2$ are the masses of the objects,
• $v_{1i}$ and $v_{2i}$ are their initial velocities,
• $v_{1f}$ and $v_{2f}$ are their final velocities.

Newton's second law of motion states that the force acting on an object is equal to the rate of change of its momentum. This can be expressed as:

F = rac{dp}{dt}

where $F$ is the force and $p$ is the momentum. When a net force is applied, it causes a change in the momentum of an object, which upholds Newton’s law. In a closed system, the net external force is zero, leading to constant total momentum, thereby aligning with the conservation of momentum.

To calculate the momentum (p) of Bagger 293:

1. Convert the mass from tonnes to kg: $14200 ext{ tonnes} = 14200 imes 1000 ext{ kg} = 14200000 ext{ kg}$
2. Use the formula for momentum: $p = mv = 14200000 ext{ kg} imes 0.17 ext{ m/s}$ $p = 2414000 ext{ kg m/s}$

When Bagger 293 picks up a stationary load of 2700 tonnes, its total mass increases. Since momentum is conserved, the increase in mass will lead to a decrease in speed to maintain the same total momentum. This is because the momentum of the system before picking up the load must equal the momentum after, resulting in a lower speed due to the increased mass.

The initial momentum of train X can be calculated using the formula:

$p = mv$

1. Convert the mass of train X from grams to kilograms: $133 ext{ g} = 0.133 ext{ kg}$
2. Calculate the momentum: $p = 0.133 ext{ kg} imes 0.05 ext{ m/s} = 0.00665 ext{ kg m/s}$

To find the speed after collision, we use conservation of momentum:

1. Initial momentum of the system: $p_{initial} = m_Xv_X + m_Yv_Y$

   • For Toy Train Y, $v_Y = 0$ (at rest). $p_{initial} = (0.133 ext{ kg} imes 0.05 ext{ m/s}) + (0.046 ext{ kg} imes 0) = 0.00665 ext{ kg m/s}$

2. After the collision, the two trains stick together: $p_{final} = (m_X + m_Y)v_f$ $0.00665 ext{ kg m/s} = (0.133 + 0.046)v_f$ v_f = rac{0.00665}{0.179} ext{ m/s} = 0.037 ext{ m/s}

After the collision, both trains move together in the direction of train X, which is to the right. This is due to the initial velocity of train X being greater than that of train Y, thus they move to the right as a combined mass.

The kinetic energy (KE) before the collision can be calculated as:

KE_{initial} = rac{1}{2}mv^2

1. For train X: KE_{X} = rac{1}{2} imes 0.133 ext{ kg} imes (0.05 ext{ m/s})^2 = 0.00016625 ext{ J}

2. For train Y: $KE_{Y} = 0 ext{ J}$

3. Total initial KE: $KE_{initial} = KE_X + KE_Y = 0.00016625 ext{ J}$

4. For both trains after collision: KE_{final} = rac{1}{2}(0.179 ext{ kg})(0.037 ext{ m/s})^2 = 0.00012365175 ext{ J}

5. Loss in kinetic energy: $ext{Loss} = KE_{initial} - KE_{final} = 0.00016625 - 0.00012365175 = 0.00004234825 ext{ J}$

The loss in kinetic energy during the collision was converted into other forms of energy such as heat and sound. This energy conversion occurs due to the inelastic nature of the collision where the two trains stick together.