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, the total momentum before an interaction is equal to the total momentum after the interaction. Mathematically, it can be represented as:

$m_1 v_1 + m_2 v_2 = m_1 v_1' + m_2 v_2'$

where $m_1$ and $m_2$ are the masses of the objects, and $v_1$, $v_2$, $v_1'$, and $v_2'$ are their respective velocities before and after the interaction.

Newton's second law states that the force acting on an object is equal to the rate of change of its momentum. This means that if no external forces are acting on a closed system, then the momentum of the system remains constant. Thus, the conservation of momentum aligns with Newton's second law because both principles maintain that momentum in a closed system cannot change without an external force.

To calculate the momentum of Bagger 293, we use the formula $p = mv$:

Given that the mass $m = 14200 ext{ tonnes} = 14200 imes 1000 ext{ kg} = 14200000 ext{ kg}$ and the maximum speed $v = 0.17 ext{ m s}^{-1}$:

$p = 14200000 ext{ kg} imes 0.17 ext{ m s}^{-1} = 2414000 ext{ kg m s}^{-1}$

When Bagger 293 picks up a stationary load, its total mass increases while its momentum must remain conserved. Since momentum is given by the product of mass and velocity, the increase in mass from picking up the load would lead to a decrease in speed to conserve momentum. Thus, the speed would decrease.

The momentum of train X can be calculated using the formula $p = mv$:

Since the mass of train X is 133 g (which is $0.133 ext{ kg}$) and the velocity is $0.05 ext{ m s}^{-1}$:

$p_X = 0.133 ext{ kg} imes 0.05 ext{ m s}^{-1} = 0.00665 ext{ kg m s}^{-1}$

After the collision, the two trains stick together. Using conservation of momentum:

$p_{initial} = p_{final}$

For the two trains:

$p_{X} = (m_X + m_Y) v_{final}$

Mass of train Y is 46 g (which is $0.046 ext{ kg}$). Thus,

$0.00665 ext{ kg m s}^{-1} = (0.133 ext{ kg} + 0.046 ext{ kg}) v_{final}$

This simplifies to:

$0.00665 = 0.179 v_{final}$

Solving for $v_{final}$:

After the collision, the two trains will move to the right since toy train X was moving to the right before the collision, and it has greater momentum than toy train Y, which was at rest.

The initial kinetic energy of train X is given by:

KE_{initial} = rac{1}{2} m v^2 = rac{1}{2}(0.133)(0.05)^2 = 0.00016625 ext{ J}

The initial kinetic energy of train Y is zero since it is at rest.

The final kinetic energy of the combined mass after the collision:

The kinetic energy that was lost in the collision was converted into heat or other forms of energy, such as sound and deformation of the trains during the impact.