3.1 On a railway shunting line a locomotive is coupling with a stationary carriage of a mass of 2 500 kg - NSC Technical Sciences - Question 3 - 2021 - Paper 1

To calculate the momentum of the locomotive before the collision, we use the formula for momentum:

$p = mv$

Here:

• Mass, $m = 5800 ext{ kg}$
• Velocity, $v = 1,5 ext{ m.s}^{-1}$

Calculating the momentum:

$p = 5800 ext{ kg} imes 1,5 ext{ m.s}^{-1} = 8700 ext{ kg.m.s}^{-1}$

The total momentum before the collision must equal the total momentum after the collision (law of conservation of momentum).

Before collision:

• Momentum of the locomotive, $p_{locomotive} = 8700 ext{ kg.m.s}^{-1}$
• Momentum of the carriage, $p_{carriage} = 0 ext{ kg.m.s}^{-1}$ (stationary)

Total momentum before collision:

$p_{total} = 8700 + 0 = 8700 ext{ kg.m.s}^{-1}$

After collision, let $v$ be the velocity of the combined system:

Total mass after collision = $m_{locomotive} + m_{carriage} = 5800 ext{ kg} + 2500 ext{ kg} = 8300 ext{ kg}$

Using conservation of momentum:

In elastic collisions, both momentum and kinetic energy are conserved. After the collision, the objects bounce off each other with no loss of kinetic energy. In inelastic collisions, momentum is conserved but kinetic energy is not; some kinetic energy is transformed into other forms of energy such as heat or sound. In the case of perfectly inelastic collisions, the objects stick together after collision.

Seatbelts save lives during collisions by providing restraint to the occupants of a vehicle, thereby preventing them from being thrown forward or ejected from their seats. They work based on Newton's first law of motion; as the car comes to a sudden stop, the inertia of the occupant continues to move them forward. The seatbelt applies a force to the occupant, decelerating them more gradually, reducing the risk of severe injury.

Impulse is defined as the change in momentum and can be calculated using the formula:

$J = riangle p = p_{final} - p_{initial}$

• Initial momentum, $p_{initial} = 24300 ext{ kg.m.s}^{-1}$
• Final momentum, $p_{final} = 0 ext{ kg.m.s}^{-1}$ (at rest)

Calculating the impulse:

$J = 0 - 24300 = -24300 ext{ kg.m.s}^{-1}$

The impulse experienced by the car is thus 24300 kg.m.s⁻¹ in the opposite direction of motion.

We know that impulse also relates to force and time as follows:

$J = F imes riangle t$

Where:

• Impulse, $J = 24300 ext{ kg.m.s}^{-1}$
• $riangle t = 1,2 ext{ s}$

Rearranging for force:

F = rac{J}{ riangle t} = rac{24300}{1,2} = 20250 ext{ N}

The wall can withstand 80 kN, which is equivalent to 80000 N. Since 20250 N is less than 80000 N, the wall will withstand the impact.