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, use the formula for momentum:

$p = mv$

Given the mass of the locomotive $m = 5800 ext{ kg}$ and its velocity $v = 1,5 ext{ m·s}^{-1}$:

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

After the collision, we can use the principle of conservation of momentum. The total momentum before the collision must equal the total momentum after the collision. Let $v_f$ be the final velocity of the combined system:

Initial momentum of the system is:

$p_i = p_{ ext{locomotive}} + p_{ ext{carriage}} = 8700 ext{ kg·m·s}^{-1} + 0 = 8700 ext{ kg·m·s}^{-1}$.

Now, after the collision, the mass of the combined system is:

$m_f = m_{ ext{locomotive}} + m_{ ext{carriage}} = 5800 ext{ kg} + 2500 ext{ kg} = 8300 ext{ kg}$

Using conservation of momentum:

$8700 ext{ kg·m·s}^{-1} = 8300 ext{ kg} imes v_f$

Solving for $v_f$:

v_f = rac{8700 ext{ kg·m·s}^{-1}}{8300 ext{ kg}} ext{ m·s}^{-1} \\ = 1,05 ext{ m·s}^{-1}

An elastic collision is one in which both momentum and kinetic energy are conserved. After the collision, the objects bounce off each other without any loss of total kinetic energy.

An inelastic collision, on the other hand, is one in which momentum is conserved, but kinetic energy is not. In inelastic collisions, the objects may stick together, and some kinetic energy is transformed into other forms of energy, such as heat or sound.

Seatbelts save lives by applying the principle of inertia as described by Newton's first law of motion. During a collision, a vehicle comes to an abrupt stop, and without a seatbelt, the occupants continue to move forward due to inertia. A seatbelt restrains the occupants, preventing them from colliding violently with the interior of the vehicle or being ejected from it. This controlled deceleration reduces the risk of serious injury or death.

Impulse is defined as the change in momentum. Here, the initial momentum ($p_i$) of the car is 24 300 kg·m·s⁻¹, and the final momentum ($p_f$) after coming to rest is 0 kg·m·s⁻¹. Hence, the impulse ($J$) can be calculated as:

$J = p_f - p_i = 0 - 24300 ext{ kg·m·s}^{-1} = -24300 ext{ kg·m·s}^{-1}$

The negative sign indicates that the impulse acts in the opposite direction to the car's initial motion.

To determine if the wall can withstand the impact, we need to find the average force experienced by the wall.

Using the impulse formula, we know that:

$J = F_{avg} imes riangle t$

From the previous part, we calculated the impulse:

$J = -24300 ext{ kg·m·s}^{-1}$

Letting $riangle t = 1.2 ext{ s}$, we can rearrange the equation to find the average force:

F_{avg} = rac{J}{ riangle t} = rac{-24300 ext{ kg·m·s}^{-1}}{1.2 ext{ s}}

Calculating:

$F_{avg} = -20250 ext{ N}$

Since the negative sign indicates direction, we take the magnitude:

The force of 20 250 N is less than the wall's capacity of 80 kN (80 000 N), thus the wall can withstand the impact.