4.1 An electrician, rushing to an urban area with a power outage, drives a truck of mass 1 350 kg towards the east travelling at 120 km.h⁻¹ - NSC Technical Sciences - Question 4 - 2020 - Paper 1

To convert the truck's speed from km.h⁻¹ to m.s⁻¹, we use the conversion factor:

$v = \frac{120 \text{ km.h}^{-1} \times 1000 \text{ m/km}}{3600 \text{ s/h}} = 33,33 \text{ m.s}^{-1}$

Thus, the velocity of the truck before the collision is 33,33 m.s⁻¹, directed east.

The initial momentum of the car can be calculated using the formula:

$p = m imes v$

where:

• $m = 1050 \text{ kg}$ (mass of the car)
• $v = 16,87 \text{ m.s}^{-1}$ (velocity of the car)

Calculating:

$p = 1050 \text{ kg} \times 16,87 \text{ m.s}^{-1} = 17 671,5 \text{ kg.m.s}^{-1}$

Thus, the initial momentum of the car is 17 671,5 kg.m.s⁻¹.

The principle of conservation of linear momentum states that in an isolated system, the total linear momentum remains constant if no external forces act on it.

To determine if the collision is elastic or inelastic, we compare the initial and final kinetic energies.

Calculating the initial kinetic energy:

$KE_i = \frac{1}{2} m_{truck} v_{truck}^2 + \frac{1}{2} m_{car} v_{car}^2$

Substituting values:

$KE_i = \frac{1}{2} (1350)(33,33)^2 + \frac{1}{2} (1050)(16,87)^2 = 895 741,68 \text{ J}$

Calculating the final kinetic energy:

$KE_f = \frac{1}{2} m_{truck} v_{truck final}^2 + \frac{1}{2} m_{car} v_{car final}^2$

Substituting the calculated values:

$KE_f = \frac{1}{2} (1350)(20,3)^2 + \frac{1}{2} (1050)(5,32)^2 = 293 019,51 \text{ J}$

Since $KE_i > KE_f$, the collision is inelastic.

The net force experienced by the car is inversely proportional to the contact time during the crash. This means that as the contact time increases, the force exerted on the car decreases, and vice versa.

The impulse experienced by the car equals the change in momentum. Hence, the relation is EQUAL TO.

Airbags increase the contact time during a crash. By extending the time over which the force is applied, the impulse received by the driver is distributed over a longer period, which reduces the peak force experienced. This decreases the risk of injury.

Using Newton's second law, the relationship between force, change in momentum (impulse), and contact time can be expressed as:

$F_{net} = \frac{\Delta p}{\Delta t}$

Substituting the known values:

• $F_{net} = -57\,500 \text{ N}$ (force applied)
• $\,\Delta p = m(v_f - v_i) = 1150 \text{ kg} (0 - 15) = -17 250 \text{ kg.m.s}^{-1}$

Thus: