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 kilometers per hour (km/h) to meters per second (m/s):

v_{truck} = 120 rac{km}{h} \times \frac{1000 m}{1 km} \times \frac{1 h}{3600 s} = 33,33 m.s^{-1}

Therefore, the velocity of the truck before the collision is 33,33 m.s⁻¹ towards the east.

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

$p_{car} = m_{car} v_{car}$

Where:

• $m_{car} = 1 050 kg$ (mass of the car)
• $v_{car} = 16,67 m.s^{-1}$ (velocity of the car)

Thus,

$p_{car} = 1 050 kg \times 16,67 m.s^{-1} = 17 503,5 kg.m.s^{-1}$

The initial momentum of the car is 17 503,5 kg.m.s⁻¹.

The principle of conservation of linear momentum states that in a closed and isolated system, the total linear momentum remains constant before and after a collision or an interaction.

To determine whether the collision is elastic or inelastic, we can compare the total kinetic energy before and after the collision.

Total kinetic energy before:

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

After the collision:

$KE_{final} = \frac{1}{2} m_{truck} v_{truck_{final}}^2 + \frac{1}{2} m_{car} v_{car_{final}}^2$

Substituting the values:

• Before: $m_{truck} = 1350 kg$, $v_{truck} = 33,33 m/s$, $m_{car} = 1050 kg$, $v_{car} = 16,67 m/s$.
• After: $v_{truck_{final}} = 20,3 m.s^{-1}$, $v_{car_{final}} = -5,32 m.s^{-1}$.

Calculating both:

Total kinetic energy before collision: $KE_{initial} = \frac{1}{2}(1350)(33,33)^2 + \frac{1}{2}(1050)(16,67)^2 = 895 741,68 J$

Total kinetic energy after collision: $KE_{final} = \frac{1}{2}(1350)(20,3)^2 + \frac{1}{2}(1050)(5,32)^2 = 293 019,51 J$

Since $KE_{initial} > KE_{final}$, the collision is inelastic.

The relationship between the net force experienced by the car and the contact time during the crash is inversely proportional. A higher net force results in a shorter contact time, while a smaller net force implies a longer contact time.

The impulse experienced by the car is EQUAL TO its change in momentum. This follows from the impulse-momentum theorem, which states that impulse is equal to the change in momentum.

Airbags increase the time over which the driver comes to a stop during a crash. By extending the contact time, airbags decrease the average force experienced by the driver (since impulse is force multiplied by time). This reduction in force decreases the likelihood and severity of injuries.

Using the formula for impulse:

$Impulse = F_{net} \times \Delta t$

Where:

• $F_{net} = 57 500 N$
• Change in momentum ($\Delta p = m(v_{final} - v_{initial}) = 1 150(0 - 15)$)

We calculate: $\Delta p = - 17 250 kg.m.s^{-1}$

Setting impulse equal to change in momentum: $57 500 N \times \Delta t = - 17 250$

Solving for $\Delta t$: $\Delta t = \frac{-17 250}{57 500} = 0,30 s$

Hence, the contact time during the crash is 0,30 seconds.