Define (a) momentum, (b) force - Leaving Cert Physics - Question 6 - 2013

The principle of conservation of momentum states that the total momentum of a closed system remains constant if no external forces act upon it. Mathematically, this is expressed as:

• Total momentum before interaction: $p_{total, before} = p_1 + p_2$
• Total momentum after interaction: $p_{total, after} = p_1' + p_2'$

Thus, $p_{total, before} = p_{total, after}$.

In the case of a jet engine moving an aircraft, the momentum of the entire system (aircraft and exhaust gases) is conserved. The momentum gained by the aircraft in a forward direction is equal to the momentum lost by the exhaust gases expelled backwards. This relationship can be understood as:

$m_{aircraft} imes v_{aircraft} + m_{gases} imes v_{gases} = 0$

where:

• $m_{aircraft}$ and $m_{gases}$ are the masses
• $v_{aircraft}$ is the forward velocity
• $v_{gases}$ is the backward velocity (resulting in a negative value).

The momentum of the truck before collision can be calculated as:

1. Calculate truck momentum: $p_{truck} = m_{truck} imes v_{truck} = 5000 ext{ kg} imes 10 ext{ m s}^{-1} = 50000 ext{ kg m s}^{-1}$

2. Calculate car momentum (stationary): $p_{car} = m_{car} imes v_{car} = 1000 ext{ kg} imes 0 ext{ m s}^{-1} = 0 ext{ kg m s}^{-1}$

3. Total momentum before collision: $p_{total, before} = p_{truck} + p_{car} = 50000 + 0 = 50000 ext{ kg m s}^{-1}$.

After the collision, the combined mass of the truck and car can be calculated as:

$m_{combined} = m_{truck} + m_{car} = 5000 ext{ kg} + 1000 ext{ kg} = 6000 ext{ kg}$.

Using the principle of conservation of momentum:

• Before collision momentum: $50000 ext{ kg m s}^{-1}$
• Therefore, momentum after collision: $p_{combined} = 50000 ext{ kg m s}^{-1}$.

To calculate the velocity after the collision, use the formula for momentum:

$p = mv$

Rearranging gives:

$v = \frac{p}{m}$

Substituting values: $v_{combined} = \frac{50000 ext{ kg m s}^{-1}}{6000 ext{ kg}} = 8.33 ext{ m s}^{-1}$.

The momentum of the truck after the collision, since it is moving with the car, is the same as the momentum of the combined system:

$ext{Momentum}_{truck, after} = p_{combined} = 50000 ext{ kg m s}^{-1}$.

The force can be calculated using the impulse-momentum theorem:

$F = \frac{\Delta p}{\Delta t}$

where:

• $\Delta p = p_{truck} - p_{car} = 50000 ext{ kg m s}^{-1} - 0 = 50000 ext{ kg m s}^{-1}$
• $\Delta t = 0.3 ext{ s}$.

Thus: $F = \frac{50000 ext{ kg m s}^{-1}}{0.3 ext{ s}} = 166666.67 ext{ N}$.

The airbag reduces the risk of injury by increasing the time over which the driver's body decelerates during a collision. This reduces the force experienced by the driver according to the impulse-momentum theorem:

$F = \frac{\Delta p}{\Delta t}$

By extending the time ($\Delta t$) of impact, the force ($F$) is lowered, lessening the potential for injury.