Example on a Car Chase
Example: Car Chase
Question
A patrol car is moving on a straight horizontal road at a velocity of \(\text{10}\) \(\text{m·s$^{-1}$}\) east. At the same time a thief in a car ahead of him is driving at a velocity of \(\text{40}\) \(\text{m·s$^{-1}$}\) in the same direction.
\(v_{PG}\): velocity of the patrol car relative to the ground \(v_{TG}\): velocity of the thief's car relative to the ground
Questions 1 and 2 from the original version in 2011 Paper 1 are no longer part of the curriculum.
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While travelling at \(\text{40}\) \(\text{m·s$^{-1}$}\), the thief's car of mass \(\text{1 000}\) \(\text{kg}\), collides head-on with a truck of mass \(\text{5 000}\) \(\text{kg}\) moving at \(\text{20}\) \(\text{m·s$^{-1}$}\). After the collision, the car and the truck move together. Ignore the effects of friction.
State the law of conservation of linear momentum in words.
(2 marks)
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Calculate the velocity of the thief's car immediately after the collision.
(6 marks)
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Research has shown that forces greater than 85 000 N during collisions may cause fatal injuries. The collision described above lasts for \(\text{0.5}\) \(\text{s}\).
Determine, by means of a calculation, whether the collision above could result in a fatal injury.
(5 marks)
[TOTAL: 13 marks]
Question 1
The total (linear) momentum remains constant (OR is conserved OR does not change) in an isolated (OR in a closed system OR in the absence of external forces).
(2 marks)
Question 2
Option 1:
Taking to the right as positive
\begin{align*} \sum p_{\text{before}} & = \sum p_{\text{after}} \\ (\text{1 000})(\text{40})+(\text{5 000})(-\text{20}) & = (\text{1 000}+\text{5 000})v_f \\ v_f & = -\text{10 m}\cdot\text{s}^{-1}\\ & = \text{10 m}\cdot\text{s}^{-1} \quad \text{left OR west} \end{align*}Option 2:
Taking to the right as positive
\begin{align*} \Delta p_{\text{car}}& = - \Delta p_{\text{truck}} \\ m_{\text{car}}(v_f -v_{i,\text{car}}) & = - m_{\text{truck}}(v_f -v_{i,\text{truck}}) \\ (\text{1 000})(v_f-(\text{40})) & = -(\text{5 000})(v_f -(-\text{20})) \\ \text{6 000}v_f & = -\text{60 000} \\ \therefore v_f & = -\text{10 m}\cdot\text{s}^{-1} \\ \therefore v_f & = \text{10 m}\cdot\text{s}^{-1} \quad \text{left OR west} \end{align*}(6 marks)
Question 3
Option 1:
Force on the car: (Taking to the right as positive)
\begin{align*} F_{\text{net}} \Delta t & = \Delta p = mv_f - mv_i \\ F_{\text{net}} (\text{0.5}) & = (\text{1 000})(-\text{10}-\text{40}) \\ \therefore F_{\text{net}} & = -10^5\text{ N} \\ & \text{OR} \\ \therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
OR
Force on the car: (Taking to the left as positive)
\begin{align*} F_{\text{net}} \Delta t & = \Delta p = mv_f - mv_i \\ F_{\text{net}} (\text{0.5}) & = (\text{1 000})(\text{10}-(-40)) \\ \therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
Option 2:
Force on the truck: (Taking to the right as positive)
\begin{align*} F_{\text{net}} \Delta t & = \Delta p = mv_f - mv_i \\ F_{\text{net}} (\text{0.5}) & = (\text{5 000})(-10-(-20)) \\ \therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
OR
Force on the truck: (Taking to the left as positive)
\begin{align*} F_{\text{net}} \Delta t & = \Delta p = mv_f - mv_i \\ F_{\text{net}} (\text{0.5}) & = (\text{1 000})(\text{10}-\text{20}) \\ \therefore F_{\text{net}} & = -10^5\text{ N} \\ & \text{OR} \\ \therefore F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
Option 3:
Force on the car: (Taking to the right as positive)
\begin{align*} v_f & = v_i + a \Delta t \\ -10 & = \text{40} + a (\text{0.5})\\ \therefore a & = -\text{100 m}\cdot\text{s}^{-2}\\ F_{\text{net}} & = ma \\ & = (\text{1 000})(-100) \\ F_{\text{net}} & = -10^5\text{ N} \quad (-\text{100 000 N})\\ F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
OR
Force on the car: (Taking to the left as positive)
\begin{align*} v_f & = v_i + a \Delta t \\ \text{10} & = -40 + a (\text{0.5})\\ \therefore a & = \text{100 m}\cdot\text{s}^{-2}\\ F_{\text{net}} & = ma \\ & = (\text{1 000})(\text{100}) \\ F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
Option 4:
Force on the truck: (Taking to the right as positive)
\begin{align*} v_f & = v_i + a \Delta t \\ -10 & = -20 + a (\text{0.5})\\ \therefore a & = \text{20 m}\cdot\text{s}^{-2}\\ F_{\text{net}} & = ma \\ & = (\text{5 000})(\text{20}) \\ F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
OR
Force on the truck: (Taking to the left as positive)
\begin{align*} v_f & = v_i + a \Delta t \\ \text{10} & = \text{20} + a (\text{0.5})\\ \therefore a & = -\text{20 m}\cdot\text{s}^{-2}\\ F_{\text{net}} & = ma \\ & = (\text{5 000})(-20) \\ F_{\text{net}} & = -10^5\text{ N} \quad (-\text{100 000 N})\\ F_{\text{net}} & = 10^5\text{ N} \quad (\text{100 000 N})\\ \therefore F_{\text{net}} & > \text{85 000 N} \end{align*}Yes, the collision is fatal.
(5 marks)
[TOTAL: 13 marks]
This lesson is part of:
Momentum and Impulse