Answer :
Answer:
Velocity of the leading vehicle would be [tex]3.75\; {\rm m\cdot s^{-1}}[/tex], same as the initial velocity of the trailing vehicle.
Velocity of the trailing vehicle would be [tex]2.60\; {\rm m\cdot s^{-1}}[/tex], same as the initial velocity of the leading vehicle.
Explanation:
Let [tex]m_{0}[/tex] and [tex]m_{1}[/tex] denote the mass of the leading vehicle and the trailing vehicle, let [tex]u_{0}[/tex] and [tex]u_{1}[/tex] denote the velocity of each vehicle before the collision, and let [tex]v_{0}[/tex] and [tex]v_{1}[/tex] denote the velocity of each vehicle after the collision.
Since the collision is elastic, both momentum and kinetic energy would be conserved. In other words, both total momentum and kinetic energy would be the same before and after the collision.
Conservation of kinetic energy:
[tex]\displaystyle m_{0}\, {u_{0}}^{2} + m_{1}\, {u_{1}}^{2} = m_{0}\, {v_{0}}^{2} + m_{1}\, {v_{1}}^{2}[/tex].
Conservation of momentum:
[tex]\displaystyle m_{0}\, u_{0} + m_{1}\, u_{1} = m_{0}\, v_{0} + m_{1}\, v_{1}[/tex].
In this question, it is given that the combined mass of the vehicle and the driver is the same for both vehicles. Let [tex]m[/tex] ([tex]m \ne 0[/tex]) denote this mass: [tex]m_{0} = m_{1} = m[/tex], and the system of equations would be come:
[tex]\displaystyle m\, {u_{0}}^{2} + m\, {u_{1}}^{2} = m\, {v_{0}}^{2} + m\, {v_{1}}^{2}[/tex].
[tex]\displaystyle m\, u_{0} + m\, u_{1} = m\, v_{0} + m\, v_{1}[/tex].
Simplify to obtain:
[tex]\displaystyle {u_{0}}^{2} + {u_{1}}^{2} = {v_{0}}^{2} + {v_{1}}^{2}[/tex].
[tex]\displaystyle u_{0} + u_{1} = v_{0} + v_{1}[/tex].
In this system of equations, [tex]v_{0}[/tex] and [tex]v_{1}[/tex] are the only two unknowns. Rearrange the second equation to find an expression for [tex]v_{1}[/tex] in terms of [tex]v_{0}[/tex]:
[tex]v_{1} = (u_{0} + u_{1}) - v_{0}[/tex].
Substitute this expression into the first equation and solve for [tex]v_{0}[/tex]:
[tex]\displaystyle {u_{0}}^{2} + {u_{1}}^{2} = {v_{0}}^{2} + \left((u_{0} + u_{1}) - v_{0}\right)^{2}[/tex].
[tex]{u_{0}}^{2} + {u_{1}}^{2} = {v_{0}}^{2} + (u_{0} + u_{1})^{2} - 2\, (u_{0} + u_{1})\, v_{0} + {v_{0}}^{2}[/tex].
[tex]2\, {v_{0}}^{2} - 2\, (u_{0} + u_{1})\, v_{0} + {u_{0}}^{2} + 2\, u_{0}\, u_{1} + {u_{1}}^{2} = {u_{0}}^{2} + {u_{1}}^{2}[/tex].
[tex]2\, {v_{0}}^{2} - 2\, (u_{0} + u_{1})\, v_{0} +2\, u_{0}\, u_{1} = 0[/tex].
[tex]{v_{0}}^{2} - (u_{0} + u_{1})\, v_{0} + u_{0}\, u_{1} = 0[/tex].
[tex](v_{0} - u_{0})\, (v_{0} - u_{1}) = 0[/tex].
Hence, either:
- [tex]v_{0} = u_{0}[/tex], meaning that velocity of the leading vehicle stays unchanged, or
- [tex]v_{0} = u_{1}[/tex], meaning that the velocity of the leading vehicle is now the initial velocity of the trailing vehicle.
However, because the trailing vehicle was initially moving faster than the leading vehicle, it is not possible for the velocity of the leading vehicle to stay unchanged (slower) while the trailing vehicle (faster) to overtake the leading vehicle. Hence, the only possible option is that the new velocity of the leading vehicle would be equal to the initial velocity of the trailing vehicle: [tex]v_{0} = u_{1}[/tex].
Substitute this value into the expression for [tex]v_{1}[/tex]:
[tex]\begin{aligned}v_{1} &= (u_{0} + u_{1}) - v_{0} \\ &= (u_{0} + u_{1}) - u_{1} \\ &= u_{0}\end{aligned}[/tex].
In other words, the new velocity of the trailing vehicle would be equal to the initial velocity of the leading vehicle.
In this question, that means:
- Velocity of the leading vehicle would be equal to the initial velocity of the trailing vehicle: [tex]v_{0} = u_{1} = 2.60\; {\rm m\cdot s^{-1}}[/tex].
- Velocity of the trailing vehicle would be equal to the initial velocity of the leading vehicle: [tex]v_{1} = u_{0} = 3.75\; {\rm m\cdot s^{-1}}[/tex].
