Answer:
The velocity with which they move together is 16.62m/s.
Explanation:
The velocity with which they move together can be determined by means of the equation for the conservation of the linear momentum:
[tex]Qb = Qa[/tex] (1)
Where [tex]Q_{b}[/tex] is the total linear momentum of the system before the collision and [tex]Q_{a}[/tex] is the total linear momentum of the system after the collision:
Remember that the total linear momentum of a system is the sum of the momentum of each member ([tex]p = m\cdot v[/tex])
[tex]m_{t}\cdot v_{t} + m_{c}\cdot v_{c} = m_{t}\cdot v_{t} + m_{c}\cdot v_{c}[/tex] (2)
Where [tex]m_{t}[/tex] is the mass of truck, [tex]v_{t}[/tex] is the velocity of the truck, [tex]m_{c}[/tex] is the mass of the car and [tex]v_{c}[/tex] is the velocity of the car.
Equation 2 establishes how the linear momentum is conserved in the system, if there is not external force acting on it or if the sum of the net force is zero.
To get the velocity it is necessary to known that since they move together after the collision, the velocity will be the same at [tex]Q_{a}[/tex], therefore it is gotten:
[tex]m_{t}\cdot v_{t} + m_{c}\cdot v_{c} = (m_{t} + m_{c})v[/tex] (3)
Finally, v can be isolated from equation 3
[tex]v = \frac{m_{t}\cdot v_{t} + m_{c}\cdot v_{c}}{(m_{t} + m_{c})}[/tex] (4)
[tex]v = \frac{(2.00x10^{3}Kg)\cdot (14.0 m/s) + (1.20x 10^{3}Kg)\cdot (21.0 m/s)}{(2.00x10^{3} Kg + 1.20x 10^{3}Kg)}[/tex]
[tex]v = \frac{53200 Kg\cdot m/s}{(2.00x10^{3} Kg + 1.20x 10^{3}Kg)}[/tex]
[tex]v = \frac{53200 Kg\cdot m/s}{3200 Kg}[/tex]
[tex]v = 16.62 m/s[/tex]
Hence, the velocity with which they move together is 16.62m/s.
