Answer:
24.084 m/s
Explanation:
From the law of conservation of linear momentum
Total momentum before collision equals to the total momentum after collision
Since momentum=mv where m is mass and v is velocity
[tex]M_{truck}V_{truck}=V_{common}*(M_{truck} +M_{standing})[/tex] where [tex]M_{truck}[/tex] is the mass of the truck, [tex]V_{truck}[/tex] is velocity of the truck, [tex]V_{common}[/tex] is the common velocity of moving and standing truck after collision and [tex]M_{standing}[/tex] is the mass of the standing truck
Making [tex]V_{truck}[/tex] the subject we obtain
[tex]V_{truck}=\frac { V_{common}*(M_{truck} +M_{standing})}{M_{truck}}[/tex]
Substituting [tex]M_{truck}[/tex] as 25000 Kg, [tex]V_{common}[/tex] as 22.3 m/s, [tex]M_{standing}[/tex] as 2000 Kg we obtain
[tex]V_{truck}=\frac { 22.3 m/s *(25000 Kg +2000 Kg)}{25000}= 24.084 m/s[/tex]
Therefore, assuming no friction and considering that after collision they still move eastwards hence common velocity and initial truck velocities are positive
The truck was moving at 24.084 m/s
