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A car with a mass of 980 kg is initially traveling east toward an intersection with a speed of vc = 21.3 m/s and a 1500 kg pickup is traveling north toward the same intersection. The car and truck collide at the intersection and stick together. After the collision, the wreckage (car and truck) moves off in a direction of 41.0° above the x-axis. Determine the initial speed of the truck and the final speed of the wreckage.

Answer :

fahadisahadam

Answer:

[tex]v_{i} = 12.09m/s[/tex]

Explanation:

x-axis

Initial velocity [tex]v_{i}=21.3m/s[/tex]

Mass: [tex]m=980kg[/tex]

y-axis

Initial velocity is unknown

Mass: [tex]m=1500kg[/tex]

conservative momentum of the system in x-axis is:

[tex]m_{c} v_{i}=(m_{c} +m_{p} )v_{f} cos[/tex](41.0°)

[tex](980)(21.3)=(980+1500)v_{f} cos(41.0)[/tex]

so find [tex]v_{f}[/tex] the subject of the formula

[tex]v_{f} =\frac{(980)(21.3)}{(980+1500)cos(41.0)}[/tex]

[tex]v_{f} =11.15m/s[/tex]

conservative momentum of the system in y-axis is:

[tex]m_{p} v_{i}=(m_{c} +m_{p} )v_{f} sin[/tex](41.0°)

[tex](1500)v_{i} =(980+1500)(11.15) sin(41.0)[/tex]

so now [tex]v_{i}[/tex] the subject of the formula

[tex]v_{i} =\frac{(980+1500)(11.15)sin(41.0)}{(1500)}[/tex]

[tex]v_{i} = 12.09m/s[/tex]

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