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Two bumper car in an amusement park ride collide elastically as on approaches the other directly from the rear. Car A has a mass of 450kg and car B 550kg, owing to differences in passenger mass. If car A approaches at 4.5m/s and car B is moving at 3.7m/s, calculate (a) their velocities after collision, and (b) the change in momentum of each.

Respuesta :

Answer:

VA = 3.62 m/s  

VB = 4.42 m /s

Change of momentum of A = -396 kgm/s

Change of momentum of B = 396 kgm/s

Explanation:

Asuming A and B move to the right after the collision

pbefore =  pafter

4.5*450 + 3.7*550 = 450VA + 550VB

In a elastis collition, the coefficient of restitution is 1

1 = (VB - VA)/(4.5 - 3.7)

VB - VA = 0.8

VB = 4.42 m/s

VA = 3.62

Change of momentum of A = 450*(3.62-4.5) = -396 kgm/s

Change of momentum of B = 550*(4.42-3.7) = 396 kgm/s

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